PHY2049R. D. Field Chapter 22 Solutions Problem 1: A +15 microC charge is located 40 cm from a +3.0 microC charge. The magnitude of the electrostatic force on the larger charge and on the smaller charge (in N) is, respectively, Answer: 2.5, 2.5 Solution: The magnitiude of the electrostatic for is given by, KQ1Q2 (8.99 × 109 Nm 2 / C 2 )(15µC )(3µC ) = = 2.53N . F= r2 (40cm) 2 Remember the force on 1 due to 2 is equal and opposite to the force on 2 due to 1. Problem 2: Two point particles have charges q1 and q2 and are separated by a distance d. Particle q2 experiences an electrostatic force of 12 milliN due to particle q1. If the charges of both particles are doubled and if the distance between them is doubled, what is the magnitude of the electrostatic force between them (in milliN)? Answer: 12 Solution: The magnitude of the initial electristatic force is Fi = Kq1q2 = 12mN . d2 The magnitude of the final electrostatic force is Ff = K (2q1 )(2q2 ) = Fi = 12mN . ( 2d ) 2 +Q Problem 3: Two identical point charges +Q are located on the y-axis at y=+d/2 and y=-d/2, as shown in the Figure. A third charge q is placed on the x-axis. At what distance from the origin is the net force on q a maximum? Solutions Chapter 22 F2 r q d θ x F1 +Q Page 1 of 6 PHY2049 R. D. Field d /( 2 2 ) Answer: Solution: The net force on q is the superposition of the forces from each of the two charges as follows: r KQq (cos θ xˆ − sin θ yˆ ) F1 = 2 r r KQq (cos θ xˆ + sin θ yˆ ) F2 = 2 r r r r 2 KQq cos θ xˆ F = F1 + F 2 = 2 r 2 KQqx = xˆ ( x 2 + (d / 2) 2 )3 / 2 where I used r2 = x2 + (d/2)2 and cosθ θ = x/r. To find the maximum we take the derivative of F and set it equal to zero as follows: 1 3x 2 dF = 2 KQq 2 − 2 3/ 2 2 2 5/ 2 + + ( ( / 2 ) ) ( ( / 2 ) ) dx x d x d ( ) 2 KQq (d / 2) 2 − 2 x 2 = =0 ( x 2 + (d / 2) 2 )5 / 2 Solving for x gives x = d /(2 2 ) . Problem 4: Two 2.0 gram balls hang from lightweight insulating threads 50 cm long from a common support point as shown in the Figure. When equal charges Q are place on each ball they are repelled, each making an angle of 10 degrees with the vertical. What is the magnitude of Q, in microC? Answer: 0.11 Solution: Requiring the x and y components of the force to vanish yields Solutions Chapter 22 θ L QE Q T Q r mg Page 2 of 6 PHY2049 R. D. Field KQ2 T sin θ = QE = 2 r . T cosθ = mg Eliminating the tension T and solving for Q2 gives mgr 2 tan θ 4 mgL2 sin 2 θ tan θ = Q = K K 2 and Q= 4 ( 2 × 10 −3 kg )( 9.8m / s 2 )( 0.5m) 2 sin 2 10 0 tan 10 0 = 0.11µC . 8.99 × 10 9 Nm 2 / C 2 Problem 5: Q -2Q Four point charges q, Q, -2Q, and Q form a square with sides of length L as shown in the Figure. What is the magnitude of the resulting electrostatic force on the charge q due to the other three charges? Answer: 0.41KqQ/L2 L 3 1 Q L q tot Solution: The electrostatic force on q is the vector superposition of the following three forces, 2 r KqQ F1 = 2 xˆ L r KqQ F2 = − 2 yˆ L r 2 KqQ 1 1 ˆ − + F3 = x yˆ ( 2 L) 2 2 2 and the net force is thus, r KqQ 1 Ftot = 2 1 − ( xˆ − yˆ ) . L 2 The magnitude is given by r KqQ Ftot = 2 L Solutions ( ) 2 − 1 = 0.41 Chapter 22 KqQ . L2 Page 3 of 6 Thus.368µC Q2 = −6 0. such that there is no net electrostatic force acting on the +5Q charge.8 m / s 2 ) x 3 mg = = 3 . Solving for the product of the two charges gives ( 0 . On the empty corner a charge is placed. D. What charge is placed on the empty corner? Answer: Solutions q +Q L +5Q L − 2 2Q Chapter 22 Page 4 of 6 +Q .01× 10 C Problem 7: A charge +Q is fixed two diagonally opposite corners of a square with sides of length L.99 × 10 Nm / C ) where I used tanθ θ = x/(2L).15 m ) 3 ( 2 × 10 −3 kg )( 9 .68 × 10−15 C 2 = 0. On another corner of the square a charge of +5Q if fixed as shown in the Figure.37 Solution: The conditions for static equilibrium on ball Q2 is KQ1Q2/x 2 KQ1Q2 x2 T cosθ = mg T sin θ = and dividing the 1st equation by the second yields tan θ = KQ1Q2 x 2 mg .0 gram charged balls hang from lightweight insulating threads 1 m long from a θ T common support point as shown in the Figure. what is the charge on the other ball (in mg microC)? Answer: 0. Q2 If one of the balls has a charge of 0. 3.PHY2049 R. Field Problem 6: θ L=1m L=1 m Two 2.01 microC Q1 x = 15 cm and if the balls are separated by a distance of 15 cm.68 × 10 −15 C 2 Q1Q 2 = 9 2 2 2 LK 2 (1m )( 8 . and 4Q form a triangle with sides of length L and 2L as shown in the Figure. The three charged +Q.1 Solution: We see from the figure that 2L 2 2 4KQ /L 4Q L θ Q 2 F 2 KQ 2 /(2 L) 2 1 tan θ = = 4 KQ 2 / L2 8 and hence θ = 7. A charge q is placed on the x-axis a distance x = d from the quadrupole. +Q. At what is the net force on q due to the quadrupole? Answer: Solutions − 0. L2 2L ( ) and solving for the magnitude of q. +Q Problem 9: Two identical point charges +Q are located on the y-axis at y=+d/2 and y=-d/2 and a third charge -2Q is located at the origin as shown in the Figure. and -2Q form an electrically neutral system called an electric quadrupole. What angle does the resulting electrostatic force on the charge Q make with the positive x-axis (in degrees)? Answer: 7. Field Solution: Setting the magnitude of the net for on the +5Q charge due to the two +Q charges equal to the magnitude of the force on the +5Q charge due to the charge q gives 2 KQ(5Q) K q (5Q) = 2 .PHY2049 R. 2Q. D.57 F2 r q x d -2Q θ F3 F1 +Q KQq xˆ 2 d Chapter 22 2 KQ /2L Page 5 of 6 .1o. 2Q Problem 8: Three point charges Q. |q| (and noting that q must be negative) yields. q = −2 2Q . Adding the forces gives r r r r 1 x xˆ − F = F1 + F2 + F3 = 2 KQq 2 2 3/ 2 2 + ( ( / 2 ) ) x d x =− 2 KQq 1 xˆ − 1 x 2 (1 + (d / 2 x) 2 )3 / 2 which at x = d yields r 2 KQq 1 xˆ − F =− 1 d 2 (1 + (1 / 2) 2 )3 / 2 =− KQq 2 KQq 1 1 − xˆ = −0.57 2 xˆ 2 3/ 2 d (5 / 4) d Problem 10: In the previous problem.3εε/2 with ε = (d/2x)2. what is the magnitude of force on the charge q when x becomes very large compared to the quadrupole separation d. Field Solution: The net force on q is the superposition of the following three vector forces: r KQq (cos θ xˆ − sin θ yˆ ) F1 = r2 r KQq (cos θ xˆ + sin θ yˆ ) F2 = r2 r 2 KQq F3 = − xˆ x2 where r2 = x2 + (d/2)2 and cosθ θ = x/r .PHY2049 R.) Answer: 3KqQd2/(4x4) Solution: From the previous problem we see that the magnitude of the quadrupole force is given by r 2 KQq 1 F = 1 − x 2 (1 + (d / 2 x) 2 )3 / 2 2 KQq 3KQqε 3KqQd 2 −3 / 2 = ≈ = 1 − (1 + ε ) x2 x2 4x4 ( ) where I used (1+εε)-3/2 ~ 1 . (Hint: take the limit of the quadrupole force on q when x << d. Solutions Chapter 22 Page 6 of 6 . D. Field Chapter 23 Solutions Problem 1: What is the magnitude (in milliN) and direction of the electrostatic force on r a -2. Problem 3: Q1 P Q2 Two charges Q1 and Q2 are separated by L/3 distance L and lie on the x-axis with Q1 at the L origin as shown in the figure.0 µ C )(100 N / C ) xˆ = − 0 .25 Solution: In order for the net electric field to be zero it must be true that Solutions Chapter 23 Page 1 of 14 . +8 nanoC and -2 6m 6m P nanoC lie on the x-axis and are separated -2 nC +8 nC by 6 meters as shown in the Figure. At a point P on the x-axis a distance L/3 from Q1 the net electric field is zero. L L ( 2 ) r r r Etot = E1 + E2 = 0 where Q = 2 nC and L = 6 m. What is the magnitude of the electric field (in N/C) at a point P on the x-axis a distance of 6 meters to the right of the negative charge? Answer: zero Solution: The electric field at the point P is the superposition of the electric fields from each of the two charged as follows: r KQ E1 = − 2 x$ L r K 4Q KQ $ $ = E2 = x 2 2 x . What is the ratio Q1/Q2? Answer: 0.PHY2049 R.2 in the -x direction Solution: The electrostatic force is given by. D. Problem 2: Two point charges.0 microC charge in a uniform electric field given by E = (100 N / C ) xˆ ? Answer: 0. r r F = q E = ( − 2 .2 mN xˆ . How far will the particle move (in m) in 2 seconds? Answer: 2 Solution: We know that q F = ma = qE so that the acceleration is a = E . How much time (in microsec) elapses before the proton strikes the sheet? Answer: 0.67x10-27 kg) is released from rest at perpendicular distance of 1.PHY2049 R. m Thus. t= 2d 2dM 4dMε 0 = = a eE eσ . We also know that if the particle starts from rest then the distance traveled.67 × 10−27 kg )(8.2 microC/m2. Solving for t gives.0 m from the sheet.55 Solution: We know that F = eE = Ma so that a = eE/M. which implies that Q1/Q2 = 1/4.1 C/kg starts from rest in a uniform electric field with magnitude. 4(1m)(1. 55 s (1.6 × 10−19 C )(1. A proton (with a mass of 1. the distance traveled is d= 1 2 1 q 2 at = Et 2 2m 1 = (0. Field KQ1 KQ2 = ( L / 3) 2 (2 L / 3) 2 . d = at2/2.1C / kg )(10 N / C )(2s ) 2 = 2m 2 Problem 5: A large insulating sheet carries a uniform surface charge of density -1. Solutions Chapter 23 Page 2 of 14 .85 × 10−12 C 2 /( Nm 2 )) = = µ 0 . Problem 4: A particle with a charge to mass ratio of 0. E = 10 N/C.2 × 10−6 C / m 2 ) where I used E = σ/(2εε0). D. At what distance (in m) from the charge is the electric field strength equal to 50 N/C? Answer: 1.PHY2049 R.6x1010 Solution: We know that F = qE = ma. is placed.99 × 109 Nm 2 / C 2 Problem 8: A neon sign includes a long neon-filled glass tube with electrodes at each end across which an electric field of 20. Problem 7: The magnitude of the electric field 300 m from a point charge Q is equal to 1. Field Problem 6: At a distance of 1 meter from an isolated point charge the electric field strength is 100 N/C. What is the charge Q (in C)? Answer: 0. r2 = 2m = 141 .000 N/C. D.000 N / C )(300m) 2 = ≈ 0. what is their acceleration (m/s2)? Answer: 9.35x10-26 kg) are singly ionized (i.41 Solution: We know that E1=KQ/r12 and E2=KQ/r22 so that E 100 N / C E1 r22 = 2 or r22 = r12 1 = (1m)2 = 2m2 .e. have charge e=1.01 Solution: We know that E= KQ r2 and solving for r gives Er 2 (1. Q= K 8.000 N/C. E2 50 N / C E2 r1 Hence. m. so that Solutions Chapter 23 Page 3 of 14 . Assuming that some of the neon ions (mass 3.6x10-19 C) and are accelerated by the field.01C . This large field accelerates free electrons which collide with and ionize a portion of the neon atoms which emit red light as they recombine. 6 × 1010 m / s2 . D. If the rods have equal and opposite uniform charge density. Field qE (16 . × 10−19 C)(20.000 N / C) = 9. -Q.PHY2049 R.35 × 10 kg m a d c b e Problem 9: Of the five charge configurations (a)-(e) shown in the Figure which one results in the maximum magnitude of the electric field at the center of the square? Each configuration consists of a square with either +Q. = a= − 26 3. Problem 10: Two non-conducting infinite parallel rods are a distance d apart as shown in the Figure. λ. what is the magnitude of the electric field along a line that is midway between the two rods? Solutions Chapter 23 +λ d -λ Page 4 of 14 . or no charge at each corner? Answer: d Solution: The magnitude of the electric field at the center of each square is given by 2 KQ L2 KQ (b) E = 2 2 2 L 4 KQ (c ) E= 2 L KQ (d ) E = 4 2 2 L (e) E = 0 (a ) E= where L is the length of the side of the square. The square with the largest E-field at the center is (d). the magnitude of the electric field along the line midway between the two rods is given by E= 2 Kλ 2 Kλ 8Kλ + = d /2 d /2 d . Problem 11: 6m P +2 nC Two point charges. at the point x =2 m (on the x-axis). lie on 1m the x-axis and are separated by a E = 13 N/C distance of 1 meter as shown in the q Q Figure. If the x-component of the electric field from the two charges. Field Answer: 8Kλ λ/d Solution: The magnitude of the electric field from a single infinite rod is given by E = 2Kλ λ/r. The charge Q lies at the E=0 origin and the net electric field (on at x = 0. D. (3m) 2 9m 2 Problem 12: 2m Two point charges.PHY2049 R.99 × 109 Nm 2 / C )(6 × 10 −9 C ) | Etot |= = = 6N / C . What is the magnitude of the electric field (in N/C) midway between them? Answer: 6 Solution: The electric field at the point P is the superposition of the electric fields from each of the two charged as follows: r K (8 nC ) E1 = xˆ (3 m ) 2 r K ( 2 nC ) E2 = − xˆ (3m ) 2 r r r K ( 6 nC ) E tot = E1 + E 2 = xˆ (3 m ) 2 The magnitude of the net electric field is r K (6nC ) (8. Ex = 13 N/C. what is Q (in nanoC)? Answer: 4 Solutions Chapter 23 Page 5 of 14 . Q and q.75 m the x-axis) from the two charges is zero at the point x = 0. +8 nanoC and +2 +8 nC 3m nanoC are separated by 6 meters.75 m. Hence. at what point (or 1m points) on the x-axis is the net electric field zero? -1 µC +4 µC Answer: x = 2.25m) 2 which implies The second condition (E = 13 N/C at x = 2 m) gives KQ Kq KQ 1 1 + = + = 13N / C (2m) 2 (1m) 2 (1m) 2 4 9 and solving for Q yields 36(1m) 2 (13N / C ) ≈ 4nC . lie on the x-axis and are separated by a distance of 1 meter as shown in the Figure. (0. If the +4 microC x lies at the origin. Solutions Chapter 23 Page 6 of 14 .0 m Solution: The first condition that must be satisfied for the net electric field to be zero is that the magnitude of the E-field from the +4 mC charge must be equal to the magnitude of the E-field from the –1 mC charge. ( x − 1m) The second condition that must be satisfied is that the vectors must point in opposite directions and is true only at the point x = 2 m (at x = 2/3 m they point in the same direction).75 m) gives KQ Kq = q = Q / 9. Field Solution: The first condition (net electric field zero at x = 0. +4 microC and -1 microC. D.PHY2049 R.75m) 2 (0. Q= 9 2 2 13(8.99 × 10 Nm / C ) Problem 13: Two point charges. This implies that 4 1 = x 2 ( x − 1m) 2 Solving for the distance x yields x = ±2 or x = 2 m and x = 2/3 m. The rods have charge uniformly distributed along their length with a linear density λ.564 m (at x = 0.56 m Solution: The first condition that must be satisfied for the net electric field to be zero is that the magnitude of the E-field from the infinite sheet must be equal to the magnitude of the E-field from the –2 C point charge.564)m . Field Problem 14: A -2 C point charge lies on the x-axis a distance of 1 meter from an infinite nonconducting sheet of charge with a uniform surface charge density σ = 1 C/m2 which 1m -2 C x-axis lies in the yz-plane as shown in the Figure.2Kλ λ/L Solutions Chapter 23 Page 7 of 14 .PHY2049 R. One rod lies from x = L/2 to x = infinity and the other rod lies from x = -L/2 to and x = -infinity. At what point (or points) along σ = 1 C/m the x-axis is the net electric field zero? Answer: x=1. D. dE θ P θ y=L/2 r λ λ x dQ L Problem 15: Two non-conducting semi-infinite rods lies along the x-axis as shown in the Figure. The second condition that must be satisfied is that the vectors must point in opposite directions and is true only at the point x = 1.436 m they point in the same direction). What is the magnitude of the electric field at a point P located on the y-axis at y = L/2? Answer: 1. This implies that 2 |σ | K |Q| 2 Kε 0 | Q | |Q| 2 = ( 1 ) x − m = = 2ε 0 ( x − 1m) 2 or |σ | 2π | σ | Solving for the distance x yields x = (1 ± 0. z. E y = 2∫ dE y = 2∫ KdQ cosθ r2 y 2 Kλ dx = 1− = 2 Kλ ∫ 2 2 3/ 2 y ( x y ) + L/2 ∞ y 2 + ( L / 2) 2 L/2 where I used dQ = λdx and cosθ θ = y/r with r2 = x2 + y2. Solutions Chapter 23 Page 8 of 14 . If the ring has a charge -Q located at its center (see Figure). z = R / 3 . Field Solution: From the symmetry in the problem we see that the x-component of the electric field at the point P is zero and that the y-component of the electric field from the two rods is twice that from one of the rods as follows. at what point . along the axis of the ring (other than infinity) is the net electric field zero? -Q R z-axis R/ 3 Answer: Solution: The magnitude of the net electric field on the z-axis will be zero provided K (8Q) z (z 2 + R ) 2 3/ 2 = KQ z2 8z 3 = ( z 2 + R 2 ) . D. Plugging in y = L/2 yields 2 Kλ L/2 1− Ey = L/2 ( L / 2) 2 + ( L / 2) 2 4 Kλ 1 Kλ = 1 − = 1.17 L L 2 Total Charge +8Q on ring Problem 16: A positive charge 8Q is distributed uniformly along a thin circular ring of radius R. 3/ 2 or Taking the 2/3 power of both sided yields 4z 2 = z 2 + R 2 and solving for z gives.PHY2049 R. as shown in the Figure. D. Field Problem 17: Two identical point charges +Q are located on the y-axis at y=+d/2 and y=-d/2. (Hint: take the limit of the electric field when x >> d)? Answer: 2KQ/x2 Solution: From the previous problem we see that r | E |= Solutions 2 KQx 2 KQ 2 KQ = → ( x 2 + (d / 2) 2 )3 / 2 x 2 (1 + (d / 2 x) 2 )3 / 2 x2 .43KQ/d2 Solution: The net electric field at the point P is the superposition of the forces from each of the two charges as follows: +Q E2 r x d θ P E tot E1 +Q r KQ (cos θ xˆ − sin θ yˆ ) E1 = r2 r KQ (cos θ xˆ + sin θ yˆ ) E2 = 2 r r r r 2 KQ E = E1 + E 2 = cos θ xˆ r2 2 KQx = xˆ ( x 2 + (d / 2) 2 )3/ 2 θ = x/r. What is the magnitude of the electric field on the x-axis a distance x = d from the origin? Answer: 1. Setting x = d yields where I used r2 = x2 + (d/2)2 and cosθ r | E |= 2 KQ KQ = 1 . Problem 18: In the previous problem.PHY2049 R. Chapter 23 Page 9 of 14 . 43 (5 / 4)3 / 2 d 2 d2 . what is the magnitude of the electric field on the xaxis when x becomes very large compared to the separation d between the charges. If two quarters of the circle carry a uniform charge density +λ λ and one quarter of the circle carries a uniform charge density -λ λ. Field L Semi-infinite Rod L L E1 P E2 Charge on Rod = Q Problem 19: A rod carrying a total charge Q uniformly distributed along its length L (constant λ) lies on the x-axis a distance L from a semi-infinite rod with the same uniform charge density λ as the finite rod as shown in the Figure.PHY2049 R. what is the magnitude of the electric field at the center? 10 K λ / R Answer: Solution: First I calculate the electric field at the center of a semi-circle as with charge density λ as follows: Solutions Chapter 23 R E1 E2 + - +λ - + -λ Semi-Circle λ dE θ R Page 10 of 14 . D. What is the magnitude of the net electric field at the point P on the x-axis a distance L from the end of finite rod? Answer: 5KQ/(6L2) Solution: The net electric field at the point P is the superposition of the forces from each of the two rods (finite and semi-infinite) as follows: r r E 1 = E rod = KQ xˆ x(x + L) r r Kλ xˆ = E 2 = E semi = 3L r r r 1 KQ 1 E = E1 + E 2 = + 2 3 L 2 KQ xˆ 2 L2 KQ xˆ 3 L2 5 KQ xˆ xˆ = 2 6L = where I used λ = Q/L and I used the fact that the point P was a distance 3L from the end of the semi-infinite rod. + +λ + Problem 20: A thin plastic rod is bent so that it makes a three quarters of a circle with radius R as shown in the Figure. PHY2049 R. at a point P located a distance y above the end of the rod ? Answer: Kλ λ/y Solutions Chapter 23 θ P θ r dEx y λ dQ -x Page 11 of 14 . D. The rod has charge uniformly distributed along its length with a linear density λ. where I have used dE = KdQ/R2 with dQ = λds θ. Ex. What is the x-component of the electric field. Now I calculate the = λRdθ θ and dEx = dEcosθ electric field at the center of a quarter-circle with charge density λ as follows: Quarter-Circle λ dE θ π /4 Kλ 2 Kλ Ex = d cos θ θ = R −π∫/ 4 R . Field π /2 Kλ 2 Kλ Ex = d cos θ θ = R −π∫/ 2 R . R The net electric field at the center of the threequarter circle is the superposition of the fields from the semi-circle and the quarter-circle as follows: r E2 r r 2Kλ xˆ E 1 = E semi = R r 2Kλ 1 1 = E quarter = xˆ − yˆ R 2 2 r r r Kλ ( 3 xˆ − 1 yˆ ) E = E1 + E 2 = R where I used sin 45 o = cos 45 o = 1 / 2 . dE Problem 21: A non-conducting semi-infinite rod lies along the x-axis as shown in the Figure (one end at x=0 and the other at x=-infinity). The magnitude of the net electric field is thus r Kλ Kλ | E |= 9 + 1 = 10 R R . d d 2 2 2 2 Solutions Chapter 23 Page 12 of 14 . D. as shown in the figure.737 2 . Solving for the magnitude yields. Field Solution: From the figure we see that KdQ sin θ r2 0 x Kλ = − Kλ ∫ 2 = dx ( x + y 2 )3 / 2 y −∞ E x = ∫ dEx = ∫ where I used dQ = λdx and sinθ θ = -x/r with r2 = x2 + y2. Problem 22: d An electric dipole is placed on the y-axis with +Q at y=0 and -Q at y=-d. 2 2 KQ 1 1 KQ E = 2 1 − + = 0.PHY2049 R. What is the magnitude of the electric field at a point P located at x=d on the x-axis? Answer: 0.74KQ/d2 Solution: We superimpose the two Efields as follows: P +Q E1 θ E2 d -Q r KQ E1 = 2 xˆ d r KQ 1 1 E2 = 2 − xˆ − yˆ 2d 2 2 r r r KQ 1 1 E = E1 + E2 = 2 (1 − ) xˆ − d 2 2 2 2 yˆ o o where I have used sin 45 = cos 45 = 1 / 2 . where a=13. a ball of mass M=2 kg and charge Q=3 C is suspended on a string of negligible mass and length L=1 m in a non-uniform θ L T θ r x electric field E ( x ) = ax xˆ . If the ball hangs at a Mg non-zero θ from the vertical. Problem 24: Two semicircular uniform nonconducting loops. Eliminating the tension T gives sin θ QE Qax QaL sin θ = = = . cos θ = QaL (3C )(13 .PHY2049 R.8 m / s 2 ) Mg = = 0 . each of radius R.) Answer: 60o Solution: Requiring the x and y components of the force to vanish yields E T sin θ = QE T cosθ = Mg . Field y-axis Problem 23: x-axis As shown in the figure. what is θ? (Hint: gravity pulls the ball down with acceleration g=9.5 . D. What is the magnitude of the electric field at the center of the circular loop? Answer: 4KQ/(π π R2) Solutions Chapter 23 +Q R -Q E Page 13 of 14 . are charged oppositely to +Q and -Q. The loops are now joined end to end to form a circular loop as shown in the figure. cos θ Mg Mg Mg and thus ( 2 kg )( 9 .07 N /( Cm ))(1m ) so that θ = 60o.8 m/s2.07 N/(Cm) is a constant. respectively. PHY2049 R. Now for two semi-circles of opposite charge. the two electric fields add giving E = E1 + E2 = 2 KQ 2 KQ 4 KQ + = πR 2 πR 2 πR 2 . What is the magnitude of the net electric field at P on the x-axis a distance x = L/3 from the end of the rod? Answer: zero Solution: We superimpose the two E-fields (rod and point) as follows: r E rod = KQ xˆ x( x + L) r KQ E po int = − xˆ 2 − ( ) L x r r r E = E rod + E po int 1 1 = KQ − 2 x(x + L) (L − x) xˆ At x = L/3 we have r KQ 9 9 E = 2 − xˆ = 0 . L 4 4 Solutions Chapter 23 Page 14 of 14 . L L Q Epoint Erod Charge on Rod = Q x Problem 25: L-x A rod carrying a total charge Q uniformly distributed along its length L (constant λ) and a point charge Q both lie on the x-axis and are separated by a distance L as shown in the Figure. R θ dE where I have used dE = KdQ/R2 with dQ = λds = λRdθ θ and dEx = dEcosθ θ and Q = πRλ λ. Field Solution: First I calculate the electric field at the center of a semi-circle as follows: Ex = dQ π /2 Kλ 2 Kλ 2 KQ cos d θ θ = = ∫ R −π / 2 R πR 2 . D. Problem 3: A solid insulating sphere of radius R has charge distributed uniformly throughout its volume. Problem 2: In the previous problem. since sphere with radius r such that r < R then we get E(r)4π Qenclosed = 0. within the radius r < R is given by Solutions Chapter 24 Page 1 of 8 . If we choose our Gaussian surface to be a Surface πr2 = 0. Without touching or disturbing it. what is the electric field inside shell S1 (r < R)? Answer: zero Solution: Gauss' Law tells us that r r Qenclosed Φ E = ∫ E ⋅ dA = ε0 .PHY2049 R. Field Chapter 24 Solutions Problem 1: Consider a spherical conducting shell S1 of radius R on which charge +Q is placed. this shell is now surrounded concentrically by a similar shell S2 of radius 2R on which charge -Q is placed (see Figure). If we choose our Gaussian surface to be a Surface πr2 = Q/εε0 and sphere with radius r such that R < r < 2R then we get E(r)4π solving for E gives E = KQ/r2. What is the magnitude of the electric field in the region between the two shells (R <r < 2R)? Answer: KQ/r2 Solution: Gauss' Law tells us that Shell S2 has charge -Q 2R R Shell S1 has charge +Q d t r r Qenclosed Φ E = ∫ E ⋅ dA = ε0 . Q(r). and solving for E gives E = 0. What fraction of the sphere’s total charge is located within the region r < R/2? Answer: 1/8 Solution: The amount of charge. D. How much electric charge (in nanoC) is located inside a shell with an inner radius of 0. D. Field r r 4 Q(r ) = ∫ ρdV = ∫ ρ 4πr 2 dr = πr 3 ρ .6 Solution: Gauss' Law tells us that Qenclosed = ε 0 ∫ E ⋅ dA = ε 0 (4πr22 E (r2 ) − 4πr12 E (r1 )) = 4πε 0 (r22 E (r2 ) − r12 E (r1 )) a 3 3 (150 N /(Cm))((1m)3 − (0. where a = 150N/(Cm).5 meters and an outer radius of 1. What is the total charge Q of the insulating sphere? Answer: πaR4 Solution: The amount of charge. Problem 4: A solid insulating sphere of radius R has a non-uniform volume charge distribution given by ρ(r) = ar. Thus.5m)3 ) = (r2 − r1 ) = = 14. For r = R/2.0 meters? Answer: 14. 2 0 0 0 Problem 5: In a certain region of space within a distribution of charge the electric field r is given by E ( r ) = ar rˆ . 3 Q(r ) r = . Q. where a is a constant.99 × 109 Nm 2 / C 2 Solutions Chapter 24 Page 2 of 8 . within the sphere is given by R R R Q(r ) = ∫ ρ (r )dV = ∫ (ar )4πr dr = 4πa ∫ r 3dr = πaR 4 . Q R where Q = 4π πR3ρ/3 is the sphere’s total charge.PHY2049 R.6nC K 8. 3 0 0 where ρ is the volume charge density and dV = 4π πr2dr. It points radially away from the origin and has a magnitude E(r) = ar. (r/R)3 = 1/8. D.5 meters? Answer: 5 Solution: Gauss' Law tells us that r r Qenclosed Φ E = ∫ E ⋅ dA = ε0 .5m) = = = 5nC .99 × 10 Nm / C y-axis Problem 7: L In a certain region of space within a distribution of r charge the electric field is given by E ( x ) = ax xˆ . How much electric charge (in nanoC) is located inside the cube with sides of length L = 2 m shown in the Figure? Answer: 10.PHY2049 R.6nC Solutions Chapter 24 Page 3 of 8 . At the surface of a sphere of radius R this Surface implies that E(R)4π πε0) πR2 = Q/εε0 and solving for Q and using K=1/(4πε gives. How much electric charge (in nanoC) is located inside a sphere with radius R = 0. Q= 9 2 2 K K 8.6 Solution: Gauss' Law tells us that x-axis O L L z-axis Qenclosed = ε 0 ∫ E ⋅ dA = ε 0 ( Ex ( L) L2 − Ex (0) L2 ) = ε 0 aL3 = (8. E ( R) R 2 aR (90 Nm / C )(0. It points radially away from the origin and has a magnitude E(r) = a/r. where a = 90Nm/C. It points in the x direction and has a magnitude Ex(x) = ax. where a = 150N/(Cm). Field Problem 6: r In a certain region of space the electric field is given by E ( r ) = ( a / r ) rˆ .85 × 10−12 C 2 / Nm 2 )(150 N / Cm)(2m)3 = 10. 5 N /(Cm))(2m)3 (8. Field Problem 8: y-axis Consider a cube of sides L=2 m.4 Solution: Gauss' Law tells us that ΦE = x-axis O L L z-axis r r Qenclosed ⋅ dA = E ∫ ε0 . what is the magnitude of the electric field at the point r = 4R outside the conductor? Answer: KQ/(16R2) Solution: Gauss' Law tells us that Solutions Chapter 24 Page 4 of 8 .85 × 10−12 C 2 /( Nm)) = 35. What is the total net charge within the cube (in picoC)? Answer: +35. D. The insulating Insulator sphere is surrounded by a solid spherical charge +Q conducting shell with inner radius 2R and outer radius 3R as shown in the Figure.PHY2049 R. where a=1 N/C and b=0. At the surface of the cube Surface φE = (E(L)-E(0))L2.5 N/(Cm). Conductor The conductor is in static equilibrium and has a zero net charge. and hence Q = ( E ( L) − E (0)) L2ε 0 = (a + bL − a) L2ε 0 = bL3ε 0 = (0. Suppose that a non-uniform electric field is L r E present and is given by ( x ) = ( a + bx ) xˆ . Net charge zero 3R A solid insulating sphere of radius R has a on conductor charge +Q distributed uniformly 2R throughout its volume (the volume charge R density ρ is constant). What is the magnitude of the electric field at the point r = 5R/2 inside the conductor? Answer: zero Solution: The electric field inside a conductor in static equilibrium is zero! Problem 10: In the previous problem. as shown in the figure.4 pC Problem 9: . For a Gaussian sphere of radius r < 3R this Surface πε0) implies that E(r)4π πr2 = Q/εε0 and solving for E(r) and using K=1/(4πε gives E(r) = KQ/r2 and plugging in r = 4R yields E = KQ/(16R2). ER 2 ( −1. ΦE Qenclosed 6. Hence.5 Solution: Gauss' Law tells us that r r Qenclosed Φ E = ∫ E ⋅ dA = ε0 .85×10 C / Nm ) Solutions Chapter 24 Page 5 of 8 .0 ×10−6 C Φ face = = = = 1.000 N / C)(20m) 2 = −44. Measurements indicate that there is a uniform electric field at the surface of the cloud that has a magnitude of 1.13×105 Nm2 / C .PHY2049 ΦE = R.13x105 Solution: The total flux through the cube is φE = Q/εε0 and symmetry tell us that the same amount of flux goes through each of the six faces of the cube. What is the electric flux (in Nm2/C) through one face of the cube? (Hint: think symmetry and don’t do an integral. −12 2 2 6 6ε 0 6(8. Problem 11: A thunderstorm forms a mysterious spherical cloud with a radius of 20 m.5µC .0 microC charge is at the center of a cube 10 cm on a side. What is the total net charge within the cloud (in microC)? Answer: -44. Field r r Qenclosed E ⋅ dA = ∫ ε0 .) Answer: 1. At the surface of the cloud Surface E(R)4π πR2 = Q/εε0 and solving for Q gives. Q= = K 8.99 × 109 Nm2 / C 2 Problem 12: A 6. D.000 N/C and is directed radially inward toward the center of the cloud. D. For r = 3R we get E = 4π πKRρ ρ/27. For a spherical surface inside the insulator (r < ε0 Surface πr3/(3εε0) and solving for E(r) gives. The insulating sphere is surrounded by a solid spherical conductor with inner radius R and outer radius 2R as shown in the figure.PHY2049 R. R) we have E(r)4π πr2 = Q(r)/εε0 = ρ4π Solutions Chapter 24 Page 6 of 8 . Field Problem 13: 2R A solid insulating sphere of radius R has charge distributed uniformly throughout its volume (the volume charge density ρ is constant). Problem 14: In the previous problem. what is the magnitude of the electric field at the point r=R/2 inside the insulator? Answer: 2π πKRρ ρ/3 Solution: Gauss' Law tells us that r r Q Φ E = ∫ E ⋅ dA = enclosed . For a spherical surface outside the Surface πR3/(3εε0) and solving for conductor (r > 2R) we have E(r)4π πr2 = Q/εε0 = ρ4π E(r) gives. The net charge on the conductor is zero. what is the magnitude of the electric field at the point r=3R/2 inside the conductor? Answer: zero Solution: The electric field is zero inside a conductor in static equilibrium. 4πKρR 3 E (r ) = 3r 2 . What is the magnitude of the electric field at the point r=3R outside the conductor? Answer: 4π πKRρ ρ/27 Solution: Gauss' Law tells us that ΦE = R Insulator ρ = charge/unit volume Conductor Zero net charge r r Qenclosed ⋅ dA = E ∫ ε0 . This is our definition of a conductor! Problem 15: In the previous problem. Field 4 E (r ) = πKρr . For r = R/2 we get E = KQ/(2R2). Solutions Chapter 24 Page 7 of 8 . The insulating sphere is surrounded by a solid spherical conductor with inner radius R and outer radius 2R as shown in the Figure. D. 4 KQr E (r ) = πKρr = 3 . Problem 17: In the previous problem. For a spherical surface inside the insulator Surface πr3/(3εε0) and solving with radius r (r < R) we have E(r)4π πr2 = Q(r)/εε0 = ρ4π for E(r) gives. 3 πKRρ ρ/3. For r = R/2 we get E = 2π Problem 16: A solid insulating sphere of radius R has a charge +Q distributed uniformly throughout its volume (the volume charge density ρ is constant). 3 R where I used Q = 4π πR3ρ/3. what is the magnitude of the electric field at the point r = 3R/2 inside the conductor? Answer: zero Solution: The electric field inside a conductor in static equilibrium is always zero.PHY2049 R. What is the magnitude of the electric field at the point r = R/2 inside the insulating sphere? Answer: KQ/(2R2) Solution: Gauss' Law tells us that ΦE = 2R R Insulator ρ = charge/unit volume Conductor Zero net charge r r Qenclosed ⋅ dA = E ∫ ε0 . The conductor is in static equilibrium and has a net charge +Q. and solving for E gives E = KQ/r2. Solutions Chapter 24 Page 8 of 8 . D. the conducting shell has a total net charge of +Q. How much charge is located on the outer surface (r = 2R) of the conducting shell? Answer: 2Q Solution: We know that the net charge on the conductor is given by the sum of charge on the inner and outer surfaces. what is the magnitude of the electric field in the region r < R inside the hole in the conducting shell? Answer: KQ/r2 Solution: Gauss' Law tells us that r r Qenclosed Φ E = ∫ E ⋅ dA = ε0 . Field Problem 18: Net charge +Q A point charge +Q is located at the 2R on conductor center of a solid spherical conducting shell with inner radius of R and outer R radius of 2R as shown in the Figure. If we choose our Gaussian surface to be a Surface sphere with radius r such that r < R then we get E(r)4π πr2 = Q/εε0. If we choose our Gaussian surface to be a Surface sphere with radius r such that R< r < 2R then E = 0 (inside conducting material) which implies that Qenclosed = Qin+Q = 0. In +Q addition. Problem 19: In the previous problem. Qin = -Q and Qout = 2Q. Gauss' Law tells us that r r Qenclosed Φ E = ∫ E ⋅ dA = ε0 .PHY2049 R. Thus. Qnet = Qin + Qout. D.24 d L .79 L . What is the electric potential at the center of the cube? (Take V = 0 at infinity as the reference point. Field Chapter 25 Solutions Problem 1: If the potential difference between points A and B is equal to VA-VB = 3x104 Volts. 3 L 3L / 2 . Solutions Chapter 25 Page 1 of 12 . Q. how much work (in milliJoules) must be done (against the electric force) to move a change particle (Q = 3 microC) from point A to point B without changing the kinetic energy of the particle? Answer: -90 Solution: WAB = Q(VB-VA) = (3x10-6 C)(3x104 V) = 9x10-2 J = 90 microC. Problem 3: What is the electric potential energy of the configuration of eight identical point charges. 2 i =1 i i 2 2 3 L where Vcorner is the electric potential at the corner of the cube due to the other 7 charges. We know that the work done against the electric field must be negative since the particle will fall from high potential to low potential.PHY2049 R.) Answer: 9. Problem 2: Eight identical point charges.2KQ/L Solution: The potential at the center of the cube is given by Vcenter = where I used d = KQ 8 KQ 16 KQ = = 9. Q.8KQ2/L Solution: The total electric potential energy is given by U cube KQ 2 1 8 8 3 1 KQ = ∑ QV = QVcorner = 4Q 3 + + = 22. at the corners of the cube with sides of length L in the previous problem? Answer: 22. are located at the corners of a cube with sides of length L. 10 2 L y-axis Problem 5: Two point charges Q1 = Q and Q2 = Q are on the x-axis at x = +L and x = -L and a third point charge Q3 = Q is on the yaxis at the point y = L as shown in the previous Figure. What is the electric potential energy of this charge configuration? Answer: 1. x = -L ∆V = V2-V1 between the point P2 at the midpoint of the line between Q1 and Q3 and the point P1 at the origin? Answer: 0.5KQ/L Solution: The electric potential at P1 is given by. Q 1=Q What is the electric potential difference.PHY2049 R. 2 KQ 4 KQ + − 3 = 0. is given by.46 V2 − V1 = L . U1. D. At P2 the electric potential is V2 = where d = 2 KQ KQ 4 2 KQ + = + 2L / 2 d 2 10 L ( 2 L / 2) 2 + ( 2 L) 2 = 10 L / 2 . Solutions Chapter 25 Page 2 of 12 .9KQ2/L Q1=Q x = -L Q3=Q Q2=Q x = L x-axis Solution: The electric potential energy of the charge configuration. Field Problem 4: y-axis Two point charges Q1 = Q and Q2 = Q are on the x-axis at x = +L and x = -L and a third point charge Q3 = Q is on the y-axis at the point y = L as shown in the Figure. V1 = 3 Q 3=Q y=L 2 d Q 2=Q 1 x = L x-axis KQ L . Thus. What is the electric potential at the point P that bisects the hypotenuse? (Take V = 0 at infinity as the reference point. how much work must be done (against the electric force) to move the charge Q3 from the point y = L on the y-axis to the origin resulting in the configuration shown in the Figure? Answer: 0.) Answer: 4.91 U1 = L . KQ 2 2 KQ 2 5 KQ 2 U2 = + = 2L L 2 L . P L Q Q L KQ KQ KQ =3 2 = 4. are located at the corners of a right triangle with sides of length L as shown in the Figure.6KQ2/L Solution: The potential potential energy of this new charge configuration. is given by. where I used r = 2 L / 2 = L / 2 . D. 2 2 Problem 7: Q Three identical point charges.24KQ/L Solution: The electric potential at the point P that bisects the hypotenuse is given by. 2L 2L 2 L Problem 6: In the problem 4.59 W = U 2 − U1 = − − 2 L L L .24 r L L . and thus the work required is ( ) 2 KQ 2 KQ 2 5 1 KQ = 2− 2 = 0. Q. V =3 Solutions Chapter 25 Page 3 of 12 .PHY2049 R. Field 2 KQ 2 2 KQ 2 1 KQ 2 KQ + = + 2 = 1. U2. PHY2049 R.71 = 2 + L L 2 Problem 9: Q In problem 7. is given by. Q.828 L L 2 Solutions Chapter 25 Page 4 of 12 . is given by.71KQ2/L Solution: The potential potential energy of the charge configuration. U1. D. 1 = = + L 2 2 L 2 and thus the work required is 1 KQ 2 5 −2− W = U 2 − U1 = 2 L 2 KQ 2 4 KQ 2 = − 2 = 0. L Q L Q KQ 2 KQ 2 U2 = +2 2L 2L / 2 4 KQ 2 5 KQ 2 . KQ 2 KQ 2 U1 = 2 + L 2L KQ 2 1 KQ 2 = 2. Field Problem 8: What is the electric potential energy of the configuration of three identical point charges.828KQ2/L Solution: The potential potential energy of this new charge configuration. located at the corners of a right triangle with sides of length L in the previous problem? Answer: 2. how much work is required to move the charge located at the 90o angle of the equilateral triangle to the point P that bisects the hypotenuse? Answer: 0. U2. PHY2049 R. D. Field Problem 10: Three identical charged particles (Q = 3.0 microC) are placed at the vertices of an equilateral triangle with side of length L= 1 m and released simultaneously from rest as shown in the Figure. What is the kinetic energy (in milliJ) of one of the particles after the triangle Q has expanded to four times its initial area? Answer: 40.5 Solution: Here we use energy conservation as follows: Q L L Q L KQ2 Ei = 0 + 3 L KQ 2 , E f = 3KE + 3 2L where KE is the kinetic energy of one of the particles and where Li = L and Lf = 2L (since the area increased by factor of 4). Setting Ei = Ef yields KQ2 1 KQ2 (8.99 × 109 Nm2 / C 2 )(3 × 10−6 C)2 KE = 1 − = = = 40.5mJ . L 2 2L 2(1m) Problem 11: E 2m A uniform electric field E=10 V/m points in the x direction as shown in the Figure. Point A has coordinates (0,1m) and point B has coordinated (2m,2m), where points are labeled according to P=(x,y). What is the potential difference VB-VA (in Volts)? Answer: -20.0 Solution: The potential difference is equal to the following line integral, C B A C r r r r B r r VB − V A = − ∫ E ⋅ dl = − ∫ E ⋅ dl − ∫ E ⋅ dl , B A Solutions A Chapter 25 C Page 5 of 12 PHY2049 R. D. Field where I have chosen the path shown in the Figure. Evaluating the line integral along AC gives zero since E is perpendicular to the path and along the path CB E is parallel to the path so that, r r ∫ E ⋅ dl = 0 C A r r ∫ E ⋅ dl = Ed = (10V / m)(2m) = 20V B C and hence VB-VA = -20V. Problem 12: A uniform electric field E=400 V/m crosses the x-axis at a 40o angle as shown in the Figure. o B A 5 cm 40 Point B lies on the x-axis 5 cm to the right of the origin A. What is the potential difference E VB-VA (in Volts)? Answer: -15.3 Solution: We know that B r r V B − V A = − ∫ E ⋅ dr and hence VB-VA = -E cosθ L = -(400 V/m) cos(40o) A (0.05 m) = -15.3 V. E(x) Problem 13: A non-uniform electric field is given by r E ( x ) = ax 2 xˆ , where a = 10 V/m3 as shown in the Figure. Point B lies at x=y=3 m and point A is at the origin (x=y=0). What is the potential difference VB-VA (in Volts)? Answer: -90 Solution: The potential difference is equal to the following line integral, Solutions Chapter 25 B C A 3m Page 6 of 12 PHY2049 R. D. Field C r r r r B r r VB − V A = − ∫ E ⋅ dl = − ∫ E ⋅ dl − ∫ E ⋅ dl , B A A C where I have chosen the path shown in the Figure. Evaluating the line integral along CB gives zero since E is perpendicular to the path and along the path AC E is parallel to the path so that, r r − ∫ E ⋅ dl = 0 B C r r 1 − ∫ E ⋅ dl = − ∫ ax 2 dx = − a(3m)3 = −90V 3 0 A 3 C and hence VB-VA = -90V. Problem 14: The potential along the x-axis is given by V(x) = ax-bx2, where a=2 V/m and b=1 V/m2. At what value(s) of x (in m) is the electric field equal to zero? Answer: 1 Solution: The electric field is minus the derivative of the potential, Ex = − dV = −( a − 2bx ) = 0 . dx Solving for x gives x= a 2V / m = = 1m . 2b 2(1V / m2 ) Problem 15: The potential is given by V(x,y) = a/(x2+y2), where a = 2 Vm2. What is the magnitude of the electric field (V/m) at the point, P = (x,y) = (1m,1m)? Answer: 1.4 Solution: The three components of the electric field are given by, Solutions Chapter 25 Page 7 of 12 b = -4 V/m .and c = 5 V/m2.0)? Answer: 5 Solution: The three components of the electric field are given by.PHY2049 R.y. Solutions Chapter 25 Page 8 of 12 . ∂V = −a ∂x ∂V Ey = − = −b ∂y ∂V Ez = − = −2cz ∂z Ex = − and the magnitude of the electric field is E = E x2 + E y2 + Ez2 = a 2 + b 2 + 4c 2 z 2 .y.0. where a = 3 V/m. P=(x. What is the magnitude of the electric field (in V/m) at the origin. At the origin z = 0 and the magnitude of the electric field is E = a 2 + b 2 = (3V / m) 2 + (−4V / m) 2 = 5V / m .z)=(0. D.41V / m . At the origin x = 1m and y = 1m the magnitude of the electric field is 2(2Vm 2 ) = 2V / m = 1. E= 3 ( 2 m) Problem 16: The potential is given by V(x. Field ∂V = 2 xa /( x 2 + y 2 ) 2 ∂x ∂V = 2 ya /( x 2 + y 2 ) 2 Ey = − ∂y ∂V =0 Ez = − ∂z Ex = − and the magnitude of the electric field is E = Ex2 + E y2 + E z2 = 2a ( x 2 + y 2 )3 / 2 .z) = ax+by+cz2. D. V= +λ -λ R 2R KQ1 KQ2 K 2πr1λ K 2πr2 λ + = − = 2πK (λ − λ ) = 0 .PHY2049 R.293KQ/R Solution: The net potential is the sum of the potential of the ring plus the potential of the point charge as follows: KQ KQ − V ( z) = 2 z which at z =R becomes z + R2 V ( R) = Solutions KQ 1 KQ − 1 = −0.293 R 2 R . A negatively -Q charged particle (charge -Q) is at the center of the R ring as shown in the Figure. distributed along their circumferences as shown in the Figure. (Take V=0 at infinity z-axis as the reference point. Problem 18: A positive charge Q is distributed uniformly along Total Charge +Q on ring a thin circular ring of radius R. r1 r2 r1 r2 where the charge on a circular rod is Q = 2π πrλ λ. Chapter 25 Page 9 of 12 . What is the electric potential at a point on the z-axis of the ring a distance R from the center. respectively.) Answer: zero Solution: The electric potential at the center of the two circles is given by. Field Problem 17: Two thin concentric circular plastic rods of radius R and 2R have equal and opposite uniform charge densities -λ λ and +λ λ. What is the electric potential at the center of the two circles? (Take V = 0 at infinity as the reference point.) Answer: -0. 4kV Solutions Chapter 25 Page 10 of 12 . If point P1 = (a. 2 1 1 3 where I have chosen the path shown in the Figure. Evaluating the line integral along 32 gives zero since E is perpendicular to the path and along the path 13 E is parallel to the path so that.b). x-axis 3 r r 2 r r r r ∆V = V2 − V1 = − ∫ E ⋅ dl = − ∫ E ⋅ dl − ∫ E ⋅ dl .4 Solution: We know that the electric field from and infinite rod points away from the rod and has a magnitude given by E(r) = 2Kλ λ/r. what is the 3 1 potential difference ∆V = V2-V1 (in a kiloVolts)? E = 2Kλ/r Answer: -37. D.0) and point P2 = b (2a. r r − ∫ E ⋅ dl = 0 2 3 2a r r 2 Kλ − ∫ E ⋅ dl = − ∫ dx = − 2 Kλ ln(2) x 1 a 3 and hence ∆V = V2 − V1 = −2 Kλ ln(2) = −2(8.99 × 109 Nm 2 / C 2 )(3 × 10−6 C / m) ln(2) = −37. where P = (x.PHY2049 R.y). The potential difference is equal to the following line integral. Field y-axis Problem 19: An infinitely long thin rod lies along the y-axis and has a uniform linear charge 2 2a = 3 C/m density λ = 3 microC/m as shown in the λ µ Figure. )? Answer: -2KQ/R Solution: The electric potential at the center of the circle is given by. 2L V= where I used dQ = λdx. Field L L dQ x P Charge Q on Rod Problem 20: A positive charge Q is distributed uniformly along a thin rod of length L as shown in the Figure. D.693KQ/L Solution: The electric potential at the point P is given by Kλdx KQ KQ = λ = = K ln( 2 ) ln( 2 ) 0 . Calculate the electric potential (in Volts) at the center of the conducting sphere.) Answer: 2 Solutions Chapter 25 Page 11 of 12 . What is the electric potential at a point P on the axis of the rod a distance L from the end of the rod? (Take V = 0 at infinity as the reference point.PHY2049 R. 693 ∫L x L L . Problem 22: The electric field at the surface of a solid spherical conductor (radius R = 20 cm) points radially outward and has a magnitude of 10 V/m. Problem 21: A circular plastic rod of radius R has a -3Q positive charge +Q uniformly distributed R along one-quarter of its circumference and a negative charge of -3Q uniformly distributed +Q along the rest of the circumference as shown in the Figure.) Answer: 0. V= KQ 3KQ KQ − = −2 R R R . What is the electric potential at the center of the circle? (Take V = 0 at infinity as the reference point. (Take V = 0 at infinity as the reference point. R r r KQr 1 KQ = − ∫ E ⋅ dr = − ∫ 3 dr = − R 2 R . Hence. what is the potential at the surface r = R (in Volts)? Answer: 8 Solution: The electric field inside the insulator (r < R) is given by E(r) = KQr/R3 and the potental difference between the center and the surface is. Field Solution: At the surface of the conductor we know that. surface 2 R and 3 Chapter 25 Page 12 of 12 . D. 0 0 R Vsurface − Vcenter We know that Vsurface = KQ/R which means that Vcenter = Solutions 3 KQ 2 = V Vcenter = 8V . Problem 23: A solid insulating sphere of radius R has charge a total charge Q distributed uniformly throughout its volume (the volume charge density ρ is constant).PHY2049 R. If the electric potential is zero at infinity and 12 Volts at the center of the sphere. Vsurface = KQ R Esurface = KQ R2 . Vcenter = Vsurface = Esurface R = (10V / m)(0.2m) = 2V . 0x10-3 Joules is stored by a capacitor that has a potential difference of 85 Volts across it. where r = 2R is the surface of the conductor.5 Solution: We know that 1 U = C (∆V ) 2 . Hence. C= (∆V ) 2 (85V ) 2 Problem 3: Capacitor #1 with capacitance C and initial charge Q is connected in parallel across an initially uncharged capacitor with capacitance 2C (#2). After the system comes to equilibrium. 2 and hence 2U 2(9 × 10−3 J ) = = 2. Field Chapter 26 Solutions Problem 1: What is the capacitance C of a solid spherical conducting shell with inner radius R and outer radius 2R? Answer: 2R/K Solution: The electric potential of the solid spherical conductor is given by V= KQ KQ = r 2R . Problem 2: An energy of 9. C= Q 2R = V K .PHY2049 R. D. what is the charge on capacitor #1? Answer: Q/3 Solution: After the two capacitors are connected their charges must satisify Solutions Chapter 26 Page 1 of 4 . Determine the capacitance of the capacitor (in microF). Answer: 2.5µF . Solving for Q1 yields Q1 = Q 3.PHY2049 R. D.5 Solution: At the surface of the conductor we know that.1 J/m3. Vsurface = KQ R Esurface = KQ R2 and the electric field energy density is Solutions Chapter 26 Page 2 of 4 . Field Q1 + Q2 = Q Q1 Q2 = C 2C whre the first condition comes from charge conservation and the second comes from the fact that the potential drop across each capacitor is the same since they are connected in parallel. Vsurface = KQ R Esurface = KQ R2 .11 Solution: At the surface of the conductor we know that.11J / m3 −2 2 2(5 × 10 m) Problem 5: The electric field energy density near the surface of an isolated charged conducting sphere with a radius of 5 cm is 0. the electric field energy density is 1 1 V2 2 u = ε0E = ε0 2 2 2 R (8. What is the electric field energy density (in J/m3) at a point near the surface outside the conductor? Answer: 0. Hence. What is the electric potential (in kiloVolts) at the center of the conductor? Answer: 7.85 × 10−12 C 2 /( Nm 2 ))(8 × 103V ) 2 = = 0. Problem 4: A charged conducting sphere with a radius of 5 cm has a potential of 8 kilovolts relative to infinity. E = Ayx$ . The electric field points O in the x direction and has a constant L magnitude given by E(y)=Ay.05m) = 7. Problem 6: z-axis In a certain region of space the electric field r is given by. D.85x10-12 C2/(Nm2))(100 V/m2)2(20 m)5/6 = 47.PHY2049 R.2 Solution: The energy stored in the electric field lines is given by L L 1 1 1 2 2 2 2 U = ∫ udV = ∫ ε0 E L dy = ε0 A L ∫ y 2 dy = ε0 A 2 L5 . What is the total amount of electric potential energy (in milliJoules) contained in the electric field L within the cube with sides of length L=20 x-axis meters shown in Figure? Answer: 47. Solutions Chapter 26 Page 3 of 4 .5kV 8. 2 2 6 0 0 y-axis and plugging in the numbers gives U = (8. Vcenter = Vsurface = R 2u ε0 2(0. Field 1 u = ε0E2 .1J / m3 ) = (0. 2 Thus. L where A=100 V/m2.85 × 10−12 C 2 /( Nm 2 ) where I have used the fact that the potential is constant throughout the conductor.2 mJ. b<d) of d the same area A is now carefully inserted between the plates parallel to them without -Q touching. the conducting material will be pulled in by the electrostatic force with the difference in stored energy doing the work. and what is the new capacitance? Answer: ε0A/(d-b). After the conducting material is inserted. If the charge Q on the capacitor was unchanged during this operation. Ci = ε0A/d. must the metal slice be forced between the plates or will it be pulled in by electrostatic forces. we have two capacitors in series with C= ε0 A 2ε0 A = d b (d − b ) − 2 2 and Ceff = ε A C = 0 2 ( d − b) . Since the final stored energy is less than the initial stored energy. Field Problem 7: Given an open-air parallel-plate capacitor +Q of plate area A and plate separation d as shown in the Figure. and stored energy Q2 Q2d = Ui = 2Ci 2ε0 A . Solutions Chapter 26 Page 4 of 4 b . pulled Solution: Initially the capacitor has E = Q/(εε0A). The stored energy is now Q2 Q 2 (d − b ) = < Ui Uf = 2Ceff 2ε0 A . A flat slice of conducting material (thickness b.PHY2049 R. D. ∆V = Ed. and the two wires have the same volume. where V is the volumn.e. RB = VB VA VA 1Ω Problem 2: How much current passes through the 1 Ω resistor in the circuit shown in the Figure? Answer: 6I/11 Solution: Since the potential drop across each resistor is the same (i. parallel) we know that I1 I I2 2Ω 3Ω I3 I1R1 = I 2 R2 = I 3 R3 and in addition I1 + I 2 + I 3 = I . Hence. Field Chapter 27 Solutions Problem 1: Wire B is three times longer than wire A. and resistivity ρ is given by ρL ρL2 = R= A V . I1 + R R R1 I1 R1I1 + = I1 1 + 1 + 1 = I R2 R3 R2 R3 and solving for I1 gives I1 = Solutions I R R 1 + 1 + 1 R 2 R3 = I 6 = I 1 1 11 1 + + 2 3 Chapter 27 Page 1 of 3 . If both wires are made of the same substance and if the resistance of wire A is 9 Ohms. D.PHY2049 R. ρL2B ρ (3LA ) 2 ρL2A = =9 = 9 RA = 9(9Ω) = 81Ω . what is the resistance of wire B (in Ohms). V = LA. Answer: 81 Solution: We know that the resistance of a wire of length L. cross sectional area A. Thus. How much total current I passes through the wire? Answer: πaR4/2 Solution: The current I is the “flux” associated with the current density as follows: R r r R 2 1 I = ∫ J ⋅ dA = ∫ ar 2πrdr = 2πa ∫ r 3dr = πaR 4 . If a = 2 A/m. D.PHY2049 R.1 m is carrying the non-uniform current density which points along the axis of the wire and has a magnitude.26 Solution: The current I is the “flux” associated with the current density as follows: r r R a I = ∫ J ⋅ dA = ∫ 2πrdr = 2πaR = 2π (2 A / m)(0.1m) = 1.26 A . 2 0 0 ( ) Problem 4: Consider a long straight cylindrical wire with radius R = 0. r 0 Solutions Chapter 27 Page 2 of 3 . where r is the distance from the center axis of the wire. J(r)=ar2. where r is the distance from the center axis of the wire. J(r) = a/r. Field Problem 3: Consider a long straight cylindrical wire with radius R carrying the nonuniform current density. how much total current I passes through the wire (in A)? Answer: 1. PHY2049 R. Solutions Chapter 27 Page 3 of 3 . We then combine the three resistors in series and then combine r and R+2r in parallel giving. 1 1 1 = + R r R + 2r . Answer: 0. which implies that ( ) R 2 + 2rR − 2r 2 = 0 and R = − 1 + 3 r = 0. Calculate the effective resistance. D. (in Ohms) of the network between the terminals A and B given that each of the resistors has resistance r = 1 Ohm. R. Field A r R r r r r r r r r r r r B cut Problem 5: Consider the infinite chain of resistors shown in the Figure.7 Solution: We first A cut the infinite chain A r at the point shown in the figure and R R R+2r R r replace the infinite r chain to the right of r the cut by its B B effective resistance R.73r. Field Chapter 28 Solutions R0 = 3 Ω Problem 1: Consider the circuit consisting of an EMF and three resistors shown in the Figure. Solutions Chapter 28 Page 1 of 5 . how much power is dissipated in the transmission line (in W) when the power station delivers an EMF of 20 Volts (assuming the same power is supplied in both cases)? Answer: 50 Solution: The powers supplied by the station is given by Pstation = I1V1 = I 2V2 .PHY2049 R. is constant so that I 2 V1 = I1 V2 . 24 V R1 = 4 Ω R2 = 12 Ω How much current flows through the 4 Ω resistor (in Amps)? I1 I2 I Answer: 3 Solution: The original circuit is equivalent to a circuit with and EMF and one resistor with resistance Rtot given by Rtot = 3Ω + Reff = 3Ω + 3Ω = 6Ω . Thus. If the power dissipated in the transmission line is 200 Watts when the power station delivers an EMF of 10 Volts. Ω from where Reff = 3Ω 1 1 1 1 = + = Reff 4Ω 12Ω 3Ω . the current I = 24V/6Ω Ω = 4A. Now. R1 + R2 16 Problem 2: A constant power P is supplied to a transmission line of resistance R by a power station. D. we know the I = I1 + I2 and that ∆V1 = I1R1 = ∆V2 = I2 R2 and hence I1 = R2 12 I = (4 A) = 3 A . R 12V After the switch is closed (EMF = 12 V). where I used Qmax = εC. how long does it take (in s) for it to loose 90% of its initial charge? Answer: 13. D. Field where V1 = 10 V and V2 = 20 V.PHY2049 R. Problem 5: A charged capacitor C has a resistance R connected across its terminals to form the RC circuit shown in the Figure. Now the power disapated in the line is P = I2R so that 2 2 V I P P2 = 2 P1 = 1 P1 = 1 = 50W . what is the magnitude of the potential difference across the C capacitor (in V) when the charge on the capacitor reaches one-third of its maximum value? Answer: 4 Solution: The potential difference across the is ∆VC = Q/C = εC/(3C) = ε/3 = 4 V.1τ . where τ = RC and where I used Q(t)/Q0 = 1/3. Problem 4: The capacitor in the Figure is initially uncharged.1 Solution: For an RC circuit with the charge Q(t) given by Q(t ) = Q0e− t /τ and t = −τ log(Q(t ) / Q0 ) = −τ log(1 / 3) = 1. How many time constants τ = RC must elapse in order for the capacitor to lose 2/3 of its charge? Answer: 1.3 Solutions Chapter 28 C R Page 2 of 5 . 4 V2 I1 The power station loses less energy in the line if it transmits power at a higher voltage! Problem 3: A capacitor C with an initial charge Q discharges through a resistor R. If it takes 2 seconds for the capacitor to loose one-half of its stored energy. Q(t). and solving for τ gives τ = RC = − 2t1 − 2t1 = ln(U (t ) / U 0 ) ln(0. Thus. as a function of time is given by Q(t ) = Q0 e −t / RC = Q0 e −t /τ Q2 = U 0 e −2t / RC = U 0e −2t /τ .5) ln(0. U (t ) = 2C where U0 = Q02/(2C) and Q0 = CV0. If ta is the time when the stored energy reaches 50% of its initial value.5) Problem 6: A capacitor.3s . in the circuit (in A) when the stored energy in the capacitor has dropped to 50% of its initial value? Answer: 3. and stored energy. Field Solution: The stored energy in the RC circuit is given by Q( t ) 2 = U 0 e − 2 t /τ U (t ) = 2C .5) . I. then e −2 ta / τ = 1 1 −ta / τ = e and taking the square-root gives .5 Solution: The charge. U(t). Here V0 is the potential across the capacitor at t = 0 (equal to the EMF that fully charged the capacitor). ln(0. where t1 = 2 s. C.PHY2049 R. is (fully) charged by connecting it I to a 10 Volt EMF and then is allowed to discharge through a 2 Ω resistor as shown in the Figure? R=2Ω C What is the current. t= 2t1 ln(01 . D. The charge as a function of time is Q(t ) = Q0e− t /τ and t = −τ ln(Q(t ) / Q0 ) .) = = 13. ) 2(2s) ln(01 . 2 2 The current at time ta is given by Solutions Chapter 28 Page 3 of 5 . 2 R 2 2Ω 2 where I used I0 = V0/R. Suppose that a charged leaky 2. Field 5 V 1 10V 1 I (t a ) = I 0 e −ta /τ = 0 = = A = 3. in megaOhms? (Hint: the leaky capacitor forms an RC circuit.0 microF capacitor's potential difference drops to one-fourth its initial value in 2.721MΩ . C ln(Q(t ) / Q0 ) (2 × 10−6 F ) ln(0. while the capacitor on the right discharges through two equal resistors with resistance R connected in parallel.72 Solution: The leaky capacitor forms an RC circuit with the charge Q(t) given by Q(t ) = Q0e − t /τ and τ = RC = −t ln(Q(t ) / Q0 ) . D. If the switches are closed simultaneously.5 A . Problem 7: A "leaky" capacitor's faulty insulation allows charge to pass slowly from one plate to the other. how much charge is Solutions Chapter 28 Page 4 of 5 .0 sec.25) R C R C left R right Problem 8: Consider the discharging of the two capacitors shown in the Figure. What is the equivalent resistance between the capacitor plates.) Answer: 0. Both capacitors have equal capacitance C and both have an initial charge of 8 microC. When the switches are closed the capacitor an the left will discharge through a single resistor with resistance R. Solving for the resistance R gives R= −t − (2s) = = 0.PHY2049 R. when the left side Q(t)/Q0 = 4/8 = 1/2 the right side has Q(t)/Q0 = 1/4 so that Qright = 2 µC.PHY2049 R. Solutions Chapter 28 Page 5 of 5 . Hence. = Q Q 0 right 0 left Thus. where I used the fact the the effective resistance on the right side is Reff = R/2. D. Field on the right capacitor (in microC) when the left capacitor’s charge is 4 microC? Answer: 2 Solution: The charge as a function of time on the two capacitors is given by Qleft (t ) = Q0e −t / RC Qright (t ) = Q0e −2t / RC . 2 Q(t ) Q(t ) . Problem 3: One end of a straight wire segment is located at (x. 2 −1 0 Note that v is in the same direction as B and hence the magnetic force is zero. Q = 0.y. Problem 2: A charged particle.2m. if its velocity is v = 4 xˆ − 2 yˆ (in m/s)? Answer: zero Solution: We know that the magnetic force on a particle is xˆ yˆ zˆ r r r F = Qv × B = Q 4 − 2 0 = 0 . Q=0. What is the magnitude of r the magnetic force on the particle (in N). and mass M = 200 grams enters a region with r a uniform magnetic field B = 2 xˆ − 1 yˆ (in Tesla).6 Solution: We know that the magnetic force on a particle is xˆ r r r F = Qv × B = Q 2 2 yˆ zˆ 3 3 2 = Q (6 xˆ − 4 yˆ ) . 4 The magnitude of the force is F = Q 36 + 16 = ( 0 .5 ) 52 N = 3 . D. A current of 2 Amps flows through the r segment. enters a region with a uniform magnetic field r r B = 2 xˆ + 3 yˆ + 4 zˆ (in Tesla).0.PHY2049 R.5 C. what is the magnitude of the magnetic force on the particle (in N)? Answer: 3. 6 N .0) and the other end is at (1m.5 C. Field Chapter 29 Solutions Problem 1: A charged particle. If its velocity is given by v = 2 xˆ + 3 yˆ + 2 zˆ (in m/s). The segment sits in a uniform magnetic field B = 2 xˆ − 1 yˆ (in Solutions Chapter 29 Page 1 of 5 .z) = (0.3m). 73 N . Problem 4: An electric power transmission line located an average distance of 20 m above the earth's surface carries a current of 800 Amps from east to west. 2 −1 0 with the magnitude of the force given by r F = I 9 + 36 + 25 Tm = ( 2 A )8 . L Problem 5: A charged particle. What is the magnitude of the force per meter on the line (in milliN/m)? Answer: 64 Solution: We know that the force on a straight wire is given by r r r F = IL × B . What is the magnitude of the magnetic force (in N) on the wire segment? Answer: 16.2 Tesla. traveling in the x-direction with velocity r v = v 0 xˆ enters a region of space that has an electric field in the y-direction r given by E = E 0 yˆ with E0 = 10 Volts and a magnetic field in the zr B direction given by = B0 zˆ with B0 = 0.8 gauss due north at 60o below the horizontal. what is its speed v0 (in m/s)? Answer: 50 Solution: The electric force is given by Solutions Chapter 29 Page 2 of 5 . In this case L and B are perpendicular and hence.1 C. 8 × 10 − 4 T ) = 64 × 10 − 3 N / m . F = IB = (800 A )( 0 .PHY2049 R. 367 Tm = 16 . Q=0.7 Solution: We know that the magnetic force on a wire segment is xˆ yˆ zˆ r r r F = IL × B = I 1 2 3 = I (3 xˆ + 6 yˆ − 5 zˆ )Tm . If the particle experiences no net force and continues with the same speed and direction. in a region where the earth's magnetic field is 0. D. Field Tesla). 01 kg m/s enters a region of space that has a uniform 1 Tesla magnetic field in the z-direction as shown the Figure. Solutions Chapter 29 Page 3 of 5 .PHY2049 R. We know that the sign of the charge must be negative since the particle moves to the left. If 0. B0 0 . mv 2 | Q | vB = . RB dB ( 0 . what is the charge of the particle (in milliC)? Answer: -40 Solution: The radius of the circular path followed by the charged particle calculated by setting the force equal to the mass times the centripetal acceleration as follows. 2T Problem 6: A charged particle traveling in the y-direction B-out with a momentum of 0. D. R and solving for the magnitude of the charge gives | Q |= mv 2 p 2 ( 0 . Field r FE = QE 0 yˆ and the magnetic force is r r r r r FE = Q v × B = − Qv 0 B 0 yˆ and setting FE + FB = 0 implies that v0 = E0 10 V = = 50 m / s .5 m )(1T ) where I used p = mv and d = 2R.5 m to the left of the point A.01 kgm / s ) = = = 40 mC .5 m the particle enters the magnetic field at the point A and then exits the magnetic field at the point B located a distance of 0. QB R Thus. mv mv 2 QvB = . what is the difference in their mass. 2 1 and both have the same speed. ∆M = M2-M1 (in grams)? Answer: 3 Solution: The radius of the circular path followed by a charged particle is calculated by setting the force equal to the mass times the centripetal acceleration as follows. ∆x v If the particles are a distance ∆x = x2-x1 = 0. Both x-axis particles have the same charge. D.1 C. Field Problem 7: A charged particle traveling in the y-direction B-out enters a region of space that has a uniform 2 R Tesla magnetic field in the z-direction as shown in the Figure.1 Solution: The radius of the circular path followed by the charged particle calculated by setting the force equal to the mass times the centripetal acceleration as follows.14 s .2 kg how long v (in seconds) does it take for the particle to reverse direction and exit the region? Answer: 3. If the particle has a charge of 0.3 meters apart when they exit the region. v QB ( 0 . Mv Mv 2 QvB = .PHY2049 R. 1C )( 2T ) Problem 8: y-axis Particle #1 and particle #2 travel along the y-axis and enter a region of space that has a B-out uniform 2 Tesla magnetic field in the zdirection as shown in the Figure. v = 10 m/s. and solving for the radius gives R = . Solutions Chapter 29 Page 4 of 5 .2 kg ) = = = 3 .1 C and a mass of 0. Q=0. and solving for the radius gives R = . QB R The time is given by the distance traveled divided by the speed as follows: t= πR πm π ( 0 . PHY2049 R. perpendicular to the axis of the loop).4 Solution: The magnetic field on the z-axis of the loop a distance d = 0. D.5 meters from the loop is 2 kIπ R 2 ( 2 × 10 −7 Tm / A )( 2 A )π (1m ) 2 −7 8 99 10 . Bz = 2 T. What is the I magnitude of the magnetic force (in milliN) on the particle due to the loop? Answer: 5.e. = × / 3 2 2 3/ 2 = 2 2 (z + R ) (( 0.3 m ) = 3 g 2 (10 m / s ) where I used ∆ x = x 2 − x1 = 2 R 2 − 2 R1 = 2 ∆ R . Solutions Chapter 29 Page 5 of 5 .1C )( 2T ) ( 0 .4 mN . Field ∆M = M 2 − M 1 = QB QB ∆x ( R 2 − R1 ) = v 2v ( 0 . A particle with charge Q = 3x10-3 C is on the axis of the loop d (z-axis) a distance of d = 0.5m) + (1m) ) The magnitude of the magnetic force on the particle is r F = QvB = ( 3 × 10 − 3 C )( 2 × 10 6 m / s )( 8.99 × 10 − 7 T ) ≈ 5.5 meters away from the loop and is moving with a speed of 2x106 m/s along the x-axis R (i. = Problem 9: z-axis v A circular loop of wire of radius R = 1 m is carrying a current of 2 Amps as shown in the Figure. If the wires carry a current I.PHY2049 R. 62 R R . If extrapolated to their intersection point straight wires would form a 30o angle. out of page v d l × rˆ = 0 and wire B Solution: Wire A does not contribute at P since (semi-infinite wire) contributes kI/d (one half of 2kI/d). Field Chapter 30 Solutions I Problem 1: d P A The bent wire carries current I. R Chapter 30 Page 1 of 11 . D. at distance d B o directly opposite the 90 bend (k = µ0/4π)? Answer: kI/d. as shown the Figure. What is the magnetic field at point P. what is the magnitude of the magnetic field at the center of the circular arc (k = µ0/4π)? Answer: 4.62 ) = 4 . Problem 2: Two semi-infinite straight pieces of wire are connected by a circular loop segment of radius R as shown I 30o in the Figure.62kI/R Solution: The magnetic field at the center of the circular loop segment is the superposition of three terms as follows: kI kI 150 o π Bz = + + R R 180 o Solutions kI kI kI = ( 2 + 2 . If the I two straight pieces of wire are perpendicular to each other and if the circuit carries a current I. Solving for I1 gives I1 = 2I. clockwise Solution: If we define the z-axis to be out of the paper then the net magnetic field at the center of the three circles is given by Bz = − 3I 2I 2π kI 1 2π k ( 2 I ) 2π k ( 3 I ) + + = 0. 2 R 2 R R R R Chapter 30 Page 2 of 11 . what is the magnitude and direction of the current in the loop with radius R? Answer: 2I. Problem 4: Two semi-infinite straight pieces of wire are connected by a circular loop segment of radius R as shown in the Figure. what is the I magnitude of the magnetic field at the center of the circular arc (k = µ0/4π)? Answer: 6.71kI/R Solution: If we define the z-axis to be out of the paper then the net magnetic field at the center of the circular arc is given by Bz = Solutions R 3π kI kI kI 3π kI kI + + = 2 + = 6 . Field Problem 3: Three concentric current loops have radii R. 2R 3R R where I1 is the current (clockwise) in the smallest loop.71 . and 3R. as shown in the Figure.PHY2049 R. D. If the net magnetic field at the center of the three loops is zero. The loop with radius 2R carries a current of 2I and the loop with radius 3R carries a 3R current of 3I in the same direction R 2R (counter-clockwise). 2R. the net magnetic field is given by B=2 Solutions 2 kI kI kI =4 2 = 5 .67 .7kI/L Solution: From the Figure that the magnetic fields from wires 1 and 3 cancel and the magnetic fields from wires 2 and 4 add.PHY2049 R. y and if the net magnetic field at the center of the circular loop is zero. Thus. Field Problem 5: I 2 An infinitely long straight wire carries a current I1. L L L/ 2 Chapter 30 Page 3 of 11 . 27 R . If I1 = 4I2. how far is the center of the loop I1 from the straight wire? Answer: 1. If three of 1 4 the wire carry the current I in the same direction L (out of the page) and one of the wires carries 3 the current I in the opposite direction (into the page). what is the net magnetic field at the I-out 2 I-out 3 L center of the square (k = µ0/4π)? Answer: 5. y R Solving for y gives y= I1 4 R = R = 1 .27R Solution: Setting magnitudes of the magnetic field from the loop and wire equal to each other yields 2 kI 1 2π kI 2 = . D. πI 2 π Problem 6: I-out 1 I-in 4 Four infinitely long parallel wires each carrying a current I form the corners of square with sides 2 of length L as shown in the Figure. and a wire loop R of radius R carries a current I2 as shown in the Figure. − 2x2 1 = 0. + = 4 kI 2 2 2 2 2 dx x d x d ( ( / 2 ) ) ( ( / 2 ) ) + + dB y which implies that − 2 x 2 + x 2 + ( d / 2 ) 2 = 0 and x = d / 2 . D. what is the distance x to the point P on the positive x-axis where the magnitude of the magnetic field is maximum? Answer: d/2 Solution: The magnetic field at the point P is the vector superposition of the magnetic field from each of the two wires as follows: y-axis I-out θ r d P θ x-axis x I-out r 2 kI (cos θ xˆ + sin θ yˆ ) B1 = r r 2 kI (− cos θ xˆ + sin θ yˆ ) B2 = r where r2 = x2 +(d/2)2. If the wires are located on the y-axis at y = d/2 and y = -d/2. Field Problem 7: Two infinitely long parallel wires are a distance d apart and carry equal parallel currents I in the same direction as shown the Figure. the net magnetic field is in the positive y direction with magnitude By = 4 kI 4 kIx 4 kIx sin θ = 2 = 2 . Hence. r r ( x + (d / 2) 2 ) where I used sinθ θ = x/r.PHY2049 R. To find the point where B is maximum we take the derivative as follows. Solutions Chapter 30 Page 4 of 11 . If the wires are located on the y-axis at y = 0 and y = d. Field Problem 8: Two infinitely long parallel wires are a distance d apart and carry equal antiparallel currents I as shown in the Figure. If the wires are located on the y-axis at y=+d/2 and y=-d/2. r r2 ( x 2 + (d / 2) 2 ) x2 where I used cosθ θ = d/(2r). what is the magnitude of the magnetic field at the point x = d on the x-axis? 2kI / d Answer: Solution: The magnetic field at P is the superposition of the fields from each wire as follows: Solutions Chapter 30 I-out d B1 P I-in 45 o x-axis x=d B2 Page 5 of 11 . D. Hence. what is the magnitude of the magnetic field at a point x on the x-axis (assume that x >> d and let k = µ0/4π)? y-axis I-out θ r d P θ x-axis x Answer: 2kId/x2 I-in Solution: The magnetic field at the point P is the vector superposition of the magnetic field from each of the two wires as follows: r 2 kI (cos θ xˆ + sin θ yˆ ) B1 = r r 2 kI (cos θ xˆ − sin θ yˆ ) B2 = r where r2 = x2 +(d/2)2.PHY2049 R. the net magnetic field is in the positive x direction with magnitude Bx = 4 kI 2 kId 2 kId 2 kId = → cos θ = . Problem 9: y-axis Two infinitely long parallel wires are a distance d apart and carry equal antiparallel currents I as shown in the Figure. as shown in the Figure. D.PHY2049 R. Field 2 kI 1 1 x$ + y$ 2d 2 2 r 2 kI B2 = − y$ d r r kI = B1 + B2 = ( x$ − y$ ) d r B1 = r B to t r with Btot = 2 kI / d . as shown in the Figure. If current I1 goes through the top loop and current I2 through the lower loop (with I = I1+I2) and if I2 = 2I1 what is the magnitude of the magnetic field at the center of the circle (k = µ0/4π)? π kI /( 3 R ) Answer: Solution: The magnetic field at the center of the loop is Bz = π kI 2 π kI 1 π k ( I 2 − I 1 ) π kI − = = . If current I1 goes through the top loop (and I-I1 through the lower loop) and if the magnetic field at the center of r the loops is B = πkI / (2 R ) z$ where the z-axis is out of the paper. I1 Problem 10: I A single infinitely long straight piece of wire carrying R a current I is split and bent so I2 that it includes two half circular loops of radius R. R R R 3R where I used I1 = I/3 and I2 = 2I/3. Problem 11: I1 A single infinitely long I straight piece of wire R carrying a current I is split and bent so that it includes I-I1 two half circular loops of radius R. what is the current I1? Solutions Chapter 30 Page 6 of 11 . Field Answer: I/4 Solution: The magnetic field at the center of the loops is Bz = π k ( I − I 1 ) π kI 1 π k ( I − 2 I 1 ) π kI − = = R R R 2R . If the centers of 1 2 3 the current loops are a distance R apart what is the magnitude of the magnetic field at the center of the middle current loop (k = µ0/4π)? Answer: 1. 2 2 Chapter 30 Page 7 of 11 . D. Hence.84kI/R Solution: The magnetic field at the center of the middle loop is the superposition of the magnetic fields from each of the three loops as follows: r B1 = − 2π kIR 2 π kI ˆ zˆ z = − ( R 2 + R 2 )3 / 2 R 2 r 2π kI B2 = zˆ R r 2π kIR 2 π kI ˆ zˆ B3 = − 2 z = − 2 3/2 (R + R ) R 2 The net magnetic field is given by Bz = Solutions π kI R kI 1 1 − 2 − = 1 . 84 R .PHY2049 R. The center loop carries the z-axis R R current in the opposite direction from R R R the outer two loops. I − 2 I1 = I 2 I1 = I 4. Problem 12: Three coaxial current loops of radius R I I each carrry a current I as shown in the I Figure. PHY2049 R. D. Field Problem 13: I Three infinitely long parallel d P straight wires lie in the xy-plane as shown in the Figure. The middle I y wire carres a current I to the left d x while the top and bottom wires each I carry a current I to the right. If the wires are a distance d apart, what is the magnitude of the magnetic field at the midpoint P between the top and middle wire (k = µ0/4π)? Answer: 6.67kI/d Solution: The magnetic field at midpoint P between the top and middle wire is the superposition of the magnetic fields from each of the three wires as follows: r 2 kI 4 kI Btop = − zˆ = − zˆ d /2 d r 2 kI 4 kI B midddle = − zˆ = − zˆ d /2 d r 2 kI 4 kI Bbottom = zˆ = zˆ 3d / 2 3d The net magnetic field is given by Bz = − 2 kI d kI 2 2 + 2 − = − 6 . 67 . d 3 Problem 14: In the previous problem, what is the y-component of the force per unit length exerted on the top wire due to the other two wires (k = µ0/4π)? Answer: kI2/d Solution: The magnetic field at the top wire due to the other two wires is given by r kI 2 kI 1 zˆ B=− 1 − zˆ = − d d 2 and the force on a length L of the top wire due to this magnetic field is Solutions Chapter 30 Page 8 of 11 PHY2049 R. D. Field r r r kI 2 L F = IL × B = yˆ . d The y-component of the force per unit length is thus, Fy/L = kI2/d. Problem 15: The Figure shows the cross section of a solid cylindrical conductor with radius R. The conductor is carrying a uniformly distributed current I (out of page). Find the magnitude of the magnetic field, B, inside the conductor a distance r = R/4 from the center (k = µ0/4π). Answer: kI/(2R) Solution: If we draw a loop at radius r (r < R), then Ampere's Law tells us that ∫ I-out r R r r B ⋅ dl = 2 π rB ( r ) = µ 0 I enclosed , with I enclosed = π r 2 J and Loop I tot = π R 2 J , where J is the (uniform) current density. Hence, 2 kI enclosed 2 krI tot B (r ) = = . r R2 Plugging in r = R/4 gives B = kI/(2R). Problem 16: Itot The Figure shows the cross section of a solid cylindrical conductor with radius 0.2 m. The conductor is carrying r a uniformly distributed current I (out of page). If the R = 0.2m magnitude of the magnetic field inside the conductor a distance 0.1 m from the center is 0.1 microTesla, what is the total currrent, I, carried by the cylindrical conductor (in Amps)? Answer: 0.2 Solution: If we draw a loop at radius r (r < R), then Ampere's Law tells us that Solutions Chapter 30 Page 9 of 11 PHY2049 ∫ R. D. Field r r B ⋅ dl = 2 π rB ( r ) = µ 0 I enclosed , with I enclosed = π r 2 J and Loop I tot = π R 2 J , where J is the (uniform) current density. Hence, 2 krI tot kI tot B (r ) = ( / 2 ) B R = and . R2 R Solving for Itot gives I tot RB ( R / 2 ) ( 0 .2 m )( 0 .1 × 10 − 6 T ) = = = 0 .2 A . k 1 × 10 − 7 Tm / A y Problem 17: An infinite conducting sheet of thickness T lies in the xz-plane and carries a uniformly r distributed current density J in J B T x B L the –z direction as shown in the z Figure. What is the magnitude and direction of the magnetic field at distance d above the sheet (in the +y direction, k = µ0/4π)? Answer: + 0.5µ 0 JTxˆ Solution: If we draw an “Amperian” loop (square with sides of length L), then Ampere's Law tells us that r r ∫ B ⋅ d l = 2 LB = µ 0 I enclosed = µ 0 JTL . Loop Solving for B gives r B = 0 .5 µ 0 JT xˆ . Problem 18: J-out A long straight cylindrical wire with radius R (see the Figure) is carrying a non-uniform current density which points along the axis of the wire and has a magnitude, J(r) = a/r, where a is a constant and r is the distance from the center axis of the wire. What is the magnitude of the Solutions Chapter 30 r R Page 10 of 11 B (r ) = µ 0 I enclosed = µ 0a . at a distance r = 2R from the center. where J is the (uniform) current density. 2 kI enclosed 2 k ( r 2 − R 2 ) I tot = B (r ) = . Loop with I enclosed r r r a = ∫ J ⋅ d A = ∫ 2π rdr = 2π ar . 8 rR 2 r Plugging in r = 2R gives B = 3kI/(8R). B. Answer: 3kI/(8R) Solution: If we draw a loop at radius r (R < r < 3R). Hence. Solutions Chapter 30 Page 11 of 11 . Find the magnitude of the magnetic field. then Ampere's Law tells us that ∫ r r B ⋅ dl = 2 π rB ( r ) = µ 0 I enclosed . then Ampere's Law tells us that ∫ I-out R r 3R r r B ⋅ dl = 2 π rB ( r ) = µ 0 I enclosed . The conductor is carrying a uniformly distributed current I (out of page). Field magnetic field inside the wire a distance r from the center (0 < r < R)? Answer: µ0a Solution: If we draw a loop at radius r (r < R). D. with I enclosed = ( π r 2 − π R 2 ) J Loop 2 2 2 and I tot = ( π ( 3 R ) − π R ) J = 8π R J . 2π r Problem 19: The Figure shows the cross section of a solid cylindrical conductor with radius 3R with a cylindrical hole of radius R. r 0 Hence.PHY2049 R. away from the wire Solution: Pushing the wire loop toward the current carrying wire causes the magnetic flux to increase in the loop and thus the induced current will flow in a counterclockwise direction so that the induced magnetic field will be up (opposite to the increasing downward external magnetic field). If the loop is moved toward the wire. P=I R= R 1 . R R and power N 2 A2a 2 (100 )π 2 ( 0 .e.PHY2049 R. If the coil's total resistance is 1. as shown in the Figure. away from the straight wire). D.2 m. how much power (in Watts) is dissipated at time t = 1 s? Answer: 1 Α and from Solution: The magnetic flux through the loop is ΦB = NBΑ Faraday's Law we have. dt dt At t = 1 second (actually at any time t) this gives I = ε NAa = . Nearby and lying in the same plane is a circular loop. Field Chapter 31 Solutions Problem 1: A long straight wire carries current I. ε =− dΦ B dB = − NA = − NAa . what will be the direction of the I current induced in the loop (if any) and what will be the direction of any electromagnetic force exerted on the loop? Answer: counterclockwise. Problem 2: A magnetic field given by B(t) = at+b with a = 1 T/s and b = -1 T is directed perpendicular to the plane of a circular coil of 10 turns and radius 0.58 Ω 2 Solutions Chapter 31 Page 1 of 12 .58 Ohms.2 ) 4 (1T / s ) 2 = = 1W . The force on the loop is in a direction that opposes the push (i. PHY2049 R. What is the induced current in the coil (in milliA) at t = 1 second? Answer: 1.4 ) Solutions Chapter 31 .3T / s ) (10 s ) 3 ( 0 .33 Α and from Solution: The magnetic flux through the loop is ΦB = NBΑ Faraday's Law we have.75 J Ω 3( 0 . If the total resistance of the square wire is 0. 2T / s ) 25 (8 × 10 − 4 A )( 0 .4t . Its plane is perpendicular to a magnetic field given by B(t) = 0. dt dt At t = 1 second this gives ε NA ( 0 . ε =− dΦ B dB = − NA = − NA ( 0 .0. Page 2 of 12 . Hence.5 m.6 t ) . D.3 T/s2 is directed perpendicular to the plane of a square wire loop with sides of length 0.3t2 (where B is in Tesla and t is in seconds). I = R 3Ω 3Ω Problem 4: A uniform time dependent magnetic field given by B(t) = at2 with a = 0.4 − 0 .8 Solution: The magnetic flux through the loop is ΦB = BA = BL2 and from Faraday's Law we have. I = |εε|/R = 2atL2/R. 2T / s ) = = = 1 . and the power dissipated by heat in the wire is P = dU/dt = I2R = 4a2t2L4/R. ε =− dΦ B dB = − L2 = − 2atL 2 .5 m ) 4 = = 18 .33 mA . dt dt The induced current in the wire is.4 Ohms. how much total energy (in Joules) is dissipated by heat in the wire during the time from t = 0 to t = 10 seconds? Answer: 18. Field Problem 3: A 25-turn coil of resistance 3 Ohms has area of 8 cm2. 10 s U = ∫ 0 10 s Pdt = ∫ 0 4 a 2 t 2 L4 4 a 2 (10 s ) 3 L4 dt = R 3R 2 2 4 ( 0 . If the period of the rotation is one second. Field Problem 5: A single wire square loop with area A = 100 cm2 rotates with a constant angular velocity in B Square Loop the magnetic field of the earth (about 100 microT). The Solutions Chapter 31 Square Loop Axis B θ Axis Page 3 of 12 .3 B Solution: The magnetic flux through the loop is ΦB = BΑ Αcosθ θ and from Faraday's Law we have. D. what is the maximum EMF generated in the loop (microV)? Answer: 6.28 µ V 1s ε max = ω BA = where I used ω = 2π π/T. ε =− dΦ B d cos θ = − BA = BA ω sin θ . The axis of rotation goes through Axis one side of the square loop (in the plane of the loop) and is perpendicular to the magnetic field as shown in the Figure.PHY2049 R. dt dt where ω = dθ θ/dt is the angular velocity. dt dt θ Axis where ω = dθ θ/dt is the angular velocity. ε =− dΦ B d cos θ = − BA = BA ω sin θ . Problem 6: B How will the answer in the previous problem change if one moves the axis of rotation of the loop from the side to the center of the loop (in the plane of the loop) as shown in the Figure? Answer: no change Solution: The magnetic flux through the loop is still given by ΦB = BΑ Αcosθ θ and the induced EMF is. The maximum EMF occurs when sinθ θ = 1 and is given by 2π BA T 2π (100 × 10 − 6 T )(100 × 10 − 4 m ) = = 6 . 5 m/s.4 Solutions Chapter 31 B-out Page 4 of 12 . If the radius of the circular loop is increasing at a constant rate given by r(t) = vt with v = 0.5 Tesla and if the bar was at x = 0 at t = 0.PHY2049 R. dt Problem 8: A wire circular loop with a resistance per unit length of 0. with B0 = 2 Tesla. | ε |= dΦ B = v 2 tB = (3 m / s ) 2 ( 2 s )( 0 . If the system is immersed in a uniform magnetic field (out of the paper) with magnitude B = 0. Problem 7: A moveable (massless and frictionless) bar is being moved at a constant velocity of 3 m/s B-out from left to right along two conducting rails as shown in the Figure. Now from Faraday's Law we have. as shown in the Figure (z is out of the paper).5T ) = 9V .5 Ohms/meter is placed in r = vt a uniform magnetic field given by r B = B0 zˆ . what is the magnitude of the induced EMF in the right triangle (in Volts) at t = 2 seconds? Answer: 9 Solution: The magnetic flux through the right triangle is Φ B = BA = 1 2 1 x B = v 2t 2 B 2 2 where I used A = x2/2 and x = vt. Field maximum EMF occurs when sinθ θ = 1 and is the same as in the previous problem. D. The two conducting v rails make an angle of 45o with each other o 45 and together with the moving bar form a x right triangle. what is the magnitude of the induced EMF (in Volts) at t = 10 seconds? Answer: 31. dt and | ε |= 2π B 0 v 2 t = 2π ( 2T )( 0 . Faraday's Law we have. Problem 9: In the presious problem.5 Ω / m The direction of the current is clockwise (opposite my orientation). clockwise Solution: The magnitude of the induced current is I = |ε | 2π B 0 rv ( 2T )( 0 . ε = − dt ε ( 2 T )( 1 m )( 1 0 m / s ) BLv I = = = = 4A. what is the magnitude (in Amps) and direction of the induced current in the loop at t = 10 seconds? Answer: 2. Field Solution: If I choose my orientation to be counterclockwise then the magnetic flux through the loop is ΦB = B0πr2 = B0πv2t2 and from Faraday's Law we have. R 2π r ( 0 .5 Ω / m ) 0 .PHY2049 R.5 m / s ) 2 (10 s ) = 31 . If the system is immersed in a uniform magnetic field (out of the x paper) with magnitude B=2 T. Problem 10: A moveable (massless and frictionless) bar with a length of L = 1 meter is being moved at a R = 5Ω L constant velocity of 10 m/s from left to right v along two conducting rails by an external force. as shown in the Figure. 5 m / s ) = = = 2A . ε =− dΦ B = − 2π B 0 v 2 t . what is the induced current (in Amps) in the 5 Ohm resistor? Answer: 4 Solution: If I choose my orientation to be counterclockwise then the magnetic flux through the loop is ΦB = BΑ Α = BxL=Β =ΒLvt and from =Β dΦ B = − B Lv and hence. 5Ω R R Solutions Chapter 31 B-out Page 5 of 12 . D. 4V . F. ε =− dΦ B 1 dθ = − ω R 2 B . dt 2 dt Plugging in the numbers yields. The rotating rod is connected at the center of the semicircular loop to a stationary rod that is also in contact with the semicircular loop.4 Solution: If I choose my orientation to be counterclockwise then the magnetic flux through the circuit is ΦB = 1 θR 2 B .PHY2049 R. and if it takes 1 second for the angle between the two rods to go from θ = 0 to θ = 90o. 2 and Faraday's Law implies. Solutions Chapter 31 Page 6 of 12 .5 m force so that it rotates with a constant angular velocity along a semicircular loop of wire with radius of r = 0. where ω = . If the entire system is placed in a uniform 2 Tesla magnetic field (out of the page). D. | ε |= 1 1 π /2 2 ωR 2 B = ( 0 .39 V . what is the magnitude of the induced emf (in Volts) in the circuit? Answer: 0.5 m as shown θ in the Figure. 2 2 1s where I used ω = (π π/2)/1s.5 m ) 2T = 0 . Field Problem 11: A moveable (massless and frictionless) F B-out conducting rod is pulled by an external r = 0. 0 milliAmps and is changing at a rate dI/dt = -1.0 A/s. C dt and IR = (0. and two inductors in parallel with L1 = 2 Henry and L2 = 4 Henry as shown in the Figure. L1 I R L2 A B Problem 13: A current I flows through a resistor with R = 333 Ω.0 A/s. At a time when Q=8 milliC and the current I=0. D.PHY2049 R.2 × 10 −3 A)( 25 × 10 3 Ω) = 5V Q 8mC = = 2V C 4mF dI = (7 H )(1A / s ) = 7V L dt Thus. a capacitor with C=4 milliF. what is the potential difference VB-VA (in Volts)? Answer: 1 Solution: Moving in the direction of the current flow we have Solutions Chapter 31 Page 7 of 12 . At a time when the current I = 1. and an inductor with L=7 Henry as shown in the Figure. VB-VA = -5V -2V -7V = -14V.2 milliAmps and is changing at a rate dI/dt=1. what is the potential difference VB-VA (inVolts)? Answer: -14 Solution: Moving in the direction of the current flow we have V B − V A = − IR − Q dI − L . Field C R I - + L A B Q Problem 12: A current I flows through a resistor with R=25 kΩ Ω. 000 V/m electric field? Answer: 1x10-5 Solution: Setting the magnetic field energy density equal to the electric field energy density yields uB = 1 1 B 2 = uE = ε0E 2 2µ0 2 and solving for B gives B= µ 0ε 0 E = E 3. and and two inductors in parallel (L1 = 1 milliHenry and L2 = 3 milliHenry) as shown in the Figure. VB-VA = -0. 8 c 3 × 10 m / s L1 C I R - + A B Q L2 Problem 15: A current I flows through a resistor with R = 2 Ω. D.33V = 1V. 000 V / m = = 1 × 10 − 5 T . Thus. Problem 14: What is the magnitude of the uniform magnetic field (in Tesla) that contains as much energy per unit volume as a uniform 3.33 H )( −1A / s ) = 1.6 microC and Solutions Chapter 31 Page 8 of 12 . dt where 1 1 1 1 1 3 = + = + = L eff L1 L2 2H 4H 4H .333V +1.33V dt so that. Field V B − V A = − IR − L eff dI .33 H and − IR = − (1 × 10 −3 A)(333Ω ) = −0. At a time when Q = 0. since for inductors in parallel you add the inverses. a capacitor with C = 3 milliF. Leff = 1.333V dI − Leff = − (1.PHY2049 R. 75 ) = 11 .2mV C 3mF dI 3 − Leff = − mH ( −0.2mA)( 2Ω ) = −0.6 mV = 0.PHY2049 R.4mV Q 0 .6 mV dt 4 Thus.1 Solution: The current is given by I (t ) = ε ( 1 − e − t / τ ) = I max (1 − e − t / τ ) R and thus e − t / τ = 1 − I / I max and solving for t gives t = −τ ln (1 − I / I max ) = − (8 ms ) ln (1 − 0 . 1ms .8 A / s ) = +0.4 mV –0. where I have used τ = L/R = 8ms. Problem 16: In an LR circuit where L = 120 milliH and R= 15 ohms.2 mV + 0. Field the current I = 0.8 Amps/s.2 milliAmps and is decreasing at a rate of 0.6 µ C − =− = −0. Solutions Chapter 31 Page 9 of 12 . what is the potential difference VB-VA (milliVolts)? Answer: zero Solution: Moving in the direction of the current flow we have V B − V A = − IR − Q dI − L eff . if the switch is closed at t = 0 how long (in milli-seconds) does it take for the current to reach 75% of its final value? Answer: 11. D. VB-VA = -0. C dt where 1 L eff = 1 1 + L1 L2 or L eff = L1 L 2 (1 mH )( 3 mH ) 3 = = mH L1 + L 2 1 mH + 3 mH 4 and − IR = −(0. D.6 Solution: Using the “Loop Rule” for circuits we see that I2 L R dI 1 − I1 R = 0 dt dI ε − L 2 − I2R = 0 dt ε−L and adding these two equations yields 2ε − L Solutions dI − IR = 0 .PHY2049 R. R.8 ms . Field Problem 17: In an LR circuit where L = 120 milliH and R = 15 ohms. if the switch is closed at t = 0 how long (in milli-seconds) does it take for the stored energy in the inductor to reach 50% of its maximum value? Answer: 5. where I have used τ = L/R = 8ms.5 Solution: The current and the stored energy are given by I (t ) = U (t ) = ε ( 1 − e −t / τ ) R ( 1 LI 2 = U max 1 − e − t / τ 2 ) 2 and thus e − t / τ = 1 − U / U max and solving for t gives ( ) ( ) t = −τ ln 1 − U / U max = − (8 ms ) ln 1 − 0 . L. how long (in seconds) does it take for the total current in the circuit to reach 90% of its final value? Answer: 4. If L = 4 milliHenry and R= R 2 milliOhms and if the switch is closed at t = 0. Problem 18: I I1 Consider the LR circuit shown in the Figure which consists of two equal L EMF inductors. 5 = 9 . and an EMF. and two equal resistors. dt Chapter 31 Page 10 of 12 . eff Reff R1 R 2 or 6Ω 3 R1 + R 2 The current I1 is given by I 1 (t ) = R2 ε I (t ) = 1 − e −t / τ ). ( R eff Ω) = 2 s. ( R eff where τ = L/Reff and 1 1 1 ( 2 Ω )( 4 Ω ) 4 R1 R 2 = + R = = = Ω. dt where Leff = L/2 and Reff = R/2 and we know the solution is given by I ( t ) = I max (1 − e − t / τ ) = ε 1 − e −t / τ ).605 s Problem 19: I Consider the LR circuit shown in the Figure which consists a 8 Volt EMF. when (in seconds) is the current through the 2 Ω resistor equal to 1 R2 = 4Ω Amp? I2 Answer: 0.86 Solution: The total current in the circuit. If the switch is closed at R1 = 2Ω I1 t = 0. Solving for the time gives with t = Leff/Reff = L/R = (4 mH)/(2 mΩ t = −τ ln (1 − I ( t ) / I max ) = − ( 2 s ) ln(1 − 0 . I = I1 + I2. The above equation for I is equivalent to the following ε − Leff dI − IR eff = 0 . an inductor. ( R1 + R 2 R1 and solving for t yields Solutions Chapter 31 Page 11 of 12 .PHY2049 R. is given by I ( t ) = I max (1 − e − t / τ ) = L = 4H ε 1 − e −t / τ ). Field where the total current I = I1 + I2. L = 4 H. D. and two 8V resistors.9 ) = 4 . D.PHY2049 R.863 s 4 Solutions Chapter 31 Page 12 of 12 . Field RI ( 2 Ω )(1 A ) 4H t = −τ ln 1 − 1 1 = − ln 1 − V 4 / 3 8 Ω ε 1 = − ( 3 s ) ln 1 − = − ( 3 s ) ln( 0 .75 ) = 0 . πr2 and If I choose a loop of radius r within the capacitor then ΦE = Eπ 2π rB ( r ) = µ 0 ε 0π r 2 I dE = µ 0 ε 0π r 2 2 dt ε 0π R .PHY2049 R. D. A = πR2. If the switch is closed at t = 0. where I have used E = Q/(εε0A). R2 which for r = R/2 becomes B = kI/R. Problem 2: A parallel-plate capacitor whose plates have radius r is ε being charged through a resistor R by an EMF ε as shown C in the Figure. and I = dQ/dt. What is the magnitude of the magnetic field (in Tesla) between the plates a distance r = R/2 from the center? Answer: kI/R Solution: We know from Ampere's Law (with J = 0 inside the capacitor) that ∫ r r dΦ E B ⋅ dl = µ 0 ε 0 dt . Field Chapter 32 Solutions Problem 1: A parallel-plate capacitor whose plates have radius R is being charged by a current I. what is the maximum magnitude of the magnetic field between the plates of the capacitor a distance r/2 from the center? (Note: k = µ0/4π) Answer: kεε/(rR) Solution: We know from Ampere's Law (with J = 0 inside the capacitor) that R ∫ Solutions r r dΦ E B ⋅ dl = µ 0 ε 0 dt . Solving for B gives B (r ) = 2 kIr . Chapter 32 Page 1 of 5 . where I have used E = Q/(εε0A). where I have used E = Q/(εε0A). A = πR2. VC. Solving for B gives 2 k dQ 2 k dV = C R dt R dt 2 (10 − 7 Tm / A )( 3 F )(10 V / s ) = = 300 µ V 2 × 10 − 2 m B(R) = where I used Q = CV which implies that dQ/dt = C dV/dt with dV/dt = a. What is the magnitude of the magnetic field (in microT) between the plates of the capacitor a distance R from the center? Answer: 300 Solution: We know from Ampere's Law (with J = 0 inside the capacitor) that ∫ r r dΦ E B ⋅ dl = µ 0 ε 0 dt . A = πr2. The maximum magnetic field occurs when the current is maximum and solving for Bmax at ra = r/2 gives B max = kI max kε = . and I = dQ/dt. and I = dQ/dt. with a = 10 V/s. across a 3 Farad parallel-plate capacitor whose plates have radius R = 2 cm varies with time and is given by VC(t) = at. Problem 3: The potential difference. D. Solutions Chapter 32 Page 2 of 5 . Field πra2 and If I choose a loop of radius ra within the capacitor then ΦE = Eπ 2π ra B ( ra ) = µ 0 ε 0π ra2 I dE = µ 0 ε 0π ra2 2 dt ε 0π r . πR2 and If I choose a loop of radius R within the capacitor then ΦE = Eπ 2π RB ( R ) = µ 0 ε 0π R 2 I dE = µ 0ε 0π R 2 2 dt ε 0π R . r rR where I used Imax = ε/R (which occurs at t = 0).PHY2049 R. D. dt dt d dt where I have used C = ε0A/d with A = πR2. A uniform electric field points in the z direction (out z-out of the paper) with a value given by Ez(t)=a . where a = 18 V/m and b = 9 V/(m s) and is confined to a circular area with a radius R = 4 meters. 5 × 10 − 2 m B= Problem 5: Consider the situation illustrated in the figure.PHY2049 R.bt. clockwise Solution: We know from Ampere's Law (with J = 0) that Solutions Chapter 32 r B Page 3 of 5 . 2 (10 − 7 Tm / A )( 2 × 10 − 3 F )(150 V / s ) 12 T = = µ 0 . Hence. 2π RB ( R ) = µ 0 ε 0π R 2 dE dV 1 dV = µ 0 ε 0π R 2 = µ 0C .5 cm is connected to an EMF-source that increases uniformly with time at a rate of 150 V/s. If I choose a loop of radius R around the capacitor then ΦE = Eπ πR2 and the potential difference across the capacitor is V = Ed. What is the magnitude and direction of the magnetic field (in picoTesla) at a point P on the circumference of the circle at the time t = 2 seconds? Answer: 2x10-4. Solving for B gives 2 k dV C R dt . What is the magnitude of the magnetic field (in microTesla) between the plates of the capacitor a distance r = R from the center? Answer: 12 Solution: We know from Ampere's Law (with J = 0 inside the capacitor) that ∫ r r dΦ E B ⋅ dl = µ 0 ε 0 dt . where d is the seperation of the plates. Field Problem 4: A 2 milliFarad parallel-plate capacitor whose plates are circular with radius R = 0. dt RA ω cos( ω t ) . D. Problem 6: z-out Consider the situation illustrated in the Figure. A uniform electric field points in the z direction (out of the paper) with a value given by Ez(t) =Asinω ωt. where A = R 90 V/m and ω =1. 2 8 2 2c 2 ( 3 × 10 m / s ) Since B is negative it points opposite the direction of my orientation (clockwise). If I choose my orientation to by counter-clockwise then ΦE = Eπ πr2 and dE π r 2b 2π rB ( r ) = µ 0 ε 0π r =− 2 .PHY2049 R. If I choose my orientation to by counter-clockwise then ΦE = Eπ πR2 and 2π RB ( R ) = µ 0 ε 0π R 2 and solving for B gives B(R) = and hence B max = Solutions dE = µ 0 ε 0π R 2 A ω cos( ω t ) .000 radians/sec and is confined to a circular area with a radius R = 2 meters. What is the maximum magnitude of the magnetic field (in picoTesla) at a point P on the circumference of the circle? Answer: 1 Solution: We know from Ampere's Law (with J = 0) that ∫ r r dΦ E B ⋅ dl = µ 0 ε 0 dt . 2c 2 2 ( 3 × 10 8 m / s ) 2 Chapter 32 Page 4 of 5 . dt c 2 and solving for B gives B (r ) = − rb ( 4 m )( 9V / ms ) =− = − 2 × 10 − 4 pT .000 / s ) = = 1 pT . 2c 2 RA ω ( 2 m )( 90V / m )(1. Field ∫ r r dΦ E B ⋅ dl = µ 0 ε 0 dt . Field Problem 7: Consider the situation illustrated in the Figure. A uniform electric field points in the z-direction (out z-out of the paper) with a value given by Ez(t) = at. ∫ dt πR2 and If I choose my orientation to by counter-clockwise then ΦB = Bπ 2π RE ( R ) = −π R 2 dB = −π R 2 b . 2 Thus. 2c 2 Now from Faraday's Law we know that r r dΦ B E d l ⋅ = − . D. what is the magnitude induced magnetic field (in picoTesla) at the point P at the time t = 2 seconds? Answer: 0. dt c2 2 and solving for the induced magnetic field B gives B(R) = Ra .PHY2049 R. bc 2 (1T / s )( 3 × 10 8 m / s ) 2 Chapter 32 Page 5 of 5 . dt and solving for the magnitude of the induced electric field E gives E (R) = Rb .1 pT . Both fields are confined R to a circular region with a radius R. B(R) = Solutions aE ( R ) (1V / ms )( 9 . and a uniform magnetic field points in the z-direction with P a value given by Bz(t) = bt.000 V/m. If a = 1 V/(m s) and b = 1 T/s and if the magnitude of the induced electric field at the point P on the circumference of the circle at the time t = 2 seconds is 9.000 V / m ) = = 0 . πR2 and If I choose my orientation to by counter-clockwise then ΦE = Eπ dE πR 2 a 2π RB ( R ) = µ 0 ε 0π R = .1 Solution: We know from Ampere's Law (with Ienclosed = 0) that ∫ r r dΦ E B ⋅ dl = µ 0 ε 0 dt . 1 C the current is 0. Field Chapter 33 Solutions Problem 1: If the total energy in an LC circuit is 5. If it the period of the LC oscillations is 2 seconds what is the maximum charge on the capacitor (in C)? Answer: 0. Problem 3: A charged capacitor is connected across an inductor to form an LC circuit. then what is the maximum current (in milliA)? Answer: 20 Solution: The maximum stored energy in the inductor is U max = 1 2 LI max 2 and solving for Imax yields I max = 2U max = L 2 (5 µ J ) = 20 mA 25 mH . If the circuit is completed at time t = 0.0 microJoules and L=25 milliH. 89 ms .5 Amps.5 A. D. Energy is consrved in an LC circuit which implies that at any other time t. When the charge on the capacitor is 0.19 Solution: We know that at some time (say t = 0) that Q0 = 0. Problem 2: A 10 microF capacitor with an initial charge Q0 is connected across an 8 milliH inductor.PHY2049 R. how soon (in millisec) afterward will the charge on the capacitor reverse sign and become equal to -Q0? Answer: 0. Solutions Chapter 33 Page 1 of 4 .89 Solution: For this LC circuit Q = Q0cos(ω ωt) which is equal to –Q0 when ωt = π so that t= π =π ω LC = π (8 mH )(10 µ F ) = 0 .1 C and I0 = 0. PHY2049 R.1C ) 2 + (( 2 s )( 0 . Thus. Problem 5: A 1 milliFarad capacitor with an initial charge Q0 is connected across a 9 milliHenry inductor to form an LC circuit.14 Solution: We know that Solutions Chapter 33 Page 2 of 4 . 2C 2 2C 2 The maximum charge occurs when I = 0 and hence.188 C where I used ω = 1 2π = . Starting from when the current is maximum. 2 Q max Q 02 1 = + LI 02 . how long will it take (in millisec) for the capactor to become fully charged? Answer: 0.7 Solution: When the current is maximum the charge on the capacitor is zero and it will take one-quarter of a period for it to become fully charged.5 A ) /( 2π )) 2 = 0 . T LC where T is the period of the oscillations. Problem 4: A current flows in an LC circuit with L = 50 milliH and C = 4 microF. D. If the circuit is completed at time t = 0.7 ms . 2C 2C 2 Solving for Qmax gives Q max = = Q 02 + ( I 02 / ω 2 ) = Q 02 + (TI 0 /( 2π )) 2 ( 0 . t= T π = 4 2 LC = π 2 (50 × 10 − 3 H )( 4 × 10 − 6 F ) = 0 . how soon (in millisec) afterward will the charge on the capacitor decrease to one-half of its initial value? Answer: 3. Field Q2 1 Q 02 1 2 + LI = + LI 02 . where I used T = 2π π/ω ω and ω = 1 / LC . Hence. Field Q (t ) = Q0 cos( ω t ) where ω = 1 / LC . 5 ) = ( 3 ms )( π / 3 ) = 3 . Imax. ω Problem 7: A capacitor (C=8 milliFarads) initially charged by connecting it to a 2 Volt battery is connected to an inductor (L=2 milliHenry) to form an LC circuit (the resistance of the circuit is negligible). I max = ( 3 × 10 − 3 A )( 3 ms ) = 9 × 10 − 6 C . if the maximum current is 3 milliAmps what is Q0 (in microC)? Answer: 9 Solution: We know that the amplitude of the current oscillations. A short Solutions Chapter 33 I C Q L Page 3 of 4 .01 C. Solving for t gives t = cos − 1 ( Q / Q 0 ) / ω = = LC cos − 1 ( Q / Q 0 ) ( 9 × 10 − 3 H )(1 × 10 − 3 F ) cos − 1 ( 0 . where ω = 1 / Q0 = LC . At a certain instant of time when the current I = 2 Amps the charge on the capacitor is Q = 0. what is the maximum energy stored in the magnetic field of the inductor (in milliJoules)? Answer: 16 Solution: Since energy is conserved the maximum stored energy in the inductor is eual to the initial stored energy as follows: 1 1 U m ax = U 0 = C ( ∆ V ) 2 = ( 8 m F )( 2V ) 2 = 1 6 m J . 14 ms Problem 6: In the previous problem.PHY2049 R. D. is given by Imax = Q0ω. 2 2 Problem 8: A capacitor C and an inductor L form an LC circuit as shown in the Figure. During subsequent oscillations. D. Solving for ω yields 1 ω f = = 2π 2π 1 = 2π Solutions I 22 − I 12 Q12 − Q 22 (1 A ) 2 − ( 2 A ) 2 100 = = 15 . What is the frequency of the oscillations (in Hz)? Answer: 15. Field time later when the current I = 1 Amps the charge on the capacitor is Q = 0.9 Solution: From energy conservation we know that Q12 1 Q 22 1 2 + LI 1 = + LI 22 2C 2 2C 2 and multiplying by 2/L gives ω 2 Q12 + I 12 = ω 2 Q 22 + I 22 .9 Hz 2 2 ( 0 . 01C ) − ( 0 .PHY2049 R.02 C.02 C ) 2π Chapter 33 Page 4 of 4 . where ω = 1 / LC . where E0 = 300 V/m and k = 1. Hence. 77 × 10 14 Hz . sin( kx − ω t ) = − 0 = − c 3 × 10 8 m / s c where I have used sin( kx − ω t ) = − sin( ω t ) = − sin( 2π ft ) = − sin( π / 2 ) = − 1 . where E0 = 300 V/m and a vacuum is given by Ey = E0 sin (kx -ω k=107/m. +x direction Solution: The angular frequency ω = 2π πf and the speed of light in a vacuum is c = ω/k. f = 2π 2π 2π Problem 2: What is the value of the z-component of the magnetic field (in microT) of the electromagnetic wave in the previous problem at x = 0 and t = 5.PHY2049 R. Field Chapter 34 Solutions Problem 1: The electric field component of an electromagnetic wave traveling in ωt).77x1014/s)(5.25x10-16s) = 0. Problem 3: The electric field component of an electromagnetic plane wave traveling in r a vacuum is given by E = E 0 sin( kx + ω t ) yˆ . What is the magnetic field component of the electromagnetic wave? r Answer: B = − ( E 0 / c ) sin( kx + ω t ) zˆ Solutions Chapter 34 Page 1 of 13 . D.25x10-16 seconds? Answer: -1 Solution: The z-component of the magnetic field is given by Bz = E 300 V / m E0 = − 1µ T .25.77x1014.25x107/m. What are the frequency of the oscillations (in Hz) and the direction of propagation? Answer: 4. ω ck ( 3 × 10 8 m / s )(10 7 / m ) = = = 4 . since ft = (4. what is the power output of the source (in Watts)? Answer: 60 Solutions Chapter 34 Page 2 of 13 .PHY2049 R.8 Solution: For total reflection we have 2 IA 2 A E 02 AE 02 = F = = c c 2 µ 0 c µ 0 c 2 (1m 2 )( 300 V / m ) 2 = = 0 . kc (1 . Field Solution: We know that sin(kx+ω ωt) corresponds to a wave traveling in the –x direction and we know that the direction of propagation of an r r electromagnetic wave is given by E × B and hence. where I also used B = E/c. Hence. r B = − ( E 0 / c ) sin( kx + ω t ) zˆ .25 × 10 7 / m )( 3 × 10 8 m / s ) = = 6 × 10 14 Hz . Problem 4: What is the frequency of the electromagnetic wave in the previous problem (in Hz)? Answer: 6x1014 Solution: We know that c = ω/k and ω = 2π πf. f = 2π 2π Problem 5: How much force (in microN) will the electromagnetic wave in the previous problem exert on one square meter of a wall when it hits the wall (normal to the surface) and is completely reflected? Answer: 0.796 µ N ( 4π × 10 − 7 Tm / A )( 3 × 10 8 m / s ) 2 Problem 6: If the amplitude of the electric field oscillations 6 meters from a point source of electromagnetic radiation is 10 V/m. D. 6 Solution: We know F I E 02 B02 P= = = = 2 µ0 . 4π ( 2 m ) 2 Problem 8: Light from a distant point source of electromagnetic radiation exerts a radiation pressure of one microN per meter squared when it is completely absorbed by the surface S.PHY2049 R. Chapter 34 Page 3 of 13 . Field Solution: We know P E 02 I = = 4π r 2 2 µ 0 c .6 µ T . What is the maximum value of the magnetic field (in microT) within the radiation at S? Answer: 1. Solving for B0 gives B0 = Solutions 2 µ0 P = 2 ( 4 π × 10 − 7 Tm / A )(1µ N / m 2 ) = 1. P= 2µ 0c 2 ( 4π × 10 − 7 Tm / A )( 3 × 10 8 m / s ) Problem 7: What is the amplitude of the electric field oscillations (in V/m) 2 meters from a 60 Watt point source of electromagnetic radiation (assume the radiation is a plane wave)? Answer: 30 Solution: We know P E 02 I = = πr2. where P is the power of the source. Solving for P gives 4π r 2 E 02 4π ( 6 m ) 2 (10V / m ) 2 = = 60W . where P is the radiation pressure and A c 2 µ0 c 2 where I have used E = cB. D. A 2 µ 0 c . where P is the power of the source and where A = 4π Solving for E0 gives E0 = 2 µ 0 cP = A 2 ( 4π × 10 −7 Tm / A )( 3 × 10 8 m / s )( 60W ) = 30V / m . 2 × 10 3 W / m 2 )(1m 2 ) = 8µN . F = = 3 × 10 8 m / s c Problem 10: A radio station delivers 75 MegaWatts of power which is distributed uniformly in all directions to the radio waves. What is the maximum solar radiation force (in N) on a perfectly reflecting square piece of foil with sides of length L = 1 meter located one million miles from the sun (Note: 1 mile = 1. perfect absorber placed perpendicular to these waves a distance of 10 meters from the transmitter? (Note: 1 Pa = 1 N/m2) Answer: 199 Solution: We know that for absorption.99x1030 kg and a radiation power of 3.2 kW/m2 falls normally on a surface with an area of 1 m2 and is completely reflected. D.08 Solution: The intensity of the solar radiation a distance R from the sun is given by I sun = Psun 4π R 2 . The force of the radiation on the surface (in microN) is Answer: 8 Solution: We know that 2 IA 2 (1.PHY2049 R. Field Problem 9: Light with an intensity of 1. = Power = c 4π r 2 c 4π (10 m ) 2 ( 3 × 10 8 m / s ) Problem 11: The sun has a mass of 1. What is the radiation pressure exerted by these waves (in MicroPascals) upon a flat. and the radiation force for total reflection (normal to the surface) is Solutions Chapter 34 Page 4 of 13 . PPr essure = I P 75 × 106 W = 199 µ Pa .609 meters)? Answer: 0.9x1026 Watts. 609 × 10 6 m ) 2 ( 3 × 10 8 m / s ) F = Problem 12: The sun has a mass of 1.000 kg )(1 .67x10-11 Nm2/kg2. 3 .1 Solutions Chapter 34 Page 5 of 13 . using a large sail made of foil (sailing on the solar wind).3). D. What is the diameter (in m) of the largest circle on the pool's surface through which light coming directly from the source can emerge? Answer: 24.67 × 10 −11 Nm 2 / kg 2 )(1.000 kg and if the sail is a perfectly reflecting square with sides of length L.PHY2049 R. 9 × 10 26 W )(1m ) 2 = 0 . Field 2 I sun 2 Psun 2 A= L 4π R 2 c c 2 ( 3 .08 N = 4π (1.90x1026 Watts. GmM sun 2π c L= Psun ( 6 .99x1030 kg and a radiation power of 3.9 × 10 26 W = 801 m = Problem 13: A point source of light is located 10 meters below the surface of a large lake (n=1. Solving where I have used I = Psun/(4π for L yields. c r2 4π r 2 c πr2) with r is the distance to the sun. what is the smallest value of L (in meters) that will allow the ship to escape the sun's gravitational attraction? (The gravitational constant G = 6.) Answer: 801 Solution: To find the minimum sail size we set the magnitude radiation force equal to the magnitude of the gravitational force as follows Fradiation = mM sun 2I 2 2 Psun 2 L = L G = = Fgravity .99 × 10 30 kg ) 2π (3 × 10 8 m / s ) . If the ship has a mass of 1. A spaceship might be propelled in our solar system by the radiation pressure from the sun. 0 Answer: 48.33) strikes a piece of θmin Water n1 = 1. D. Problem 15: A beam of light travelling through water (n=1.16 cm . 3) − 1 2 = 24 .5) at an incident angle θ as shown in the Figure. What is the beams actual path length (in cm) through the glass? Answer: 2.1m .33 glass (n=1. where T is the n2sinθ thickness of the glass and L is the path length.5) at an incident angle of 35 degrees.7 Solution: The critical angle (total internal reflection in the glass) is given by Solutions Chapter 34 Page 6 of 13 . What is θc the minimum angle θ (in degrees) such that the light will not reach Glass n2 = 1. Field Solution: The critical angle (angle of total internal reflection) is given by sinθ θc = 1/n and tanθ θc = R/d. D = 2 R = 2 d tan θ c = 2d n −1 2 = 2(10m) (1 . Thus. where I used n1 = 1 and θ1 = 35o.5 θ1 ( 2 cm )(1 . 574 ) 2 = 2 .5 the air (n=1) on the other side of the glass? Air n3 = 1. L= T = cos θ 2 Tn 2 n 22 − sin 2 θ 1 = θ2 L T n=1.5 ) 2 − ( 0 . Problem 14: A beam of light strikes the surface of a 2cm thick parallel-sided sheet of glass (n = 1.PHY2049 R. Thus. 5 ) (1 .16 θ1 = Solution: We know that n1sinθ θ2 and cosθ θ2 = T/L. 75 n1 1 . Field n1 sin θ min = n 2 sin θ c = n3 sin( 90 o ) Solving for θmin gives sin θ min = 1 n3 = = 0 . 25 × 10 8 m / s ) = 1 .PHY2049 R.59 × 10 8 m / s = o sin( 45 ) Solutions Chapter 34 Page 7 of 13 .33 and thus θmin = 48. Problem 16: 45 o A A beam of light travelling through medium A at a speed of 2.59x108 Solution: From Snell’s law we have n A sin θ A = n B sin θ B and since nA = c/vA and nB = c/vB this implies o sin θ A sin θ B = vA vB . If the refracted angle of the light is 30 degrees.7o. D. what is the speed of the light in medium B (in m/s)? Answer: 1. Solving for vB gives vB = sin θ B vA sin θ A sin( 30 o ) ( 2 .25x108 m/s strikes medium B B at an incident angle of 45 degrees as shown in 30 the Figure. 1 .22 Solution: Requiring total internal reflection at the point B implies o Air A Glass θ1 θ1 o 90 -θ1 B n sin( π / 2 − θ 1 ) = n cos( θ 1 ) = 1 and Snell’s Law applied at the point A gives n sin( θ 1 ) = sin( θ 0 ) . Thus.PHY2049 R.5 n1 Solutions Chapter 34 θ0=45 o 1m o 45 45 o 1m θ1 1m water x d Page 8 of 13 . where I used sin 2 (θ 0 ) = sin 2 (45o ) = 1 / 2. Field Problem 17: θ0=45 A light ray enters a rectangular glass slab at point A at an incident angle of 45 degrees and then undergoes total internal reflection at point B as shown in the Figure.22 . What minimum value of the index of refraction of the glass can be inferred from this information? Answer: 1. air Problem 18: A fiber optic cable (n = 1. D.33). Squaring the first equation and using the second equation gives 1 = n 2 cos 2 (θ 1 ) = n 2 (1 − sin 2 (θ 1 )) = n 2 − sin 2 (θ 0 ) . What is the critical angle (in degrees) for light rays to stay inside the cable? Answer: 62 Solution: The critical angle (angle of total internal reflection) is given by 1 . where n is the index of refraction of the glass and θ0 is the incident angle. 33 n sin( θ c ) = 2 = = 0 .50) is submerged in water (n = 1.5 = 1 . n = 1 + sin 2 (θ 0 ) = 1 + 0 .887 . 63 Solution: From Snell’s law we have 1 sin( 45 o ) sin( θ 1 ) = sin( θ 0 ) = . What is the maximum value of θ (in degrees) for which the refracted ray will undergo total internal reflection at the left vertical face of the block? Answer: 90 Solution: From the Figure we see that θ3 = π −θ2 2 and sin( θ 3 c ) = θ1 n = 1.0 meter long vertical pole extends from the bottom of a swimming pool to a point 1. is given by d = 1m + x = 1m + (1m ) tan θ 1 = 1m + (1m ) tan( 32 o ) = 1m + 0 . 671 .33) as shown in the Figure. We also know that sin( θ 2 ) = 1 sin( θ 1 ) . Problem 19: A 2.63 m Problem 20: Light of wavelength 589 nm is incident at an angle theta on the top surface of a polystyrene block (n = 1.15o. Field and thus θ = 62. the critical angle for θ3 is θ3c = 42. 1 .84o > θ3c which means the maximum angle for θ1 is 90o (any ray that enters the block with θ1 > 0 will be totally internally reflected at the left face).33 and hence θ1 = 32o. Now the length of the shadow.49 θ2 θ3 1 = 0 .63 m = 1 . n 1 .46o. If sunlight is incident 45 degrees above the horizon.49 Thus. Solutions Chapter 34 Page 9 of 13 . d. what is the length of the shadow of the pole (in m) on the level bottom of the pool? Answer: 1. 1 . D.49 which for θ1 = 90o yields θ2 = 42.PHY2049 R.49) as shown in the Figure.15o which implies that θ3 = 47.0 meter above the surface of the water (n = 1. If the prism is immersed in Solutions Chapter 34 water glass o b 90 θc φ c Page 10 of 13 .7 as shown in the Figure. Air The glass slab has an with index of refraction n = 1.PHY2049 R. 77 o ) Problem 22: a A ray of light is incident normally of face ab of a glass prism in the shape of a right triangle and with index of refraction n = 1.77 o ) T = 0 .6 and a thickness T. Field Problem 21: A light ray enters a rectangular glass slab at an incident angle of 60 degrees.54T Solution: From Snell’s Law we know that θ0-θ1 θ0=60 o x θ1 Glass d n1 sin θ 1 = n 0 sin θ 0 and solving for θ1 gives n 0 sin θ 0 sin 60 o o = sin θ 1 = or θ 1 = 32 . 1 .54 T = cos( 32 . What is the displacement d? Answer: 0. 77 . D. T The beam emerging from the other side of the glass slab is parallel to the incident beam but displaced by a d distance d as shown in the Figure.6 n1 Now if we let x be the distance the light ray travels within the glass we see that sin( θ 0 − θ 1 ) = d T and cos θ 1 = . x x Solving for d gives d = x sin( θ 0 − θ 1 ) = sin( θ 0 − θ 1 ) T cos θ 1 sin( 60 o − 32 . D.PHY2049 R.5 θc o Eye θ=45 L o n C L θ n2 = 1.88 o = 40 . Field water (n = 1.33 Solutions Chapter 34 Page 11 of 13 .1 Solution: Total internal reflection at the face ac implies n1 sin θ c = n 0 sin 90 o and solving for the critical angle θc gives sin θ c = n0 1 .4 Solution: We see that n sinθ θ = sin90o = 1 so that n = 1/sinθ θ= 1/sin45o = 1.4. an observer with eyes level with the top of the tank can just see the corner C.3).12 o . what is the largest value of the angle φ (in degrees) so that the ray is totally reflected at face ac? Answer: 40.7 o We see that φ + θ c = 90 and thus φ = 90 o − θ c = 90 o − 49 . 88 .3 o = or θ c = 49 . n1 = 1. What is the index of refraction of the liquid? Answer: 1. Problem 23: 90 When the square metal tank with sides of length L in the Figure is filled to the top with an unknown liquid. n1 1 . PHY2049 R.13 o Chapter 34 Page 12 of 13 . cos θ 2 cos 28 . 1 .50) is submerged in water (n = 1.887 .13o. that the light must travel is given by d = Solutions L 1m = = 1 . The distance.134 m . How long (in nanoSeconds) does it take the light to pass through the glass? Answer: 5.46o. d.5 Problem 25: A beam of light strikes the surface of a 1-meter thick parallel-sided sheet of glass (n = 1. o 45 θ2 L=1m d n2 = 1.5 n1 sin θ c = which implies that θc = 62. What is the maximum bending angle θ (in degrees) for light rays to stay inside the cable? Answer: 27. 54 o . Field Problem 24: A fiber optic cable (n = 1.471 .5) at an incident angle of 45 degrees as shown in the Figure . 1 .5 Solution: The critical angle for total internal reflection is given by n 2 1 .7 Solution: The refracted angle θ2 is given by sin θ 2 = 1 n1 sin θ 1 = sin 45 o = 0 . 33 = = 0 .5 n2 which implies θ2 = 28. θ bend = 90 o − θ c = 27 . D. The maximum bending angle is thus.33) as shown in the Figure. PHY2049 R. D. Field and the time is thus t= d nd (1 .5 )(1 .134 m ) = = = 5 . 67 ns , v c 3 × 10 8 m / s where I used n = c/v. Solutions Chapter 34 Page 13 of 13 PHY2049 R. D. Field Chapter 35 Solutions Problem 1: An object placed 180 cm in front of a spherical mirror produces an image with a magnification m = 0.1. What happens to the image when the object is moved 150 cm closer to the mirror? Answer: the image increases in size by a factor of 4 Solution: We know that 1 1 1 + = p i f m = − i p , and hence f = p1 180 cm = = − 20 cm , 1 − 1 / m1 (1 − 10 ) where I have used i = -m1p1, and m1 = 0.1. Now, if I move the object 150 cm closer to the mirror the new object position p2 = 30 cm (180 cm - 150 cm) and the new image position i2 will be given by 1 1 1 1 1 1 = − = − − = − i2 f p2 20 cm 30 cm 12 cm . Hence the new magnification is m2 = − i2 − ( − 12 cm ) = = 0 .4 , 30 cm p2 which is four times the original magnification. Problem 2: A converging thin lens has a focal length of 30 cm. For what two object locations (in cm) will this lens form an image three times the size of the object? Answer: 20, 40 Solution: We know that 1 1 1 + = p i f m = − i p , and hence Solutions Chapter 35 Page 1 of 11 PHY2049 R. D. Field p = f (1 − 1 / m ) . For m = 3 we get p = ( 30 cm ) (1 − 1 / 3 ) = 2 ( 30 cm ) = 20 cm , 3 and for m = -3 we get p = ( 30 cm ) (1 + 1 / 3 ) = 4 ( 30 cm ) = 40 cm . 3 Problem 3: You decide to take a picture of the moon with your 35 mm camera using its normal lens of 50 mm focal length. Taking the moon’s diameter to be approximately 3.5x106 m and its distance from the earth as 3.84x108 m, how large (in mm) is the moon's image on the film? Answer: 0.46 Solution: We know that 1 1 1 + = p i f m = − i , p and hence i ≈ f (because 1/p is very small) and since |m| = hi/ho we have 2 ho f ( 3 . 5 × 10 6 m )( 50 mm ) 2 hi = | m | 2 ho = = = 0 . 456 mm . p 3 . 84 × 10 8 m Problem 4: Which of the following types of image of a real, erect object cannot be formed by a concave spherical mirror? (A) Real, erect, enlarged (B) Real, inverted, reduced (C) Virtual. erect, enlarged (D) Enlarged, inverted, and farther away than 2f (E) Virtual, inverted, reduced? Answer: A, E Solutions Chapter 35 Page 2 of 11 Field Problem 5: A concave mirror forms a real image which is twice the size of the object. Which way and by how much should the object be moved to double the size of the image? Answer: 5 cm further from the mirror Solution: We know that 1 1 1 + = p i f m = − i p . f = R/2. 1−1/ m 3 where I have used i = -mp. where I have used i = -mp. the radius of curvature of the mirror (in cm) must be approximately: Answer: 27 Solution: We know that 1 1 1 + = p i f m = − i p . We see that the object must be moved 5 cm = 15 cm –10 cm further from the mirror. D. If the object is 20 cm from the mirror. Solutions Chapter 35 Page 3 of 11 . 7 cm . and hence 1 p = f 1 − m. and m = -2 (since i is positive). 1 p 1 = 20 cm 1 − 2 1 p 2 = 20 cm 1 − 4 = 10 cm . and = 15 cm .PHY2049 R. Thus. and hence R = 2f = 2p 4p = = 26 . Problem 6: An erect image formed by a concave mirror (f = 20 cm) has a magnification of 2. and with radii of curvature of equal to R and 2R as shown in the Figure. n = 1. what is the index of refraction n? Answer: 1. 1 − m where I have used i = -mp. and hence f = p 1 .PHY2049 R. If an object is placed a distance 3R/2 from the lens forms a virtual image that one-half the size as the object. n.5. Hence from the above equation. What is the magnitude of the focal length and type of lens? Answer: 4R. converging Solution: The lensmakers formula tell us that Solutions Chapter 35 2R 5R R Page 4 of 11 . Since m = 1/2 we see that f = -p = -3R/2. as shown in the 3R/2 Figure. we know that 1 1 1 + = p i f m = − i p .33 Solution: From the lensmakers formula we have 1 1 1 1 2 1 = ( n − 1) − − = − ( n − 1) . R. D. = ( n − 1) − f R R R R1 R 2 Also.33. Field Problem 7: A diverging thin lens is constructed object with index of refraction. n −1 = − 1 R = and n = 1. 2f 3 Problem 8: object A thin lens is constructed with index of refraction. and R with radii of curvature of equal magnitude. The magnification is given by m = − i − 20 R = = −4 . What is the focal length of the lens (in cm)? Answer: 36 Solution: We know that 1 1 1 + = p i f and hence m = − i p . i f p 4R 5R 20 R and hence i = 20R. 5 − = ( n − 1) − = .PHY2049 R. f = 4R and the lens is converging since f is positive. 3 and for m = -3 we get Solutions Chapter 35 Page 5 of 11 . D. For m = 3 we get p 1 = f (1 − 1 / 3 ) = 2 f . Field 1 1 1 1 1 1 = 0 . Hence. p 5R Thus the image is real (since i is positive) and inverted (since m is negative) and 4 times the object height (since |m| = 4). f R 2R 4R r1 r2 where I uses r1 = R and r2 = 2R. Problem 9: What type of image is formed when an object is placed a distance 5R from the lens in the previous problem? Answer: Real inverted images 4 times the object height Solution: We know that 1 1 1 1 1 1 = − = − = . p = f (1 − 1 / m ) . Problem 10: Two objects placed 24 cm apart in front of a converging thin lens both produce an image three times the size of the object. 5 m p = = 1m . 3 3 3 and thus f = 3(24 cm)/2 = 36 cm. which means i = -6 meters and m = -i/p = 4. 3 Subtracting the two equations yields p 2 − p1 = 24 cm = 4 2 2 f − f = f . f2 = ( n 1 − 1) (1 . where 1 1 1 + = p i f m = − i p .5 m 6m . and doubles in size Solution: An inverted image twice the size of the object means m = -2. D. erect. Problem 11: A thin lens constructed with index of refraction.5 meters from the lens. produces an inverted real image that is twice the size of an object placed 1. 25 − 1) The new image position given by 1 1 1 1 1 1 = − = − = − i f2 p 2 m 1 .5. n = 1. The new image is thus virtual (since i is now negative) and erect (since m is now positive) and twice the size (since |m| is now 4 instead of 2). 5 − 1) 1m = 2 m .5 meters from the lens)? Answer: becomes virtual. What happens to the image if the index of refraction of the lens is changed to n = 1.PHY2049 R.25 (and the object remains 1. Field p 2 = f (1 + 1 / 3 ) = 4 f . Solutions Chapter 35 Page 6 of 11 . = ( n − 1) − f r r 2 1 which implies that for fixed r1 and r2. (1 − 1 / m ) (1 + 1 / 2 ) We also know that 1 1 1 . and hence f1 = f = 1 . f1 = ( n 2 − 1) (1 . In order to see distant objects. which is observed to form an image three times the size of the object. and m = 1/2. If the mirror produces a 3 cm-tall upright image what kind of mirror is this and what is the magnitude of its radius of curvature? Answer: convex. and hence R = 2f = 2p 2p = = − 2 p = − 20 cm . Field Problem 12: An object 6 cm tall is located 10 cm in front of a mirror. the object-lens distance (in mm) must be changed to: Answer: 360 Solution: We know that Solutions Chapter 35 Page 7 of 11 . Problem 14: A 2-cm-tall object is placed 400 mm from a converging lens. D. 6 Solution: We know that 1 1 1 + = p i f m = − i p .PHY2049 R. 1−1/ m 1− 2 where I have used i = -mp. she should wear eyeglasses of what type and focal length (in ft)? Answer: diverging. When p is very large we get f = i and in this case we want i = -6 feet which means f = -6 feet (a diverging lens with |f| = 6 feet). To make the image five times the size of the object. Problem 13: A nearsighted person can see clearly only those objects which lie within 6 feet of her eye. f = R/2. 20 cm Solution: We know that 1 1 1 + = p i f m = − i p . Since f < 0 the mirror is convex. Field 1 1 1 + = p i f m = − i p . Problem 15: A light bulb burns in front of the center of a 40-cm wide mirror that is hung vertically on a wall as shown in the Figure. and hence f = p1 3 p1 = 1 − 1 / m1 4 . and m2 = -5 (since i is positive). where I have used i = -m1p1. A man walks in front of the mirror along a line that is parallel to the mirror and twice as far from it as the bulb. d 2d which implies x = w. Problem 16: An object placed in front of a spherical mirror with focal length f = -20 cm produces an image with a magnification m = 0. The greatest distance he can walk (in cm) and still see the image of the bulb in the mirror is: Answer: 120 Solution: From the Figure we see that tan θ = x θ θ L w =40 cm d 2d Mirror w/2 x = .1. 4 10 where I have used i = -m2p2. L = 2x+w = 3w = 120cm. and m1 = -3 (since i is positive). what happens to the image? Answer: the image increases in size by a factor of 2 Solution: We know that Solutions Chapter 35 Page 8 of 11 . D. from above I know that p 2 = f (1 − 1 / m 2 ) = 3 p1 (1 + 1 / 5 ) = 9 p 1 = 360 mm .PHY2049 R. If the the object is placed the same distance in front of a spherical mirror with focal length f = -45 cm. Now. The total path is thus. so that the image increases in size by a factor of 2. For m = 0. what is the index of refraction n? Answer: 1. since m=-1 we know that i= p so that 2 2 1 = = p 2R f . 2R R R1 R 2 Also.PHY2049 R. The new magnification is given by m = − f − ( − 45 cm ) 45 = = = 0 .2 p− f 180 cm − ( − 45 cm ) 225 . and with R radii of curvature of equal to R and 2R as shown in the Figure. p = f (1 − 1 / m ) . Problem 17: object A converging thin lens is constructed with index of refraction.67 Solution: From the lensmakers formula we have 2R image 1 1 1 1 3 1 = ( n − 1) + = ( n − 1) = ( n − 1) − f 2R . n −1= 3f 3 Solutions Chapter 35 Page 9 of 11 . 2R 2 = and n = 1. D. Field 1 1 1 + = p i f and hence m = − i p . and I have used p = 2R. and thus f = R.1 we get p = f (1 − 10 ) = − 9 f = − 9 ( − 20 cm ) = 180 cm . Hence. If an object 2R placed a distance 2R from the lens forms a real image that has the same size as the object but is inverted. n.67. 71 ) = − 2 . i2 f2 p 2 15 cm 36 cm and 2 Also. the system produces a real inverted magnified image. 14 . lens 1 produces a virtual non-inverted magnified image.71 cm to the right of lens 2. Solutions Chapter 35 Page 10 of 11 . 71 36 cm p2 and M = m 1 m 2 = ( + 3 )( − 0 . D. A luminous source is placed a distance of 10 cm in front of the first lens. which is to the left of lens 2. Answer: Image is real inverted and magnified (M = -2. 71 cm . m1 = − i1 − 30 cm = − = +3 . which is located 25. 71 cm i2 = − = − 0 .PHY2049 R. m2 = − 25 .7 cm to the right of lens 2. 10 cm p1 Thus.14) and located 25. Locate the final image and determine the overall magnification of the system. Field Image 1 Lens 1 Lens 2 f1 = +15 cm f2 = +15 cm Object 10 cm L = 6 cm Real Inverted Magnified Image Problem 18: Two identical converging lenses of focal lengths f1 = f2 = +15 cm are separated by a distance L = 6 cm. Now ignoring lens 1 and taking p2 = +(30cm+6cm)= +36 cm we get 1 1 1 1 1 = − = − i = + 25 . as shown in the Figure. Thus. Solution: Ignoring lens 2 and taking p1 = 10 cm we get 1 1 1 1 1 = − = − i = − 30 cm . i1 f1 p 1 15 cm 10 cm and 1 Also. 55)(2. Answer: Image is real inverted and magnified (M = -3. Solution: Ignoring lens 2 and taking p1 = 18 cm we get 1 1 1 1 1 = − = − i = + 36 cm . Now ignoring lens 1 and taking p2 = -(36cm-22cm)= -14 cm we get 1 1 1 1 1 = − = − i = + 24 . Find the position and the height of the final image.55) with a height of 8. Solutions Chapter 35 Page 11 of 11 . i1 f1 p 1 12 cm 18 cm and 1 Also. which is to the right of lens 2. m2 = − 24 . of focal lengths f1 = +12 cm and f2 = -32 cm.PHY2049 R. 78 ) = − 3 . separated by a distance L = 22 cm. 78 p2 − 14 cm and M = m 1 m 2 = ( − 2 )( + 1 . 55 .89 cm to the right of lens 2 with height = (3. A luminous object with a height of 2. m1 = − 36 cm i1 = − = −2 .4cm) = 8. lens 1 produces a real inverted magnified image. the system produces a real inverted magnified image 24. 89 cm i2 = − = + 1 . Field Lens 1 f1 = +12 cm Lens 2 f2 = -32 cm Object 18 cm L = 22 cm Image 1 Real Inverted Magnified Image Problem 19: The optical system shown in the Figure consists of two lenses. i2 f2 p2 − 32 cm − 14 cm and 2 Also. 18 cm p1 Thus. 89 cm . Thus.4 cm is placed 18 cm in front of the first lens.53 cm. D.9 cm to the right of lens 2.53 cm and located 24. how far apart are the two second-order bright fringes on the screen (in mm)? Answer: 1. The light passes through a double slit and falls on a screen 10 m away. Thus. θ is smaller. respectively. d A more closely spaced pattern implies that for a given m.6 nm. Which of the following changes would cause the interference pattern to be more closely spaced? (a) Use a blue filter instead of a yellow filter (b) Use a 10 Watt bulb (c) Use a 500 Watt bulb (d) Move the bulb closer to the slits (e) Move the slits closer together Answer: (a) Use a blue filter instead of a yellow filter Solution: The angular position of the bright fringes is given by sin θ = mλ . D.01 mm. at approximately 589. Problem 2: The characteristic yellow light of sodium lamps arises from two prominent wavelengths in its spectrum.0 nm and 589. we want to increase d or decrease λ and from problem 15 we see that λblue < λyellow. d d where I have set m = 2 and used the small angle approximation. If the slits are separated by a distance of 0.2 Solution: The position of the second order bright fringe (constructive interference) is given by sin θ ≈ θ = m 2λ λ = . Field Chapter 36 Solutions Problem 1: Light from a small region of a 100 Watt incandescent bulb passes through a yellow filter and then serves as the source for a Young's double-slit interference experiment.PHY2049 R. We also know that Solutions Chapter 36 Page 1 of 10 . as shown in the figure. Field ta n θ ≈ θ = y L . of the central bright spot is w = 2y = λL and d λ λ w2 1 = 2 = 0 = . 2 mm . What is the index of refraction n of the clear liquid? (Note: assume L >> d) Answer: 1. Solving for n gives Solutions Chapter 36 Page 2 of 10 . If the entire apparatus is immersed in a clear liquid with index of refraction n the width of the central bright spot shrinks to 0. The position of the second order bright fringe is thus y = 2λ L. d and y 2 − y1 = 2 (λ 2 − λ1 ) 2 ( 0 . 6 nm )( 10 m ) L = = 1 .33 Solution: We know that for double-slit interference the position of the first dark fringe is given by λ 2d y tan θ ≈ θ = L sin θ ≈ θ = and hence the width.PHY2049 R. w. D. λ1 w1 nλ0 n where I used λ1 = λ0 and λ2 = λ0/n where λ0 is the vacuum wavelength. 10 − 5 m d Problem 3: A monochromatic light placed equal distance from two slits a 1 cm distance d apart produces a central Monochromatic Light L bright spot with a width of 1 cm on a screen located a distance L away.75 cm. Thus. D. equidistant from L=20 km both antennas.5 m from the double slit. what is the wavelength of the light (in nm)? Answer: 640 Solutions Chapter 36 Page 3 of 10 . Problem 5: Monochromatic light is incident on a two slits 0. picks up an acceptable signal (see Figure). Field n= w1 1cm = = 1 . 75 cm Problem 4: Two radio antennas are 600 m apart along a north-south line and broadcast in-phase y signals at frequency 1. How far due north of the present location (in km) would the receiver again detect a signal of nearly the same intensity? Answer: 10 Solution: This is the same as “two slit interference” and the condition for constructive interference is sin θ ≈ θ = m λ λ = . We also know that ta n θ ≈ θ = y L .8 mm from the central bright spot on a screen located a distance of 1. If the first order bright fringe is 4. 33 . A receiver d=600 m θ placed 20 km to the east. w2 0 .0 MHz. y = L = d fd (10 6 / s )( 600 m ) where I used λ = c/f. d d where I have set m = 1 and used the small angle approximation. cL ( 3 × 10 8 m / s )( 20 km ) λ = = 10 km .2 mm apart.PHY2049 R. Problem 6: A soap (water!) film has refractive index 1. Field Solution: We know that for double-slit interference mλ d y tan θ ≈ θ = L sin θ ≈ θ = and hence yd ( 4 . 8 × 10 − 3 m )( 0 .5 m where I used m = 1 for the 1st order bright fringe. mL 1 .PHY2049 R. Solutions Chapter 36 Page 4 of 10 . m m which produces the following table: m Wavelength λ0 1 1424 nm 2 737 nm 3 491 nm 4 369 nm with visible light having 400 < λvisible < 700 nm.34 and is 550 nm thick. 2 × 10 − 3 m ) λ = = = 640 nm . 2 where λfilm = λ0/nsoap. D. What wavelengths of visible light (in nm) are not reflected from it when it is illuminated from directly above by sunlight? Answer: 491 Solution: Here we set the overall phase shift equal to the condition for maximum destructive interference as follows: 1 ∆ overall = 2 T + λ film / 2 = m + λ film . Solving for λ0 gives λ 0 = n soap 2T 1424 nm = . What is the thickness T (in nm) of the glass plate? Answer: 300 Solutions Chapter 36 blue T T Page 5 of 10 .5 as shown in the Figure.5 and part of it undergoes reflection between the faces of the plate as shown in the Figure. ∆ overall = 2 T + 0 + 0 = m λ film . D. What minimum nonzero T thickness T (in nm) of the glass plate will produce maximum brightness in the transmitted light? Answer: 200 Solution: For constructive interference we set the overall phase shift equal to the condition for maximum constructive interference as follows.PHY2049 R. 5 ) red A beam of red light (λ λred = 600 nm) and a beam of blue light (λ λblue = blue 450 nm) are incident on the left side of a flat glass plate with refractive index 1. The minimum thickness occurs when m = 1 giving T min = Problem 8: λ max 600 nm = = 200 nm . Field Problem 7: A beam of 600 nm light is incident on the n = 1. 2 n film 2 (1 .5 left side of a flat glass plate with refractive index 1. An observer on the other side of the glass plate sees a bright blue light (maximum constructive interference) and sees no red light (maximum destructive interference). Solving for Τ gives T = m λ film 2 = m λ max 2 n film . Solutions Chapter 36 Page 6 of 10 .PHY2049 R. Solving for Τ gives T =m λ film 2 =m λ max 450 nm =m = m (150 nm ) . 5 ) For the constructive interference of the blue light we have the following possible thickness’ of the glass: m T 1 150nm 2 300nm 3 450nm 4 600nm For the red light and destructive interference we set the overall phase shift equal to the condition for maximum destructive interference as follows. 2 n film 2 (1 . D. 5 ) For the destructive interference of the red light we have the following possible thickness’ of the glass: m T 0 100nm 1 300nm 2 500nm 3 700nm We see that T = 300 nm satisfies both conditions. Field Solution: For the blue light and constructive interference we set the overall phase shift equal to the condition for maximum constructive interference as follows. ∆ overall = 2 T + 0 + 0 = m λ film . 600 nm = ( m + 1 / 2 )( 200 nm ) 2 (1 . Solving for Τ gives T = (m + 1 / 2) = (m + 1 / 2) λ film 2 = (m + 1 / 2) λ max 2 n film . ∆ overall = 2T + 0 + 0 = ( m + 1 / 2 ) λ film . If ngas = 1.4) 1120nm = . m m Thus. If T indicates the film thickness and λ is the wavelength of the light in the film. which of the following wavelengths will be missing from the reflected beam due to destructive interference? (a) 280 nm (b) 320 nm (c) 373 nm (d) 400 nm (e) 560 nm Answer: ace Solution: For destructive interference we set the overall phase shift equal to the condition for maximum destructive interference as follows. D.4 and nwater = 1.33. Incident light ray T = 5λ/4 II. Incident light ray T = λ/2 Observer Problem 10: Three experiments involving a thin film (in air) are shown in the Figure. Incident light ray T = 3λ/2 Observer III. Field Problem 9: A thin film of gasoline with a thickness of 400 nm floats on a puddle of water. Solving for λ = λfilm/nfilm gives λ= 2Tn film m == 2(400nm)(1. ∆ overall = 2T + λ film / 2 + 0 = ( m + 1 / 2)λ film . which experiments will produce constructive interference as seen by the observer? Solutions Chapter 36 Page 7 of 10 .PHY2049 R. the following wavelengths are missing due to destructive interference: m λ 1 1120nm 2 560nm (e) 3 373nm (c) 4 280nm (a) 5 224nm Observer I. Sunlight falls almost perpendicularly on the film and reflects into your eyes. 2 which produces the following table (where λ = λfilm): III: Constructive Solutions m T 1 2 λ/2 λ Chapter 36 Page 8 of 10 . D. Solving for Τ gives 1 λ film T = m − 2 2 .PHY2049 R. Field Answer: I and III Solution: For I and II setting the overall phase shift equal to the condition for maximum constructive interference gives: ∆ o ve ra ll = 2 T + λ film / 2 = m λ film . Solving for Τ gives T = m λ film . which produces the following table (where λ = λfilm): I & II: Constructive m T 1 2 3 4 λ/4 3λ/4 5λ λ/4 7λ/4 For III setting the overall phase shift equal to the condition for maximum constructive interference gives: ∆ overall = 2 T = m λ film . Solving for Τ gives T =m λ film 2 =m λ max 450 nm =m = m (150 nm ) . 2 n film 2 (1 . ∆ overall = 2T + λ film + 0 = m λ film .PHY2049 R. ∆ overall = 2 T + 0 + 0 = m λ film . If the light is incident from the left and if T the observer on the left sees maximum constructive interference for red light (λ λred = 600 nm) and the observer on the right sees maximum constructive interference for blue light (λ λblue = 450 nm).5 as shown in the Figure. what is the thickness T (in nm) of the glass plate? Answer: 300 Solution: For the observer on the right we have blue light and constructive interference so we set the overall phase shift equal to the condition for maximum constructive interference as follows. D. Field Problem 11: Incident Light Two observers are on opposite sides blue red of a flat glass plate with refractive index 1. 5 ) For the constructive interference of the blue light on the right we have the following possible thickness’ of the glass: m T 1 150nm 2 300nm 3 450nm 4 600nm For the observer on the left we have red light and constructive interference we set the overall phase shift equal to the condition for maximum constructive interference as follows. 2 Solving for Τ gives Solutions Chapter 36 Page 9 of 10 . D. 5 ) For the constructive interference of the red light on the left we have the following possible thickness’ of the glass: m T 1 100nm 2 300nm 3 500nm 4 700nm We see that T = 300 nm satisfies both conditions. Field T = (m − 1 / 2) = (m − 1 / 2) λ film λ = ( m − 1 / 2 ) max 2 2 n film . 600 nm = ( m − 1 / 2 )( 200 nm ) 2 (1 .PHY2049 R. Solutions Chapter 36 Page 10 of 10 . 02 × 10 − 3 m Problem 2: Monochromatic light of wavelength 600 nm is incident on a single narrow slit and produces a diffraction pattern on a screen 2 meters away? If the screen is moved 1 meter further away from the slit. what wavelength of monochromatic light (in nm) will produce a diffraction pattern with all the diffraction minima at the same points on the screen? Answer: 400 Solution: The angular position of the dark fringes is given by sin θ ≈ θ = Solving for λ gives mλ y ≈ = tan θ θ .PHY2049 R.) Answer: 15 Solution: The angular position of the first dark fringe is given by sin θ ≈ θ = λ y ≈ = tan θ θ . W L Thus. and . and m we have L λ2 = 1 L2 λ1 so that Solutions Chapter 37 Page 1 of 2 . is given by 2λ 2 ( 500 × 10 − 9 m ) D = 2y = L = ( 3 m ) = 15 cm . and . Field Chapter 37 Solutions Problem 1: If monochromatic light of wavelength 500 nm is incident on a single narrow slit 0. what is the width of the central bright spot (in cm) on a screen that is 3 meters from the slit? (Hint: the width of the central bright spot is equal to the distance between the two first order minima. the width of the central bright spot. W L λ = yW mL and for fixed y. D. W 0 . D = 2y. W.02 mm wide. Field λ2 = L1 2m ( 600 nm ) = 400 nm .PHY2049 R. Page 2 of 2 . the width of the central bright spot on the screen increases by 1.18 mm and produces a diffraction pattern on a screen located a distance L from the slit. If the color of light is changed to red (λ λ = 650 nm). D. and . Solving for L gives ∆ DW (1 × 10 − 2 m )( 0 . What is the distance L to the screen (in meters)? Answer: 5 Solution: The angular position of the first dark fringe is given by sin θ ≈ θ = y λ ≈ = tan θ θ . W W where ∆D = D2 – D1 and ∆λ = λ2 – λ1. λ1 = 2 m + 1m L2 Problem 3: Blue light (λ λ = 470 nm) falls on a wall with a single slit of width 0. D = 2y. 18 × 10 = L = 2∆λ 2 (180 × 10 − 9 m ) Solutions Chapter 37 −3 m) = 5m . is given by D = 2y = 2λ 2∆λ L . and ∆ D = L. the width of the central bright spot.0 cm. W L Thus.