12/14/2015Homework # 2 Ch 26 Homework # 2 Ch 26 Due: 11:59pm on Friday, October 23, 2015 To understand how points are awarded, read the Grading Policy for this assignment. Electric Field due to Multiple Point Charges Two point charges are placed on the x axis. The first charge, q = 8.00 nC , is placed a distance 16.0 m from the origin 1 along the positive x axis; the second charge, q 2 = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis. Part A Calculate the electric field at point A, located at coordinates (0 m, 12.0 m ). Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Hint 1. How to approach the problem Find the contributions to the electric field at point A separately for q 1 and q 2 , then add them together (using vector addition) to find the total electric field at that point. You will need to use the Pythagorean theorem to find the distance of each charge from point A. Hint 2. Calculate the distance from each charge to point A Calculate the distance from each charge to point A. Enter the two distances, separated by a comma, in meters to three significant figures. ANSWER: r A1 , r A2 = 20.0,15.0 m Hint 3. Determine the directions of the electric fields Which of the following describes the directions of the electric fields E⃗ A1 and E⃗ A2 created by charges q 1 Typesetting math: 23% and q 2 at point A? ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3971761 1/34 12/14/2015 Homework # 2 Ch 26 ⃗ E A1 points up and left and E⃗ A2 points up and right. ⃗ E A1 points up and left and E⃗ A2 points down and left. ⃗ E A1 points down and right and E⃗ A2 points up and right. ⃗ E A1 points down and right and E⃗ A2 points down and left. Hint 4. Calculate the components of E ⃗ A1 Calculate the x and y components of the electric field E⃗ A1 at point A due to charge q 1 . Express your answers in newtons per coulomb, separated by a comma, to three significant figures. Hint 1. Calculate the magnitude of the total field Calculate the magnitude of the field EA1 at point A due to charge q 1 only. Express your answer in newtons per coulomb to three significant figures. ANSWER: EA1 = 0.180 N/C Hint 2. How to find the components of the total field Once you have found the magnitude of the field, use trigonometry to determine the x and y components of the field. The electric field of a positive point charge points directly away from the charge, so the direction of the electric field at point A due to charge q 1 will be along the line joining the two. Use the position coordinates of q 1 and point A to find the angle that the line joining the two makes with the x or y axis. Then use this angle to resolve the electric field vector into components. ANSWER: EA1x , EA1y = 0.144,0.108 N/C Hint 5. Calculate the components of E ⃗ A2 Calculate the x and y components of the electric field at point A due to charge q 2 . Express your answers in newtons per coulomb, separated by a comma, to three significant figures. Hint 1. Calculate the magnitude of the total field Calculate the magnitude of the field EA2 at point A due to charge q 2 only. Express your answer in newtons per coulomb to three significant figures. ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3971761 2/34 12/14/2015 Homework # 2 Ch 26 EA2 = 0.240 N/C Hint 2. How to find the components of the total field Once you have found the magnitude of the field, use trigonometry to determine the x and y components of the field. The electric field of a positive point charge points directly away from the charge, so the direction of the electric field at point A due to charge q 2 will be along the line joining the two. Use the position coordinates of q 2 and point A to find the angle that the line joining the two makes with the x or y axis. Then use this angle to resolve the electric field vector into components. ANSWER: EA2x , EA2y = 0.144,0.192 N/C ANSWER: EAx , EAy = 0,0.300 N/C Correct Part B An unknown additional charge q 3 is now placed at point B, located at coordinates (0 m, 15.0 {\rm m} ). Find the magnitude and sign of \texttip{q_{\rm 3}}{q_3} needed to make the total electric field at point A equal to zero. Express your answer in nanocoulombs to three significant figures. Hint 1. How to approach the problem You have already calculated the electric field at point A due to \texttip{q_{\rm 1}}{q_1} and \texttip{q_{\rm 2}} {q_2}. Now find the charge \texttip{q_{\rm 3}}{q_3} needed to make an opposite field at point A, so when the two are added together the total field is zero. Hint 2. Determine the sign of the charge Which sign of charge \texttip{q_{\rm 3}}{q_3} is needed to create an electric field \texttip{\vec{E}_{\rm A3}} {E_A3_vec} that points in the opposite direction of the total field due to the other two charges, \texttip{q_{\rm 1}}{q_1} and \texttip{q_{\rm 2}}{q_2}? ANSWER: positive negative https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3971761 3/34 red arrows will appear. When you are done. This simulation allows you to place multiple positive and negative pointcharges in any configuration and look at the resulting electric field. https://session. Faint red arrows indicate that the electric field is weaker than at locations where the arrows are brighter (this simulation does not use arrow length as a measure of field magnitude). You should see something similar to the figure below. ANSWER: \texttip{q_{\rm 3}}{q_3} = 0. For this problem.300 \rm{nC} Correct PhET Tutorial: Charges and Electric Fields Learning Goal: To understand the spatial distribution of the electric field for a variety of simple charge configurations. Calculating the magnitude of the new charge Keep in mind that the magnitude of the field due to \texttip{q_{\rm 3}}{q_3} is E_{\rm A3}=kq_3/r_{\rm A3}^2. Drag one positive charge and place it near the middle of the screen. showing the direction of the electric field.12/14/2015 Homework # 2 Ch 26 Hint 3. Start the simulation. right on top of two intersecting bold grid lines.com/myct/assignmentPrintView?assignmentID=3971761 4/34 .masteringphysics. and the field must be equal in magnitude to the field due to charges \texttip{q_{\rm 1}}{q_1} and \texttip{q_{\rm 2}}{q_2}. Feel free to play around with the simulation. Part A Select Show Efield and grid in the green menu. use the PhET simulation Charges and Fields. You can click and drag positive charges (red) or negative charges (blue) into the main screen. If you select Show Efield in the green menu. click Clear All before beginning Part A. com/myct/assignmentPrintView?assignmentID=3971761 5/34 . and look at the resulting field strength. Place the EField Sensor 1 \rm m away from the positive charge (1 \rm m is two bold grid lines away if going in a horizontal or vertical direction). Part B Now. would experience a force directed radially away from the original charge. and then click and drag one of the orange EField Sensors. Consider the locations to the right. You will see the magnitude of the electric field given in units of V/m (volts per meter.masteringphysics. is directed radially toward the charge at all locations near the charge. ANSWER: https://session. left.12/14/2015 Homework # 2 Ch 26 ANSWER: The electric field produced by the positive charge is directed radially away from the charge at all locations near the charge. all 1 \rm m away. let's look at how the distance from the charge affects the magnitude of the electric field. wraps circularly around the positive charge. and below the positive charge. above. if placed near the original charge. Correct This means that another positive charge. Select Show numbers on the green menu. which is the same as newtons per coulomb). greatest to the left of the charge. This is consistent with Coulombâ s law. what is the field strength \texttip{E}{E} a distance of 3 \rm m from the charge? https://session.com/myct/assignmentPrintView?assignmentID=3971761 6/34 . How to approach the problem Use an EField Sensor to determine the field strength both at 1 \rm m away and at 2 \rm m away from the charge. greatest to the right of the charge. Part D If the field strength is \texttip{E}{E} = 9 \rm V/m a distance of 1 \rm m from the charge. the magnitude of the electric field is the same.masteringphysics. The magnitude of the electric field is proportional to the inverse of the distance squared (E \propto1/r^2. Then. Correct This result implies that the strength of the electric field due to one point charge depends solely on the distance away from the charge. Correct The magnitude of the field decreases more quickly than the inverse of the distance from the charge. greatest above the charge.12/14/2015 Homework # 2 Ch 26 greatest below the charge. For these four locations. which states that the magnitude of the force between two charged particles is F = k Q_1Q_2/r^2. You should verify this by looking at the field strength 3 or 4 meters away. Mathematically. ANSWER: equal to two times The magnitude of the electric field 1 \rm m away from the positive charge is four times onehalf onequarter the magnitude of the electric field 2 \rm m away. Part C What is the magnitude of the electric field 1 m away from the positive charge compared to the magnitude of the electric field 2 m away? Hint 1. we say the electric field is spherically symmetric. take the ratio of the two field strengths. where \texttip{r}{r} is the distance from the charge). com/myct/assignmentPrintView?assignmentID=3971761 7/34 . https://session. and the electric field strength doesnâ t change. but the field strength doesnâ t change. Part E Remove the positive charge by dragging it back to the basket. and drag a negative charge (blue) toward the middle of the screen. and the magnitude of the electric field decreases at all locations. Nothing changes; the electric field remains directed radially outward. the field strength must decrease by a factor of three squared. Part F Now. placing them 1 \rm m apart. if the distance is increased by a factor of three.12/14/2015 Homework # 2 Ch 26 Hint 1. and drag two positive charges. but the electric field strength does not change. Which statement best describes the differences in the electric field due to a negative charge as compared to a positive charge? ANSWER: The electric field changes direction (now points radially inward). So if the distance is increased by a factor of three. the field produced by a negative charge is directed opposite to that of a positive charge but the magnitude of the field is the same. ANSWER: \texttip{E}{E} = 1 \rm V/m Correct Correct. Correct The electric field is now directed toward the negative charge. How to approach the problem The magnitude of the electric field is inversely proportional to distance squared (E \propto 1/r^2). as shown below. remove the negative charge. Because of the sign of the charge. The electric field changes direction (now points radially inward).masteringphysics. Determine how the electric field is different from that of the positive charge. the electric field is decreased by a factor of nine. Since E \propto 1/r^2. The electric field of a point charge is given by \vec{E}=(kQ/r^2)\hat{r}. You could use the simulation to make a measurement (you might have to drag the charge away from the center so you have enough room to get 3 \rm m away). so the net electric field is the vector sum of the electric fields due to each of the two charges. What is the magnitude of the total electric field due to both charges at this location? ANSWER: zero 25 \rm V/m 36 \rm V/m https://session.5 \rm m above the midpoint of the two charges. The electric field is roughly zero near the midpoint of the two charges. Correct Directly between the two charges. As you can verify by removing one of the positive charges.12/14/2015 Homework # 2 Ch 26 Letâ s look at the resulting electric field due to both charges. The electric field is nonzero everywhere on the screen. Part G Consider a point 0. the electric fields produced by each charge are equal in magnitude and point in opposite directions.masteringphysics.com/myct/assignmentPrintView?assignmentID=3971761 8/34 . Recall that the electric field is a vector. the electric field due to only one of the positive charges is about 18 \rm V/m. Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)? ANSWER: The electric field is zero at any location along a vertical line going through the point directly between the two charges. so the two vectors add up to zero. Part I Make a small dipole by bringing the two charges very close to each other. so the final configuration looks like the figure shown below. https://session. is also to the right. ANSWER: directed to the left. The electric field at the midpoint is zero. which is the sum of these two fields. where they are barely touching. So the net electric field.masteringphysics. The midpoint of the two charges should still be on one of the grid point intersections (see figure below).12/14/2015 Homework # 2 Ch 26 Correct Notice that this number is less than twice the magnitude of the field due to each charge. as is the electric field due to the negative charge. directed to the right. Correct The electric field due to the positive charge is directed to the right. Part H Make an electric dipole by replacing one of the positive charges with a negative charge.com/myct/assignmentPrintView?assignmentID=3971761 9/34 . This occurs because the horizontal components of the electric field due to each charge exactly cancel out (add to zero). Only the vertical components of the fields add together. Compare this value to the value you measure with an EField Sensor. say 0. Correct In fact. Part J Make a long line of positive charges. Does the strength of the electric field decrease as 1 over distance squared (1/r^2)? Hint 1. a distribution of charges produces an electric field that is very different from that of a single charge. Feel free to look at the electric field. ANSWER: No. in general. The important lesson here is that. it decreases less quickly with distance.com/myct/assignmentPrintView?assignmentID=3971761 10/34 . https://session. as it is interesting.masteringphysics.5 \rm m directly above the midpoint as well as 1 \rm m directly above. Try to place all of the charges centered along a horizontal grid line. Yes. it does.5)^2/(1/1)^2=4. similar to that shown in the figure below.5 \rm m away and 1 \rm m away. would be E_{r=0. No.5 \rm m above the midpoint. it turns out that the strength of the electric field decreases roughly as 1/r^3! So the field 1 \rm m above the midpoint is roughly eight times weaker than at 0. then the ratio of the magnitudes of the electric field measured at two distances.5}/E_{r=1} =(1/0. How to approach the problem If the strength of the field is decreasing as 1/r^2.12/14/2015 Homework # 2 Ch 26 Measure the strength of the electric field 0. it decreases more quickly with distance. Does the strength of the electric field decrease as 1 over distance squared (1/r^2)? ANSWER: No. Yes. centered at the origin. PhET Interactive Simulations University of Colorado http://phet. This is another example showing that a distribution of charges produces an electric field that is very different from that of a single charge.com/myct/assignmentPrintView?assignmentID=3971761 11/34 . it decreases less quickly with distance. The ring has radius \texttip{a}{a} and positive charge \texttip{q}{q} distributed evenly along its circumference.edu Charged Ring Consider a uniformly charged ring in the xy plane.colorado. it decreases more quickly with distance. Correct In fact. No. it turns out that the strength of the electric field decreases roughly as 1/r.5 \rm m. So the field 1 m above the midpoint is roughly half the strength at 0. https://session. it does.masteringphysics.12/14/2015 Homework # 2 Ch 26 Measure the strength of the electric field 1 \rm m directly above the middle as well as 2 \rm m directly above. the component of the electric field in the xy plane will rotate also. Now imagine that you make a copy of the ring and rotate this copy about its axis.12/14/2015 Homework # 2 Ch 26 Part A What is the direction of the electric field at any point on the z axis? Hint 1. What is this value? Does a similar argument hold for the z component of the field? ANSWER: parallel to the x axis parallel to the y axis parallel to the z axis in a circle parallel to the xy plane Correct https://session. As a result of the rotation.masteringphysics.com/myct/assignmentPrintView?assignmentID=3971761 12/34 . In what direction is the net field? What if you did this for every pair of points on opposite sides of the ring? Approach 2 Consider a general electric field at a point on the z axis. How to approach the problem Approach 1 In what direction is the field due to a point on the ring? Add to this the field from a point on the opposite side of the ring.. i. because the ring that is rotated looks just like the one that isn't. one that has a z component as well as a component in the xy plane. Now you ask a friend to look at both rings. Your friend wouldn't be able to tell them apart.e. However. they have the component of the electric field in the xy plane pointing in different directions! This apparent contradiction can be resolved if this component of the field has a particular value. Use Coulomb's law to write the magnitude of the infinitesimal \texttip{dE} {dE} at a point on the positive z axis due to the charge \texttip{dq}{dq} shown in the figure. Consider an infinitesimal piece of the ring with charge \texttip{dq}{dq}. you should use Coulomb's law to find the contribution \texttip{dE}{dE} to the electric field at the point (0. Therefore. where \large{k = \frac{1}{4 \pi \epsilon_0}}.masteringphysics. you can integrate over the ring to find the value of \texttip{E}{E}.z) from a piece of charge \texttip{dq}{dq} on the ring at a distance \texttip{r}{r} away. the net field must point along the z axis. Simplifying with symmetry By symmetry. \texttip{z}{z}. ANSWER: \texttip{dE}{dE} = \large{\frac{k dq}{z^{2}+a^{2}}} Hint 2. \large{F = k\frac{q_1q_2}{r^2}}. and \texttip{a}{a}. where \large{k = \frac{1}{4 \pi \epsilon_0}}.12/14/2015 Homework # 2 Ch 26 Part B What is the magnitude of the electric field along the positive z axis? Use \texttip{k}{k} in your answer. all we care about is the z component of each such contribution. What is the component \texttip{dE_{\mit z}} {dE_z} of the electric field caused by the charge on an infinitesimally small portion of the ring \texttip{dq}{dq} in the z direction? https://session. because the horizontal component of each contribution of magnitude \texttip{dE}{dE} is exactly canceled by the horizontal component of a similar contribution of magnitude \texttip{dE}{dE} from the other side of the ring. away from the ring. Then. Formula for the electric field You can always use Coulomb's law. In the situation below. Given the force.com/myct/assignmentPrintView?assignmentID=3971761 13/34 .0. to find the electric field (the Coulomb force per unit charge) due to a point charge. Hint 1. Use \texttip{k}{k} in your answer. the electric field say at \texttip{q_{\rm 2}}{q_2} due to \texttip{q_{\rm 1}}{q_1} is \large{E = \frac{F}{q_2} = k\frac{q_1}{r^2}}. You may also use some or all of the variables \texttip{dq}{dq}. masteringphysics. the infinitesimally small contribution to the electric field; \texttip{z}{z}. If you are not comfortable integrating \texttip{dq}{dq} over the ring. ANSWER: \texttip{dE_{\mit z}}{dE_z} = \large{\frac{dE z}{\sqrt{z^{2}+a^{2}}}} Hint 3. Integrating around the ring If you combine your results from the first two hints. change to a spatial variable.com/myct/assignmentPrintView?assignmentID=3971761 14/34 . you will have an expression for \texttip{dE_{\mit z}} {dE_z}. ANSWER: \texttip{E\left(z\right)}{E(z)} = \large{\frac{kqz}{\left(z^{2}+a^{2}\right)^{\frac{3}{2}}}} https://session. the vertical component of the field due to the infinitesimal charge \texttip{dq}{dq}. convince yourself that \large{\displaystyle{\oint}_{\rm ring}dq=\int_{0}^{2\pi}\frac{q}{2\pi}d\theta}. Since the total charge \texttip{q}{q} is distributed evenly about the ring.12/14/2015 Homework # 2 Ch 26 Express your answer in terms of \texttip{dE}{dE}. the coordinate of the point on the z axis; and \texttip{a}{a}. The total field is \vec{E}=E_z\hat{k}=\hat{k}\displaystyle{\oint}_{\rm ring}dE_z. the radius of the ring. d) and constrained to move along the z axis. For points on the positive z axis. For points on the negative z axis. it is good to see that if \texttip{|z|}{|z|} is much greater than \texttip{a}{a} the magnitude of \texttip{E\left(z\right)}{E(z)} is approximately \large{k \frac{q}{z^2}}. assume that d^2+a^2 \approx a^2. independent of the size of the ring: The field due to the ring is almost the same as that due to a point charge \texttip{q}{q} at the origin. which points toward the origin.12/14/2015 Homework # 2 Ch 26 Correct Notice that this expression is valid for both positive and negative charges as well as for points located on the positive and negative z axis. For points on the negative z axis. the electric field should point outward. For points on the positive z axis. and constants. the prime on the symbol representing the spring constant is to distinguish it from \large{k=\frac{1} {4\pi\epsilon_0}}). https://session. Newton's second law for the system states that \large{F_x= m\frac{d^2 x}{dt^2} = {k'} x}. Express your answer in terms of given charges. 0. the algebraic expression is valid for any signs of the parameters. which is outward from the origin. What will be the angular frequency \texttip{\omega }{omega} of these oscillations? Use the approximation d \ll a to simplify your calculation; that is.com/myct/assignmentPrintView?assignmentID=3971761 15/34 . Simple harmonic motion Recall the nature of simple harmonic motion of an object attached to a spring. the field points in the positive z direction. If the charge is negative. even though we obtained the above result for postive \texttip{q}{q} and \texttip{z}{z}. which also points toward the origin. As a check. the negative sign from the z coordinate and the negative sign from the charge cancel. Hint 1. Therefore. which is also outward from the origin. with no damping. what will be the ball's subsequent trajectory? ANSWER: repelled from the origin attracted toward the origin and coming to rest oscillating along the z axis between z = d and z = d circling around the z axis at z = d Correct Part D The ball will oscillate along the z axis between z=d and z=d in simple harmonic motion. the electric field should point toward the origin. and the field points in the positive z direction. If 0 < d \ll a. the field points in the negative z direction. the negative sign from the charge causes the electric field to point in the negative z direction. The ball is released from rest at the point (0. leading to oscillation at a frequency of \large{\omega=\displaystyle{\sqrt{\frac{k'}{m}}}} (here. Write an analogous equation for the ball near the charged ring in order to find the \texttip{\omega }{omega} term. The solution to this differential equation is a sinusoidal function of time with angular frequency \texttip{\omega }{omega}. If the charge is positive. Part C Imagine a small metal ball of mass \texttip{m}{m} and negative charge q_0. dimensions.masteringphysics. Three positively charged particles. F_z = qE_z. with charges q_1=q. Hint 1.0. https://session. A formula for the force on a charge in an electric field The formula for the force \texttip{\vec{F}}{F_vec} on a charge \texttip{q}{q} in an electric field \texttip{\vec{E}}{E_vec} is \vec{F} = q\vec{E}. Find the force on the charge What is \texttip{F_{\mit z}}{F_z}. Therfore.com/myct/assignmentPrintView?assignmentID=3971761 16/34 . The charge q_2 is located diagonally from the remaining (empty) corner. PROBLEMSOLVING STRATEGY 26. and \texttip{a}{a}. and show the location of the charges. You have already found \texttip{E_{\mit z}\left(z\right)}{E_z(z)} in Part B.d). ANSWER: \texttip{F_{\mit z}}{F_z} = \large{\frac{k q q_{0} d}{a^{3}}} ANSWER: \texttip{\omega }{omega} = \large{\sqrt{\frac{kqq_{0}}{a^{3}m}}} Correct PSS 26.1 for point charge problems. in particular.0. q_2=2q. Find the magnitude of the resultant electric field \texttip{\vec{E}_{\rm net}}{E_net_vec} in the empty corner of the square. \texttip{d}{d}.1 The electric field of multiple point charges MODEL: Model charged objects as point charges. and q_3=q (where q > 0). Identify the point P at which you want to calculate the electric field.12/14/2015 Homework # 2 Ch 26 Hint 2.masteringphysics. Use that expression in the equation above to find an expression for the z component of the force \texttip{F_{\mit z}}{F_z} on the ball at the point (0.1: The Electric Field of Multiple Point Charges Learning Goal: To practice ProblemSolving Strategy 26. are located at the corners of a square with sides of length \texttip{d}{d}. Don't forget to use the approximation given. \texttip{k}{k}. the z component of the force on the ball on the ball at the point (0.d)? Use the approximation d^2 + a^2 \approx a^2. \texttip{q}{q}. VISUALIZE: For the pictorial representation: Establish a coordinate system. Express your answer in terms of \texttip{q_{\rm 0}}{q_0}. Calculate the field strength of each charge's electric field. If needed. determine its distance from P and the angle of \texttip{\vec{E}_{\it i}}{E_i_vec} from the axes. is reasonable. Write each vector \texttip{\vec{E}_{\it i}}{E_i_vec} in component form. The length of your vectors will not be graded. ANSWER: https://session. Visualize Part A Below is an incomplete pictorial representation of the situation described in this problem. Make sure that all your vectors have the correct orientation. determine the magnitude and direction of \texttip{\vec{E}_{\rm net}}{E_net_vec}. For each charge. ASSESS: Check that your result has the correct units. Model Model the charged particles as point charges.com/myct/assignmentPrintView?assignmentID=3971761 17/34 . The orientation of your vectors will be graded.12/14/2015 Homework # 2 Ch 26 Draw the electric field of each charge at P. Use symmetry to determine if any of components of \texttip{\vec{E}_{\rm net}}{E_net_vec} are zero. Sum the vector components to determine \texttip{\vec{E}_{\rm net}}{E_net_vec}. SOLVE: The mathematical representation is \vec E_{\rm net}=\sum \vec E_i.masteringphysics. Complete the sketch by drawing the electric field due to each charge at point P. and agrees with any known limiting cases. Use \texttip{K}{K} for the electrostatic constant. With reference to the coordinate system shown in the previous part. Express your answer in terms of \texttip{q}{q}. ANSWER: (E_3)_x = \large{\frac{K q}{d^{2}}} https://session. the charge configuration is symmetric about the line y=x. Hint 1. \texttip{d}{d}. as you can see from the diagram you drew in Part B. Find the x component of the electric field due to charge 3 Find (E_3)_x. Solve Part C Determine the magnitude \texttip{E_{\rm net}}{E_net} of the net electric field at point P. Note that \texttip{\vec{E}_{\rm 1}}{E_1_vec} has no x component. This means that the x and y components of the net electric field must be equal. Exploit symmetry At the end of the Visualize step. if any. In particular. which component of \texttip{\vec{E}_{\rm net}}{E_net_vec}. the x component of the electric field due to \texttip{q_{\rm 3}}{q_3} at point P. consider the resultant electric field \texttip{\vec{E}_{\rm net}}{E_net_vec} at P. is zero in this problem? ANSWER: only the x component only the y component both the x and y components neither the x nor the y component Correct Although neither of the electric field components is zero in this situation. it was determined that the x and y components of \texttip{\vec{E}_{\rm net}}{E_net_vec} are equal. Use \texttip{K} {K} for the electrostatic constant.masteringphysics. Hint 2. the x component) to solve the problem.com/myct/assignmentPrintView?assignmentID=3971761 18/34 . Express your answer in terms of \texttip{q}{q}. there is still symmetry that you can exploit. \texttip{d}{d}. and \texttip{K}{K}.12/14/2015 Homework # 2 Ch 26 Correct Part B Now. So you only need to compute one of them (say. and \texttip{K}{K}. Take advantage of this fact to reduce the amount of work you need to do in the Solve step. The net electric field Recall that \vec{E}_{\rm net}=\sum \vec{E}_i. What is \texttip{\theta }{theta} in this case? ANSWER: (E_2)_x = \large{\frac{K q \sqrt{2}}{2 d^{2}}} Hint 4. Express your answer in terms of \texttip{d}{d}. Find the x component of the electric field due to charge 2 Find (E_2)_x. where \texttip{\theta }{theta} is the angle between the x axis and \texttip{\vec{E}_{\rm 2}}{E_2_vec}. \texttip{d}{d}. Use \texttip{K} {K} for the electrostatic constant. Find the magnitude of the electric field due to charge 2 What is the magnitude of \texttip{\vec{E}_{\rm 2}}{E_2_vec}. and (E_3)_x. Express your answer in terms of \texttip{q}{q}. In component form.masteringphysics. Hint 1. So to find (\vec{E}_{\rm net})_x.com/myct/assignmentPrintView?assignmentID=3971761 19/34 . simply add (E_1)_x. https://session. Hint 1. Hint 1. \texttip{d}{d}. and \texttip{K}{K}. How to calculate the x component Recall that (E_2)_x = |\vec{E}_2|\cos(\theta). (E_2)_x. ANSWER: \texttip{d_{\rm 2}}{d_2} = d \sqrt{2} ANSWER: |\vec E_2| = \large{\frac{K q}{d^{2}}} Hint 2. Express your answer in terms of \texttip{q}{q}. Find the x component of the net electric field Find the x component of the net electric field \texttip{\vec{E}_{\rm net}}{E_net_vec} at point P. and \texttip{K}{K}. and \texttip{K}{K}. the electric field due to \texttip{q_{\rm 2}}{q_2} at point P? Use \texttip{K}{K} for the electrostatic constant. Find the distance between charge 2 and point P Find the distance \texttip{d_{\rm 2}}{d_2} between q_2 and point P. this relation becomes (\vec E_{\rm net})_x=\sum (E_i)_x and (\vec{E}_{\rm net})_y=\sum (E_i)_y. Use \texttip{K}{K} for the electrostatic constant.12/14/2015 Homework # 2 Ch 26 Hint 3. \texttip{d}{d}. the x component of the electric field due to \texttip{q_{\rm 2}}{q_2} at point P. Express your answer in terms of \texttip{q}{q}. the greater the magnitude of \texttip{\vec{E}_{\rm net}}{E_net_vec} will be. Your results do make sense! It is interesting to notice that the field will not reduce to that of a point charge (of magnitude 4q) unless the point P is moved farther away from the three charges while they themselves remain at their original positions. enter AD. Find the magnitude of the electric field due to the wire at a point located a distance \texttip{d}{d} from one end of the wire along https://session. which of the following would happen to \texttip{\vec{E}_{\rm net}}{E_net_vec} if \texttip{d}{d} became very large? A. just as you would expect.com/myct/assignmentPrintView?assignmentID=3971761 20/34 . C.masteringphysics. The larger \texttip{d}{d} becomes. A straight wire of length \texttip{L}{L} has a positive charge \texttip{Q}{Q} distributed along its length. The larger \texttip{d}{d} becomes. ANSWER: C Correct Looking at your answer derived in Part C. the smaller the magnitude of \texttip{\vec{E}_{\rm net}}{E_net_vec} will be. Do not use commas. For instance. if you think that A and D are correct. D.2 The Electric Field of a Continuous Distribution of Charge Learning Goal: To practice ProblemSolving Strategy 26. you should see that the magnitude \texttip{E_{\rm net}}{E_net} will decrease as \texttip{d}{d} gets larger. B. Enter the letters of all the correct answers in alphabetical order. \texttip{\vec{E}_{\rm net}}{E_net_vec} should reduce to the field of a point charge of magnitude 4q.2 for continuous charge distribution problems. PSS 26. \texttip{\vec{E}_{\rm net}}{E_net_vec} should reduce to the field of a point charge of magnitude \texttip{q}{q}.12/14/2015 Homework # 2 Ch 26 ANSWER: (\vec E_{\rm net})_x = \large{\frac{K q \left(2+\sqrt{2}\right)}{2 d^{2}}} ANSWER: \texttip{E_{\rm net}}{E_net} = \large{{\frac{Kq}{d^{2}}}\left(1+\sqrt{2}\right)} Correct Assess Part D Intuitively. masteringphysics. 2. assume that the total charge is uniformly distributed along the wire. Divide the total charge \texttip{Q}{Q} into small pieces of charge \texttip{\Delta Q}{DeltaQ} using shapes for which you already know how to determine \texttip{\vec{E}}{E_vec}. This will help you identify distances and angles that need to be calculated. a division into point charges. assume that its diameter is much smaller than the wire's length. such as \texttip{dx}{dx}.2 The electric field of a continuous distribution of charge MODEL: Model the distribution as a simple shape. you will need to divide the total charge https://session. Carry out the integration. The integration will be over the coordinate variable that is related to \texttip{\Delta Q}{DeltaQ}. Identify the point P at which you want to calculate the electric field. 4. Let the sum become an integral. Use superposition to form an algebraic expression for each of the three components of \texttip{\vec{E}} {E_vec} at point P. SOLVE: The mathematical representation is \vec E_{\rm net}=\sum \vec E_i. y. Draw the electric field vector at P for one or two small pieces of charge. Note that the point at which you want to calculate the electric field is close to one of the ends of the wire. 5. Model No information is given on the cross section of the wire. Instead.com/myct/assignmentPrintView?assignmentID=3971761 21/34 . Replace the small charge \texttip{\Delta Q}{DeltaQ} with an equivalent expression involving a charge density and a coordinate. You may conclude that some components of \texttip{\vec{E}}{E_vec} are zero. ASSESS: Check that your result is consistent with any limits for which you know what the field should be. Also. Assume the charge is uniformly distributed. Look for symmetries in the charge distribution that simplify the field. such as a line of charge or a disk of charge. This is the critical step in making the transition from a sum to an integral because you need a coordinate to serve as the integration variable. For simplicity. This is often.) Let the (x. VISUALIZE: For the pictorial representation: 1. (Note that one or more components may be zero. Draw a picture and establish a coordinate system. Express all angles and distances in terms of the coordinates. The integration limits for this variable will depend on your choice of coordinate system. PROBLEMSOLVING STRATEGY 26. that describes the shape of charge \texttip{\Delta Q} {DeltaQ}. 3. but not always. so you cannot make use of any symmetry to simplify the problem. and model the wire as a line of charge. and simplify the result as much as possible.12/14/2015 Homework # 2 Ch 26 the line extending from the wire. z) coordinates of the point remain as variables. com/myct/assignmentPrintView?assignmentID=3971761 22/34 . if any. only the ith segment at point \texttip{x_{\it i}}{x_i} is shown in the diagram. Complete this pictorial representation by drawing the electric field \texttip{\vec{E}_{\rm i }}{E_i _vec} at P due to segment i. For clarity. Draw the vector starting at P. and the point P at which you need to calculate the electric field. a suitable coordinate system with the origin at the left end of the wire. Visualize Part A The diagram below is an incomplete pictorial representation of the situation described in this problem: It shows the positively charged wire. The total charge distributed along the wire has been divided into small segments. By modeling each segment as a point charge and using the principle of superposition. you can make a quick estimate of the direction of the net field at P. To do that. Part B With reference to the coordinate system used in the previous part.12/14/2015 Homework # 2 Ch 26 distributed along the wire into many small segments and model each segment as a pointlike charge. it is helpful to draw the electric field due to one or two more charge segments. ANSWER: Correct Each small segment will contribute to the net electric field at P. which component.masteringphysics. of the electric field due to the total charge is zero at P? https://session. each of length \texttip{\Delta x}{Deltax} and charge \texttip{\Delta Q}{DeltaQ}. Find an expression for the electric field due to segment i Find the magnitude of the electric field \texttip{E_{\rm i}}{E_i} at point P due to a segment of charge \texttip{\Delta Q}{DeltaQ} located at \texttip{x_{\it i}}{x_i} on the x axis. \texttip{L}{L}. Hint 1. and the constants \texttip{\pi }{pi} and \texttip{\epsilon _{\rm 0}}{epsilon_0}. the sum must be treated as an integral. which deals with a continuous distribution of charge rather than a pointlike distribution. as a variable. \texttip{\Delta Q} {DeltaQ}.com/myct/assignmentPrintView?assignmentID=3971761 23/34 . let \sum \vec E_i become an integral and apply your knowledge of calculus to evaluate the integral. Then. as we did in Part A. What is the distance \texttip{r_{\it i}}{r_i} between point P and a segment of charge located at \texttip{x_{\it i}}{x_i} on the x axis? https://session. How to approach the problem As explained in the strategy above. Therefore. and the constants \texttip{\pi }{pi} and \texttip{\epsilon _{\rm 0}}{epsilon_0}. \texttip{L}{L} is the length of the wire. Express your answer in terms of some or all of the variables \texttip{Q}{Q}. we simplified the problem: Now the y component of the field at P is zero and the magnitude of \texttip{\vec{E}}{E_vec} is simply given by the magnitude of its x component. the electric field due to the ith segment. and \texttip{d}{d}. and \texttip{d}{d}. After dividing the total charge into small segments. the mathematical expression for the net field is \vec E_{\rm net}=\sum \vec E_i. \texttip{L}{L}. the x coordinate of the ith segment. Because of the nature of the problem. and modeling each segment as a point charge. \texttip{\vec{E}}{E_vec} must be directed along the same line. Find the distance between point P and a segment of charge at \texttip{x_{\it i}}{x_i} Assume the origin of the x axis is at the left end of the wire and point P is located at a distance \texttip{d}{d} from the other end of the wire. Treat \texttip{x_{\it i}}{x_i}. as we did in Part A. by choosing the x axis to be parallel to the wire. Hint 1. Solve Part C Find \texttip{E_{\rm net}}{E_net}. the magnitude of the electric field at point P due to the total charge. Express the charge on the ith segment in terms of \texttip{x_{\it i}}{x_i}. Express your answer in terms of some or all of the variables \texttip{x_{\it i}}{x_i}. To set up the summation and find what to integrate. construct a mathematical expression for \texttip{\vec{E}_{\rm i}}{E_i_vec}.12/14/2015 Homework # 2 Ch 26 ANSWER: neither the x nor the y component the x component the y component both the x and y component Correct Since the point at which you want to calculate \texttip{\vec{E}}{E_vec} lies on a straight line extending from the wire.masteringphysics. Hint 2. follow the steps listed in the strategy. \texttip{L} {L}. \texttip{Q}{Q}. Find an expression for the charge of a small segment Since you will integrate with respect to a coordinate. ANSWER: \texttip{r_{\it i}}{r_i} = L+dx_{i} ANSWER: \texttip{E}{E} = \large{\frac{\Delta{Q}}{4 {\pi} {\epsilon}_{0} \left(L+dx_{i}\right)^{2}}} Hint 3. Assuming each charge segment has the same length \texttip{\Delta x}{Deltax}. \texttip{L}{L}. Express your answer in terms of \texttip{\Delta x}{Deltax} and some or all of the variables \texttip{L} {L}. what is the linear charge density \texttip{\lambda }{lambda} of the wire? Express your answer in terms of \texttip{L}{L} and \texttip{Q}{Q}. \texttip{Q}{Q}. E=\sum E_i.12/14/2015 Homework # 2 Ch 26 Express your answer in terms of \texttip{x_{\it i}}{x_i}. whose length is \texttip{L}{L}. go back and make the necessary substitutions (the previous hints may help you do that). and \texttip{d}{d}. Make sure that your expression for \texttip{E_{\rm i}}{E_i} contains only the following quantities. and \texttip{\Delta x}{Deltax}. has a total charge \texttip{Q}{Q} uniformly distributed along its length. If you assume that there is an infinite number of segments in the wire. besides the electrostatic constant 1/(4\pi\epsilon_0): \texttip{d}{d}. it is important to express the charge on a short segment of wire. Recall that the total charge on the wire is \texttip{Q}{Q} and the length of the wire is \texttip{L}{L}. that is. in terms of its length \texttip{\Delta x}{Deltax}. and \texttip{d}{d}.com/myct/assignmentPrintView?assignmentID=3971761 24/34 . you should have found an expression for the magnitude of the electric field \texttip{E_{\rm i}}{E_i} due to a small charge segment of length \texttip{\Delta x}{Deltax} located at \texttip{x_{\it i}}{x_i} along the x axis. where the sum contains as many terms as the number of charge segments. \texttip{\Delta x}{Deltax} will become infinitesimally small and https://session. \texttip{x_{\it i}}{x_i}. ANSWER: \texttip{\lambda }{lambda} = \large{\frac{Q}{L}} ANSWER: \texttip{\Delta Q}{DeltaQ} = \large{\frac{{\Delta{x}} Q}{L}} Hint 4. \texttip{\Delta Q}{DeltaQ}.masteringphysics. Find the linear charge density Assuming the wire. you can construct an algebraic expression for the magnitude of the net field using the principle of superposition. Find a mathematical expression for \texttip{\Delta Q}{DeltaQ} assuming the charge is uniformly distributed along the wire. How to make the transition from a sum to an integral At this point in the problem. If this is not the case. Hint 1. What should the limits of integration be? ANSWER: 0 to \infty 0 to \texttip{L}{L} \texttip{d}{d} to \texttip{L}{L} \texttip{d}{d} to L+d 0 to \texttip{d}{d} Hint 6. How to simplify the integral You can simplify the integral by changing the variable of integration from \texttip{x}{x} to x'=L+dx. ANSWER: \texttip{E_{\rm net}}{E_net} = \large{{\frac{1}{4{\pi}{\epsilon}_{0}}} {\frac{Q}{d\left(L+d\right)}}} Correct Assess Part D Imagine that distance \texttip{d}{d} is much greater than the length of the wire. what should the magnitude of the electric field at point P be in this case? Express your answer in terms of some or all of the variables \texttip{Q}{Q} and \texttip{d}{d}. At this point. the variable of integration will be the coordinate \texttip{x}{x}. The following formula will also be useful: \large{\int \frac{dx}{x^2} = \frac{1}{x}}.com/myct/assignmentPrintView?assignmentID=3971761 25/34 .masteringphysics. What is the magnitude of the electric field due to a point charge? https://session.12/14/2015 Homework # 2 Ch 26 can be replaced with \texttip{dx}{dx}. which will then become the integration variable \texttip{x} {x}. \texttip{L}{L}. Note that the remaining quantities \texttip{d}{d}. you can replace the sum with an integral and drop the subscript i from the variable \texttip{x_{\it i}}{x_i}. Hint 1. Envisioning the limiting case In the limiting case d\gg L. Don't forget to adjust the limits of integration appropriately when you make this change. the length of the wire is not relevant and the wire appears to be a point charge in the distance. and \texttip{Q}{Q} are constants: They are known quantities given in the problem statement. Intuitively. Hint 5. Find the limits of integration When you change the sum over charge segments into an integral. The variable \texttip{L}{L} should not appear in your answer. and the constants \texttip{\pi }{pi} and \texttip{\epsilon _{\rm 0}}{epsilon_0}. Problem 26. Not surprisingly. \texttip{L}{L} becomes negligible compared to \texttip{d}{d} and d + L \approx d. Let \texttip{r}{r} = 6. When d \gg L.4 What are the strength and direction of the electric field at the position indicated by the dot in the figure ? Part A Specify the strength of the electric field. Express your answer using two significant figures. the denominator reduces to (4\pi\epsilon_0 d^2). To see this. ANSWER: E = 4800 \rm N/C All attempts used; correct answer displayed https://session.com/myct/assignmentPrintView?assignmentID=3971761 26/34 . note that one way of writing the answer you obtained in the Solve step is: \large{E_{\rm net}=\frac{Q}{4\pi\epsilon_0 d (d+L)}}. the mathematical expression you derived for \texttip{E_{\rm net}} {E_net} reduces to that of the electric field due to a point charge.12/14/2015 Homework # 2 Ch 26 ANSWER: \texttip{E_{\rm net}}{E_net} = \large{{\frac{Q}{4{\pi}{\epsilon}_{0}}} {\frac{1}{d^{2}}}} Correct A short wire far away from point P creates a field very similar to the one created by a point charge. Thus.3 {\rm cm} .masteringphysics. as we expected. in the limiting case d \gg L. 9×104 \large{{\rm \frac{N}{C}}} Correct Part B What is the direction of the electric field \texttip{\vec{E}}{E_vec} at the midpoint between the two rings? ANSWER: To the left ring. ANSWER: \theta = 90 ^\circ below horizontal Correct Problem 26. 23 {\rm cm} apart. The left ring is charged to 18 {\rm nC} and the right ring is charged to +18 {\rm nC} . ANSWER: E = 1.com/myct/assignmentPrintView?assignmentID=3971761 27/34 .masteringphysics.12/14/2015 Homework # 2 Ch 26 Part B Specify the direction. Correct Part C What is the magnitude of the force \texttip{\vec{F}}{F_vec} on a 1. https://session. Express your answer using two significant figures.11 Two 10\rm cmdiameter charged rings face each other. To the right ring. Parallel to the plane of the rings.0 \rm nC charge placed at the midpoint? Express your answer to two significant figures and include the appropriate units. Part A What is the magnitude of the electric field \texttip{\vec{E}}{E_vec} at the midpoint between the two rings? Express your answer to two significant figures and include the appropriate units. 4×105 \large{{\rm \frac{N}{C}}} Correct Problem 26.22 An electron moving parallel to a uniform electric field increases its speed from 2. Correct Problem 26.6×107 {\rm m/s} over a distance of 1.0×107 {\rm m/s} to 3. To the right ring.com/myct/assignmentPrintView?assignmentID=3971761 28/34 . Parallel to the plane of the rings. https://session.0 \rm nC charge placed at the midpoint? ANSWER: To the left ring.8 {\rm cm} .12/14/2015 Homework # 2 Ch 26 ANSWER: F = 1. ANSWER: E = 1.masteringphysics. Both have linear charge density \lambda.36 The figure is a cross section of two infinite lines of charge that extend out of the page.9×10−5 {\rm N} Correct Part D What is the direction of the force \texttip{\vec{F}}{F_vec} on a 1. Part A What is the electric field strength? Express your answer to two significant figures and include the appropriate units. https://session.}{lambda.} \texttip{y}{y} . ANSWER: \large{{\frac{1}{4{\pi}{\epsilon}_{0}}} {\frac{y4{\lambda}}{y^{2}+\left(\frac{d}{2}\right)^{2}}}} Correct Problem 26.com/myct/assignmentPrintView?assignmentID=3971761 29/34 . \texttip{d}{d} and appropriate constants.40 The figure shows a thin rod of length L with total charge Q. Express your answer in terms of the variables \texttip{\lambda .masteringphysics.12/14/2015 Homework # 2 Ch 26 Part A Find an expression for the electric field strength E at height y above the midpoint between the lines. 44 Charge Q is uniformly distributed along a thin. Express your answer to two significant figures and include the appropriate units.0 {\rm cm} and \texttip{Q}{vQ} = 5. ANSWER: E = \large{{\frac{1}{4{\pi}{\epsilon}_{0}}} {\frac{Q}{r^{2}{\frac{L^{2}}{4}}}}} Correct Part B Verify that your expression has the expected behavior if r \gg L. \texttip{r}{r}.0 {\rm nC} .7×104 \large{{\rm \frac{N}{C}}} Correct Problem 26.masteringphysics. https://session. Express your answer in terms of the variables \texttip{L}{L}. Express your answer in terms of variables Q.3 {\rm cm} if L = 5. and appropriate constants. ANSWER: E = \large{{\frac{1}{4{\pi}{\epsilon}_{0}}} {\frac{Q}{r^{2}}}} Correct Part C Evaluate E at \texttip{r}{vr} = 4. flexible rod of length L. \texttip{Q}{Q}.com/myct/assignmentPrintView?assignmentID=3971761 30/34 .12/14/2015 Homework # 2 Ch 26 Part A Find an expression for the electric field strength on the axis of the rod at distance r from the center. \varepsilon _0. ANSWER: E = 3. The rod is then bent into the semicircle shown in the figure . r and constants \pi. 0 \times \;\;\;\;10^{6}\;{\rm m}/{\rm s} from the positive plate of the parallelplate capacitor shown in the figure . The electron lands 4. Express your answer with the appropriate units. https://session.50 An electron is launched at a 45^\circ angle and a speed of 5. Express your answer in terms of the variables Q. Hint: A small piece of arc length \Delta s spans a small angle \Delta \theta = \Delta s / R. unit vectors \texttip{\hat{i}}{i_unit}. \texttip{\hat{j}} {j_unit}.com/myct/assignmentPrintView?assignmentID=3971761 31/34 .12/14/2015 Homework # 2 Ch 26 Part A Find an expression for the electric field \vec E at the center of the semicircle. ANSWER: E = 1.0×105 \large{{\rm \frac{N}{C}}} Correct Problem 26. where R is the radius. ANSWER: \vec E = \large{{\frac{Q}{2\left(L^{2}\right)\left({\epsilon}_{0}\right)}}\hat{i}} Correct Part B Evaluate the field strength if \texttip{L}{vL} = 14 {\rm cm} and \texttip{Q}{vQ} = 36 {\rm nC} . and appropriate constants.masteringphysics.0 cm away. L. 00 {\rm cm} Correct Problem 26. Express your answer to two significant figures and include the appropriate units. ANSWER: 3550 \large{{\rm \frac{N}{C}}} Correct Part B What is the smallest possible spacing between the plates? Express your answer with the appropriate units.12/14/2015 Homework # 2 Ch 26 Part A What is the electric field strength inside the capacitor? Express your answer with the appropriate units. the electron orbits the proton in a circular orbit of radius 0. ANSWER: f = 6.55 In a classical model of the hydrogen atom.com/myct/assignmentPrintView?assignmentID=3971761 32/34 . Part A What is the orbital frequency? The proton is so much more massive than the electron that you can assume the proton is at rest.masteringphysics. ANSWER: 1.053 \rm nm.6×1015 {\rm Hz} https://session. where \alpha is called the polarizability of the molecule. ANSWER: F_{{\rm{ion\; on\; dipole}}} = \large{\left(\frac{1}{4{\pi}{\epsilon}_{0}}\right)^{2}{\frac{2q^{2}{\alpha}}{r^{5}}}} Correct Part C What is the direction of this force? ANSWER: Toward ion Away from ion https://session. Part A What are the units of \alpha? ANSWER: \rm C^2/kg \rm C^2s^2 m/kg \rm C^2s^2 m^2/kg \rm C^2s^2/kg Correct Part B An ion with charge q is distance r from a molecule with polarizability \alpha. \texttip{\alpha }{alpha} and appropriate constants.masteringphysics. \vec p = \alpha \vec E. \texttip{r}{r}. Express your answer in terms of the variables \texttip{q}{q}. That is.12/14/2015 Homework # 2 Ch 26 Correct Problem 26. The dipole moment of an induced dipole is directly proportional to the electric field. Find an expression for the magnitude of the force \vec F_{{\rm{ion\; on\; dipole}}}. A bigger field stretches the molecule farther and causes a larger dipole moment.57 An electric field can induce an electric dipole in a neutral atom or molecule by pushing the positive and negative charge in opposite directions.com/myct/assignmentPrintView?assignmentID=3971761 33/34 . masteringphysics. You received 132.com/myct/assignmentPrintView?assignmentID=3971761 34/34 .4%. https://session.17 out of a possible total of 140 points.12/14/2015 Homework # 2 Ch 26 Correct Score Summary: Your score on this assignment is 94.