Capacitors Solved Sheet for IIT



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Q.1 In four options below, all the four circuits are arranged in order of equivalent capacitance between points A & B. Select the correct order. Assume all capacitors are of equal capacitance. (1) (2) (3) (4) (A) C 1 > C 2 > C 3 > C 4 (B*) C 1 > C 3 > C 2 > C 4 (C) C 1 < C 2 < C 3 < C 4 (D) C 1 < C 3 < C 2 < C 4 Q.2 Rows of capacitors containing 1, 2, 4, 8, .........∞ capacitors each of capacitance 2µF are connected in parallel as shown in figure. The potential difference across AB is 10 Volts. (A*) Total capacitance across AB is 4µF. (B) Charge on each capacitor will be same. (C*) Charge on the capacitor in the first row is more than that on any other capacitor. (D) Total energy stored in capacitors is 50 µJ. [Sol: C eq = C + 2 C + 4 C + 8 C + ................................. = C , _ ¸ ¸ + + + + . .......... .......... 8 1 4 1 2 1 1 = 4µF Capacitance of 1st row is maximum, hence charge stored will be also maximum. (A, C) ] Q.3 In the circuit, both parallel plate capacitors are identical. Column I indicates action done on capacitor 1 and Column I I indicates effect on capacitor 2. Select correct alternative. Column I Column I I (A) Plates are moved further apart. (P) No effect (B) Plate area increased (Q) Potential difference increases (C) Left plate is earthed (R) Amount of charge on right plate decreases (D) It's Plates are short circuited (S) Amount of charge on left plate increases [Ans. (A) R, (B) Q,S, (C) P, (D) Q, S ] [Sol: 2 1 C q C q + = V V V 1 + V 2 = V V 2 = V – V 1 when plates are moved apart ⇒ C 1 ⇒ ↓ ⇒ V 1 ↑ ⇒ V 2 ↓ C eq = 2 1 2 1 C C C C + ↓ ⇒ q = C eq V ↓ right plate has –ve charge ⇒ when –q ¯ ⇒ q increases ] Q.4 Three identical large conducting plates P 1 , P 2 & P 3 each having area of either face of plate eauql to A are placed parallel to each other at very short distance d. A charge –2Q is given to plate P 1 , no charge is given to plate P 2 & +4Q is given to plate P 3 . Now plate P 2 & P 3 are connected by thin conducting wire. Then find the total amount of heat (in Joules) produced. (here d A 0 ∈ = 0.1µ coul/volt, Q = 2 × 10 –3 coul.) [Ans: 180 J] [Sol: V i = 2 1 ( ) C Q 2 3 × 2 + U outside the capacitor when they are connected the charge distribution will be as: U f = ( ) C Q 2 3 2 1 ∴ Heat produced = ( ) C Q 2 3 2 1 = 180 Joule. ] Q.5 Two identical capacitors are connected in series as shown in the figure. A dielectric slab (κ > 1) is placed between the plates of the capacitor B and the battery remains connected. Which of the following statement(s) is/are correct following the insertion of the dielectric? (A*) The charge supplied by the battery increases. (B*) The capacitance of the system increases. (C) The electric field in the capacitor B increases. (D) The electrostatic potential energy decreases. Q.6 Two large conducting plates having surface charge densities +σ and –σ, respectively, are fixed ‘d’ distance apart. A small test charge q of mass m is attached to two identical springs as shown in the adjacent figure. The charge q is now released from rest with springs in natural length. Then q will [neglect gravity] (A*) perform SHM with angular frequency m k 2 (B*) perform SHM with amplitude 0 k 2 q ∈ σ (C) not perform SHM, but will have a periodic motion (D) remain stationary [Sol: A = deformation in equilibrium state. 2KA = 0 ∈ σ ·q ∴ A = 0 2 ∈ K q σ (B) Springs are connected in parallel K eq = 2K angular frequency = m K 2 (A) Q.7 The circuit was in the shown state from a long time. Now the switch S is closed. Find the charge (in µC) that flows through the switch S (from A to B) after closing it. [Ans: 50 µC] [Sol: C AB = 4 2 4 2 + × = 6 8 = 3 4 q = CV = 3 4 × 50 = 3 200 When switch 'S' is closed, the circuit becomes as 4 2 1 2 q q · ∴ q 1 = 2q 2 4 2 1 2 q q + = 50 4 2 1 q = 50 ∴ q 1 = 2 200 = 100 q 2 = 50 charge flown = q 1 – q 2 = 100 – 50 = 50 µC ] For Problem 8 to 10 Figure shows a circuit containing a battery and three parallel plate capacitors with identical plate separation (filled with air). The capacitors lie along x-axis and a graph of the electric potential V along that axis is shown. Q.8 The charge on capacitor C 1 , C 2 & C 3 are q 1 , q 2 & q 3 respectively then (A) q 1 > q 2 > q 3 (B) q 1 < q 2 > q 3 (C*) q 1 = q 2 = q 3 (D) q 1 > q 2 < q 3 [Sol: charge would be same on capacitors connected in series, q 1 = q 2 = q 3 ] Q.9 If the area of the plates of capacitors are A 1 , A 2 & A 3 respectively then (A) A 1 < A 2 > A 3 (B*) A 1 > A 3 > A 2 (C) A 1 = A 2 = A 3 (D) A 1 > A 2 > A 3 [Sol: The graph shows that E 1 (electric field in space between plates of capacitor 1) < E 3 < E 2 & hence, V 1 < V 3 < V 2 as C 1 V 1 = C 2 V 2 = C 3 V 3 (a 1 = a 2 = a 3 ) giving us A 1 V 1 = A 2 V 2 = A 3 V 3 we get, A 2 < A 3 < A 1 ] Q.10 If the magnitude of electric field between their plates are E 1 , E 2 & E 3 respectively then (A) E 1 > E 2 > E 3 (B*) E 1 < E 2 > E 3 (C) E 1 > E 2 < E 3 (D) E 1 < E 2 < E 3 [Sol: Electric field α slope of V/x graph & hence, E 1 < E 3 < E 2 or E 1 < E 2 > E 3 ] Q.11 An isolated metallic object is charged in vacuum to potential v 0 , its electrostatic energy being W 0 . It is then disconnected from the source of potential, its charge being left unchanged and is immersed in a large volume of dielectric, with dielectric constant k. The electrostatic energy will be (A) kW 0 (B*) k W 0 (C) k W 2 0 (D) W 0 [Sol: Energy of the metallic object equals work done in charging the object against electrostatic forces. For the object completely surrounded in a medium of dielectric constant 'K', force or electric field everywhere gets reduced by the factor 'K' & hence, changed energy would be K W 0 . ] Q.12 The plates of a parallel plate capacitor are charged upto 100 volt. A 2 mm thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6 mm. The dielectric constant of the plate is (A*) 5 (B) 1.25 (C) 4 (D) 2.5 [Sol: C = d A 0 ∈ q = C × 100 = d A 0 ∈ × 100 New capacitance C 1 = ( ) , _ ¸ ¸ − + + ∈ 1 1 6 . 1 0 k t d A where t = 2 mm 100 = 1 C q = 1 0 100 C d A 1 ] 1 ¸ × ∈ Solving K = 5 ] Q.13 The figure shows a battery with emf 15 V in a circuit with R 1 = 30 Ω, R 2 = 10 Ω, R 3 = 20 Ω and capacitance (C) = 10µF. The switch S is initially in the open position and is then closed at time t = 0. What will be the final steady state charge on capacitor (A*) 75µC (B) 50µC (C) 10µC (D) none of these [Sol: i = 60 15 = 4 1 amp. V AB = 4 1 × 30 = 7.5 volt q = eV = 10 × 7.5 = 75 µe (A) ] Q.14 A point charge is placed at x in front of an earthed metal sheet y. P and Q are two points between x and y as shown. If the electric field strength at P and Q be E P and E Q . Which one of the following statements is correct (A) E P = E Q (B) E P > E Q (C*) E P < E Q (D) E P = E Q = 0 [Sol: E Q > E P (C) ] Question No. 15 & 16 (2 questions) Two identical conducting very large plate P 1 & P 2 having charges +4Q & +6Q are placed very closed to each other at separation 'd'. The plate area of either face of the plate is A. Then Q.15 Potential difference between plates P 1 & P 2 is (A) 2 1 P P V V − = 0 A Qd ∈ (B*) 0 P P A Qd V V 2 1 ∈ − · − (C) 0 P P A Qd 5 V V 2 1 ∈ · − (D) 0 P P A Qd 5 V V 2 1 ∈ − · − [Sol: Potential difference between the plates 2 1 P P V V − = , _ ¸ ¸∈ − d A Q 0 2 1 P P V V − = A Qd 0 ∈ − ] Q.16 If plate P 1 & plate P 2 are connected by thin wire then amount of heat produced is (A) d A Q 0 2 ∈ (B) d A Q 5 0 2 ∈ (C) d A Q 2 0 2 ∈ (D*) None of these [Sol: Heat produce ∆H = U i – U f ∆H = ½ d A Q 0 2 ∈ − – 0 Q.17 In the diagram shown, in steady state (A) Charge on 1 µf capacitor is 1 µC (B*) Charge on 2 µf capacitor is 2 µC (C*) Charge on 3 µf is zero (D) Potential difference across 2µf capacitor is 2 volt Q.18 Which of the following is sufficient condition for finding the electric flux E Φ through a closed surface? (A) If the magnitude of E r is known everywhere on the surface (B*) If the total charge inside the surface is specified (C) If the total charge outside the surface is specified (D) Only if the location of each point charge inside the surface is specified Q.19 The figure shows a conducting sphere 'A' of radius 'a' which is surrounded by a neutral conducting spherical shell B of radius 'b'(>a). Initially switches S 1 , S 2 and S 3 are open and sphere 'A' carries a charge Q. First the switch 'S 1 ' is closed to connect the shell B with the ground and then opened. Now the switch 'S 2 ' is closed so that the sphere 'A' is grounded and then S 2 is opened. Finally, the switch 'S 3 ' is closed to connect the spheres together. Find the heat (in Joule) which is produced after closing the switch S 3 . [Consider b = 4 cm, a = 2 cm and Q = 8 µC] [Ans. 1.8] [Sol. When outer surface is grounded charge '–Q' resides on the inner surface of sphere 'B' Now sphere A is connected to earth potential on its surface becomes zero. Let the charge on the surface A becomes q a kq – b kQ = 0 ⇒ q = Q b a Consider the figure. In this position energy stored E 1 = 2 0 Q b a a 8 1 1 ] 1 ¸ πε + b 8 Q 0 2 πε + ) Q ( Q b a b 4 1 0 − 1 ] 1 ¸ πε when 'S 3 ' is closed, total charge will appear on the outer surface of shell 'B'. In this position energy stored E 2 = 2 2 0 Q 1 b a b 8 1 , _ ¸ ¸ − πε Heat produced = E 1 – E 2 = 3 0 2 b 8 ) a b ( a Q πε − = 1.8 ] Q.20 In the figure shown, C 1 = 11µF and C 2 = 5 µF, then at steady state: (A*) the potential difference across C 1 is 5V (B) the potential difference across C 2 is 2V (C*) the potential difference between points a and b is –4V (D*) the potential difference between the terminals of 15 V vattery is 9V [Sol: At steady state I(3) + I(2) = 15 I = 3 KVL C→ D → E → a → b → c V/C – I(3) + 11 q – 7 + 5 q = V/C 11 q + 5 q = 7 + 3 × 3 = 16 q = 55 µC KVL = a → b V a – 7 + 5 q = VV b ⇒ V a – V b = 7 – 5 q = 7 – 5 55 = 5 55 = –4V P.d. across C 1 11 q = 11 55 = 5V P.d. across C 2 5 q = 11V p.d. across terminal = 15 – I(2) = 15 – 3 × 2 = 9V ] Q.21 On a parallel plate air capacitor of capacitance C 0 having plate separation d following steps are performed in the order as given in column I. (a) Capacitor is charged by connecting it across a battery of EMF V 0 . (b) Dielectric of dielectric constant k and thickness d is inserted (c) Capacitor is disconnected from battery (d) Separation between plates is doubled Column I Column I I (Steps performed Final value of Quantity (Symbols have usual meaning) in order) (A) (a)→(d)→(c)→(b) (P) Q = 2 V C 0 0 (B) (d)→(a)→(c)→(b) (Q) Q = 1 k V kC 0 0 + (C) (b)→(a)→(c)→(d) (R) C = 1 k kC 0 + (D) (a)→(b)→(d)→(c) (S) V = k 2 ) 1 k ( V 0 + [Ans.(A) P, R, S; (B) P, R, S; (C) R; (D) Q, R] [Sol. Q C V (A) C 0 V 0 C 0 V 0 2 V C 0 0 2 C 0 V 0 2 V C 0 0 2 C 0 V 0 2 V C 0 0 1 K KC 0 + K 2 ) 1 K ( V 0 + (B) 0 2 C 0 0 V 0 2 C 0 2 C 0 V 0 V 0 2 C 0 2 C 0 V 0 2 V C 0 0 1 K KC 0 + K 2 ) 1 K ( V 0 + C) 0 d KA 0 ε 0 KC 0 V 0 KC 0 V 0 KC 0 V 0 KC 0 V 0 KC 0 V 0 1 K KC 0 + (K+1)V 0 (D) C 0 V 0 C 0 V 0 KC 0 V 0 KC 0 V 0 1 K V KC 0 0 + 1 K KC 0 + V 0 1 K V KC 0 0 + 1 K KC 0 + V 0 ]
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