Page 1 31.1 CHAPTER – 31 CAPACITOR 1. Given that Number of electron = 1 × 10 12 Net charge Q = 1 × 10 12 × 1.6 × 10 –19 = 1.6 × 10 –7 C ? The net potential difference = 10 L. ? Capacitance – C = v q = 10 10 6 . 1 7 ? ? = 1.6 × 10 –8 F. 2. A = ?r 2 = 25 ?cm 2 d = 0.1 cm c = d A 0 ? = 1 . 0 14 . 3 25 10 854 . 8 12 ? ? ? ? = 6.95 × 10 –5 ?F. 3. Let the radius of the disc = R ?Area = ?R 2 C = 1 ? D = 1 mm = 10 –3 m ? C = d A 0 ? ? 1 = 3 2 12 10 r 10 85 . 8 ? ? ? ? ? ? r 2 = ? ? ? ? 85 . 8 10 10 12 3 = 784 . 27 10 9 = 5998.5 m = 6 Km 4. A = 25 cm 2 = 2.5 × 10 –3 cm 2 d = 1 mm = 0.01 m V = 6V Q = ? C = d A 0 ? = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? Q = CV = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? × 6 = 1.32810 × 10 –10 C W = Q × V = 1.32810 × 10 –10 × 6 = 8 × 10 –10 J. 5. Plate area A = 25 cm 2 = 2.5 × 10 –3 m Separation d = 2 mm = 2 × 10 –3 m Potential v = 12 v (a) We know C = d A 0 ? = 3 3 12 10 2 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 11.06 ×10 –12 F C = v q ? 11.06 ×10 –12 = 12 q ? q 1 = 1.32 × 10 –10 C. (b) Then d = decreased to 1 mm ? d = 1 mm = 1 × 10 –3 m C = d A 0 ? = v q = 3 3 12 10 1 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 12 2 ? q 2 = 8.85 × 2.5 × 12 × 10 –12 = 2.65 × 10 –10 C. ? The extra charge given to plate = (2.65 – 1.32) × 10 –10 = 1.33 × 10 –10 C. 6. C 1 = 2 ?F, C 2 = 4 ?F , C 3 = 6 ?F V = 12 V cq = C 1 + C 2 + C 3 = 2 + 4 + 6 = 12 ?F = 12 × 10 –6 F q 1 = 12 × 2 = 24 ?C, q 2 = 12 × 4 = 48 ?C, q 3 = 12 × 6 = 72 ?C 5 cm 0.1 cm 1 mm C 1 V C 2 C 3 Page 2 31.1 CHAPTER – 31 CAPACITOR 1. Given that Number of electron = 1 × 10 12 Net charge Q = 1 × 10 12 × 1.6 × 10 –19 = 1.6 × 10 –7 C ? The net potential difference = 10 L. ? Capacitance – C = v q = 10 10 6 . 1 7 ? ? = 1.6 × 10 –8 F. 2. A = ?r 2 = 25 ?cm 2 d = 0.1 cm c = d A 0 ? = 1 . 0 14 . 3 25 10 854 . 8 12 ? ? ? ? = 6.95 × 10 –5 ?F. 3. Let the radius of the disc = R ?Area = ?R 2 C = 1 ? D = 1 mm = 10 –3 m ? C = d A 0 ? ? 1 = 3 2 12 10 r 10 85 . 8 ? ? ? ? ? ? r 2 = ? ? ? ? 85 . 8 10 10 12 3 = 784 . 27 10 9 = 5998.5 m = 6 Km 4. A = 25 cm 2 = 2.5 × 10 –3 cm 2 d = 1 mm = 0.01 m V = 6V Q = ? C = d A 0 ? = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? Q = CV = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? × 6 = 1.32810 × 10 –10 C W = Q × V = 1.32810 × 10 –10 × 6 = 8 × 10 –10 J. 5. Plate area A = 25 cm 2 = 2.5 × 10 –3 m Separation d = 2 mm = 2 × 10 –3 m Potential v = 12 v (a) We know C = d A 0 ? = 3 3 12 10 2 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 11.06 ×10 –12 F C = v q ? 11.06 ×10 –12 = 12 q ? q 1 = 1.32 × 10 –10 C. (b) Then d = decreased to 1 mm ? d = 1 mm = 1 × 10 –3 m C = d A 0 ? = v q = 3 3 12 10 1 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 12 2 ? q 2 = 8.85 × 2.5 × 12 × 10 –12 = 2.65 × 10 –10 C. ? The extra charge given to plate = (2.65 – 1.32) × 10 –10 = 1.33 × 10 –10 C. 6. C 1 = 2 ?F, C 2 = 4 ?F , C 3 = 6 ?F V = 12 V cq = C 1 + C 2 + C 3 = 2 + 4 + 6 = 12 ?F = 12 × 10 –6 F q 1 = 12 × 2 = 24 ?C, q 2 = 12 × 4 = 48 ?C, q 3 = 12 × 6 = 72 ?C 5 cm 0.1 cm 1 mm C 1 V C 2 C 3 Capacitor 31.2 7. ? The equivalent capacity. C = 2 1 3 1 3 2 3 2 1 C C C C C C C C C ? ? = 30 20 40 20 40 30 40 30 20 ? ? ? ? ? ? ? = 2600 24000 = 9.23 ?F (a) Let Equivalent charge at the capacitor = q C = V q ? q = C × V = 9.23 × 12 = 110 ?C on each. As this is a series combination, the charge on each capacitor is same as the equivalent charge which is 110 ?C. ? (b) Let the work done by the battery = W ? V = q W ? W = Vq = 110 × 12 × 10 –6 = 1.33 × 10 –3 J. 8. C 1 = 8 ?F, C 2 = 4 ?F , C 3 = 4 ?F Ceq = 3 2 1 1 3 2 C C C C ) C C ( ? ? ? ? = 16 8 8 ? = 4 ?F Since B & C are parallel & are in series with A So, q 1 = 8 × 6 = 48 ?C q 2 = 4 × 6 = 24 ?C q3 = 4 × 6 = 24 ?C 9. (a) ? C 1 , C 1 are series & C 2 , C 2 are series as the V is same at p & q. So no current pass through p & q. 2 1 C 1 C 1 C 1 ? ? ? 2 1 C C 1 1 C 1 ? ? C p = 2 C 1 = 2 4 = 2 ?F And C q = 2 C 2 = 2 6 = 3 ?F ? C = C p + C q = 2 + 3 = 5 ?F (b) C 1 = 4 ?F, C 2 = 6 ?F, In case of p & q, q = 0 ? C p = 2 C 1 = 2 4 = 2 ?F C q = 2 C 2 = 2 6 = 3 ?F & C ? = 2 + 3 = 5 ?F C & C ? = 5 ?F ? The equation of capacitor C = C ? + C ?? = 5 + 5 = 10 ?F 12 V 8 ?F ? A 4 ?F ? B C 4 ?F ? A B C 1 C 2 C 1 C 2 C 1 = 4 C 2 = 6 A p B C 1 C 1 C 1 C 1 q R S C 2 C 2 C 2 C 2 30 ?F ? V = 12 V 40 ?F ? 20 ?F ? Page 3 31.1 CHAPTER – 31 CAPACITOR 1. Given that Number of electron = 1 × 10 12 Net charge Q = 1 × 10 12 × 1.6 × 10 –19 = 1.6 × 10 –7 C ? The net potential difference = 10 L. ? Capacitance – C = v q = 10 10 6 . 1 7 ? ? = 1.6 × 10 –8 F. 2. A = ?r 2 = 25 ?cm 2 d = 0.1 cm c = d A 0 ? = 1 . 0 14 . 3 25 10 854 . 8 12 ? ? ? ? = 6.95 × 10 –5 ?F. 3. Let the radius of the disc = R ?Area = ?R 2 C = 1 ? D = 1 mm = 10 –3 m ? C = d A 0 ? ? 1 = 3 2 12 10 r 10 85 . 8 ? ? ? ? ? ? r 2 = ? ? ? ? 85 . 8 10 10 12 3 = 784 . 27 10 9 = 5998.5 m = 6 Km 4. A = 25 cm 2 = 2.5 × 10 –3 cm 2 d = 1 mm = 0.01 m V = 6V Q = ? C = d A 0 ? = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? Q = CV = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? × 6 = 1.32810 × 10 –10 C W = Q × V = 1.32810 × 10 –10 × 6 = 8 × 10 –10 J. 5. Plate area A = 25 cm 2 = 2.5 × 10 –3 m Separation d = 2 mm = 2 × 10 –3 m Potential v = 12 v (a) We know C = d A 0 ? = 3 3 12 10 2 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 11.06 ×10 –12 F C = v q ? 11.06 ×10 –12 = 12 q ? q 1 = 1.32 × 10 –10 C. (b) Then d = decreased to 1 mm ? d = 1 mm = 1 × 10 –3 m C = d A 0 ? = v q = 3 3 12 10 1 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 12 2 ? q 2 = 8.85 × 2.5 × 12 × 10 –12 = 2.65 × 10 –10 C. ? The extra charge given to plate = (2.65 – 1.32) × 10 –10 = 1.33 × 10 –10 C. 6. C 1 = 2 ?F, C 2 = 4 ?F , C 3 = 6 ?F V = 12 V cq = C 1 + C 2 + C 3 = 2 + 4 + 6 = 12 ?F = 12 × 10 –6 F q 1 = 12 × 2 = 24 ?C, q 2 = 12 × 4 = 48 ?C, q 3 = 12 × 6 = 72 ?C 5 cm 0.1 cm 1 mm C 1 V C 2 C 3 Capacitor 31.2 7. ? The equivalent capacity. C = 2 1 3 1 3 2 3 2 1 C C C C C C C C C ? ? = 30 20 40 20 40 30 40 30 20 ? ? ? ? ? ? ? = 2600 24000 = 9.23 ?F (a) Let Equivalent charge at the capacitor = q C = V q ? q = C × V = 9.23 × 12 = 110 ?C on each. As this is a series combination, the charge on each capacitor is same as the equivalent charge which is 110 ?C. ? (b) Let the work done by the battery = W ? V = q W ? W = Vq = 110 × 12 × 10 –6 = 1.33 × 10 –3 J. 8. C 1 = 8 ?F, C 2 = 4 ?F , C 3 = 4 ?F Ceq = 3 2 1 1 3 2 C C C C ) C C ( ? ? ? ? = 16 8 8 ? = 4 ?F Since B & C are parallel & are in series with A So, q 1 = 8 × 6 = 48 ?C q 2 = 4 × 6 = 24 ?C q3 = 4 × 6 = 24 ?C 9. (a) ? C 1 , C 1 are series & C 2 , C 2 are series as the V is same at p & q. So no current pass through p & q. 2 1 C 1 C 1 C 1 ? ? ? 2 1 C C 1 1 C 1 ? ? C p = 2 C 1 = 2 4 = 2 ?F And C q = 2 C 2 = 2 6 = 3 ?F ? C = C p + C q = 2 + 3 = 5 ?F (b) C 1 = 4 ?F, C 2 = 6 ?F, In case of p & q, q = 0 ? C p = 2 C 1 = 2 4 = 2 ?F C q = 2 C 2 = 2 6 = 3 ?F & C ? = 2 + 3 = 5 ?F C & C ? = 5 ?F ? The equation of capacitor C = C ? + C ?? = 5 + 5 = 10 ?F 12 V 8 ?F ? A 4 ?F ? B C 4 ?F ? A B C 1 C 2 C 1 C 2 C 1 = 4 C 2 = 6 A p B C 1 C 1 C 1 C 1 q R S C 2 C 2 C 2 C 2 30 ?F ? V = 12 V 40 ?F ? 20 ?F ? Capacitor 31.3 10. V = 10 v Ceq = C 1 +C 2 [ ? They are parallel] = 5 + 6 = 11 ?F q = CV = 11 × 10 110 ?C ? 11. The capacitance of the outer sphere = 2.2 ?F C = 2.2 ?F Potential, V = 10 v Let the charge given to individual cylinder = q. C = V q ? q = CV = 2.2 × 10 = 22 ?F ? The total charge given to the inner cylinder = 22 + 22 = 44 ?F 12. C = V q , Now V = R Kq So, C 1 = ? ? 1 R / Kq q = K R 1 = 4 ?? 0 R 1 Similarly c 2 = 4 ?? 0 R 2 The combination is necessarily parallel. Hence Ceq = 4 ?? 0 R 1 +4 ?? 0 R 2 = 4 ?? 0 (R 1 + R 2 ) 13. ?C = 2 ?F ? In this system the capacitance are arranged in series. Then the capacitance is parallel to each other. (a) ? The equation of capacitance in one row C = 3 C (b) and three capacitance of capacity 3 C are connected in parallel ? The equation of capacitance C = 3 C 3 C 3 C ? ? = C = 2 ?F As the volt capacitance on each row are same and the individual is = ce tan capaci of . No Total = 3 60 = 20 V 14. Let there are ‘x’ no of capacitors in series ie in a row So, x × 50 = 200 ? x = 4 capacitors. Effective capacitance in a row = 4 10 Now, let there are ‘y’ such rows, So, 4 10 × y = 10 ? y = 4 capacitor. So, the combinations of four rows each of 4 capacitors. 6 ?F ? B 5 ?F ? 10 V A A5 B6 C A B C C C C C C C C Page 4 31.1 CHAPTER – 31 CAPACITOR 1. Given that Number of electron = 1 × 10 12 Net charge Q = 1 × 10 12 × 1.6 × 10 –19 = 1.6 × 10 –7 C ? The net potential difference = 10 L. ? Capacitance – C = v q = 10 10 6 . 1 7 ? ? = 1.6 × 10 –8 F. 2. A = ?r 2 = 25 ?cm 2 d = 0.1 cm c = d A 0 ? = 1 . 0 14 . 3 25 10 854 . 8 12 ? ? ? ? = 6.95 × 10 –5 ?F. 3. Let the radius of the disc = R ?Area = ?R 2 C = 1 ? D = 1 mm = 10 –3 m ? C = d A 0 ? ? 1 = 3 2 12 10 r 10 85 . 8 ? ? ? ? ? ? r 2 = ? ? ? ? 85 . 8 10 10 12 3 = 784 . 27 10 9 = 5998.5 m = 6 Km 4. A = 25 cm 2 = 2.5 × 10 –3 cm 2 d = 1 mm = 0.01 m V = 6V Q = ? C = d A 0 ? = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? Q = CV = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? × 6 = 1.32810 × 10 –10 C W = Q × V = 1.32810 × 10 –10 × 6 = 8 × 10 –10 J. 5. Plate area A = 25 cm 2 = 2.5 × 10 –3 m Separation d = 2 mm = 2 × 10 –3 m Potential v = 12 v (a) We know C = d A 0 ? = 3 3 12 10 2 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 11.06 ×10 –12 F C = v q ? 11.06 ×10 –12 = 12 q ? q 1 = 1.32 × 10 –10 C. (b) Then d = decreased to 1 mm ? d = 1 mm = 1 × 10 –3 m C = d A 0 ? = v q = 3 3 12 10 1 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 12 2 ? q 2 = 8.85 × 2.5 × 12 × 10 –12 = 2.65 × 10 –10 C. ? The extra charge given to plate = (2.65 – 1.32) × 10 –10 = 1.33 × 10 –10 C. 6. C 1 = 2 ?F, C 2 = 4 ?F , C 3 = 6 ?F V = 12 V cq = C 1 + C 2 + C 3 = 2 + 4 + 6 = 12 ?F = 12 × 10 –6 F q 1 = 12 × 2 = 24 ?C, q 2 = 12 × 4 = 48 ?C, q 3 = 12 × 6 = 72 ?C 5 cm 0.1 cm 1 mm C 1 V C 2 C 3 Capacitor 31.2 7. ? The equivalent capacity. C = 2 1 3 1 3 2 3 2 1 C C C C C C C C C ? ? = 30 20 40 20 40 30 40 30 20 ? ? ? ? ? ? ? = 2600 24000 = 9.23 ?F (a) Let Equivalent charge at the capacitor = q C = V q ? q = C × V = 9.23 × 12 = 110 ?C on each. As this is a series combination, the charge on each capacitor is same as the equivalent charge which is 110 ?C. ? (b) Let the work done by the battery = W ? V = q W ? W = Vq = 110 × 12 × 10 –6 = 1.33 × 10 –3 J. 8. C 1 = 8 ?F, C 2 = 4 ?F , C 3 = 4 ?F Ceq = 3 2 1 1 3 2 C C C C ) C C ( ? ? ? ? = 16 8 8 ? = 4 ?F Since B & C are parallel & are in series with A So, q 1 = 8 × 6 = 48 ?C q 2 = 4 × 6 = 24 ?C q3 = 4 × 6 = 24 ?C 9. (a) ? C 1 , C 1 are series & C 2 , C 2 are series as the V is same at p & q. So no current pass through p & q. 2 1 C 1 C 1 C 1 ? ? ? 2 1 C C 1 1 C 1 ? ? C p = 2 C 1 = 2 4 = 2 ?F And C q = 2 C 2 = 2 6 = 3 ?F ? C = C p + C q = 2 + 3 = 5 ?F (b) C 1 = 4 ?F, C 2 = 6 ?F, In case of p & q, q = 0 ? C p = 2 C 1 = 2 4 = 2 ?F C q = 2 C 2 = 2 6 = 3 ?F & C ? = 2 + 3 = 5 ?F C & C ? = 5 ?F ? The equation of capacitor C = C ? + C ?? = 5 + 5 = 10 ?F 12 V 8 ?F ? A 4 ?F ? B C 4 ?F ? A B C 1 C 2 C 1 C 2 C 1 = 4 C 2 = 6 A p B C 1 C 1 C 1 C 1 q R S C 2 C 2 C 2 C 2 30 ?F ? V = 12 V 40 ?F ? 20 ?F ? Capacitor 31.3 10. V = 10 v Ceq = C 1 +C 2 [ ? They are parallel] = 5 + 6 = 11 ?F q = CV = 11 × 10 110 ?C ? 11. The capacitance of the outer sphere = 2.2 ?F C = 2.2 ?F Potential, V = 10 v Let the charge given to individual cylinder = q. C = V q ? q = CV = 2.2 × 10 = 22 ?F ? The total charge given to the inner cylinder = 22 + 22 = 44 ?F 12. C = V q , Now V = R Kq So, C 1 = ? ? 1 R / Kq q = K R 1 = 4 ?? 0 R 1 Similarly c 2 = 4 ?? 0 R 2 The combination is necessarily parallel. Hence Ceq = 4 ?? 0 R 1 +4 ?? 0 R 2 = 4 ?? 0 (R 1 + R 2 ) 13. ?C = 2 ?F ? In this system the capacitance are arranged in series. Then the capacitance is parallel to each other. (a) ? The equation of capacitance in one row C = 3 C (b) and three capacitance of capacity 3 C are connected in parallel ? The equation of capacitance C = 3 C 3 C 3 C ? ? = C = 2 ?F As the volt capacitance on each row are same and the individual is = ce tan capaci of . No Total = 3 60 = 20 V 14. Let there are ‘x’ no of capacitors in series ie in a row So, x × 50 = 200 ? x = 4 capacitors. Effective capacitance in a row = 4 10 Now, let there are ‘y’ such rows, So, 4 10 × y = 10 ? y = 4 capacitor. So, the combinations of four rows each of 4 capacitors. 6 ?F ? B 5 ?F ? 10 V A A5 B6 C A B C C C C C C C C Capacitor 31.4 15. (a) Capacitor = 8 4 8 4 ? ? = 3 8 ?? and 3 6 3 6 ? ? = 2 ?F (i) The charge on the capacitance 3 8 ?F ? Q = 3 8 × 50 = 3 400 ? ? The potential at 4 ?F = 4 3 400 ? = 3 100 at 8 ?F = 8 3 400 ? = 6 100 The Potential difference = 6 100 3 100 ? = 3 50 ?V (ii) Hence the effective charge at 2 ?F = 50 × 2 = 100 ?F ? Potential at 3 ?F = 3 100 ; Potential at 6 ?F = 6 100 ?Difference = 6 100 3 100 ? = 3 50 ?V ? The potential at C & D is 3 50 ?V (b) ? S R q P ? = 2 1 = 2 1 = It is balanced. So from it is cleared that the wheat star bridge balanced. So the potential at the point C & D are same. So no current flow through the point C & D. So if we connect another capacitor at the point C & D the charge on the capacitor is zero. 16. Ceq between a & b = 2 1 2 1 3 2 1 2 1 C C C C C C C C C ? ? ? ? = 2 1 2 1 3 C C C C 2 C ? ? ( ?The three are parallel) 17. In the figure the three capacitors are arranged in parallel. All have same surface area = a = 3 A First capacitance C 1 = d 3 A 0 ? 2 nd capacitance C 2 = ) d b ( 3 A 0 ? ? 3 rd capacitance C 3 = ) d b 2 ( 3 A 0 ? ? Ceq = C 1 + C 2 +C 3 8 ?F ? B C A 4 ?F ? 3 ?F ? 6 ?F ? D 50 A C B D 6 ?F ? 8 ?F ? 3 ?F ? 4 ?F ? 8 ?F ? C D 50 4 ?F ? 3 ?F ? 6 ?F ? C 1 C 2 b a C 3 C 2 C 1 C 1C 2/C 1+C 2 b a C 3 C 1C 2/C 1+C 2 b C D B a A d a a b E Page 5 31.1 CHAPTER – 31 CAPACITOR 1. Given that Number of electron = 1 × 10 12 Net charge Q = 1 × 10 12 × 1.6 × 10 –19 = 1.6 × 10 –7 C ? The net potential difference = 10 L. ? Capacitance – C = v q = 10 10 6 . 1 7 ? ? = 1.6 × 10 –8 F. 2. A = ?r 2 = 25 ?cm 2 d = 0.1 cm c = d A 0 ? = 1 . 0 14 . 3 25 10 854 . 8 12 ? ? ? ? = 6.95 × 10 –5 ?F. 3. Let the radius of the disc = R ?Area = ?R 2 C = 1 ? D = 1 mm = 10 –3 m ? C = d A 0 ? ? 1 = 3 2 12 10 r 10 85 . 8 ? ? ? ? ? ? r 2 = ? ? ? ? 85 . 8 10 10 12 3 = 784 . 27 10 9 = 5998.5 m = 6 Km 4. A = 25 cm 2 = 2.5 × 10 –3 cm 2 d = 1 mm = 0.01 m V = 6V Q = ? C = d A 0 ? = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? Q = CV = 01 . 0 10 5 . 2 10 854 . 8 3 12 ? ? ? ? ? × 6 = 1.32810 × 10 –10 C W = Q × V = 1.32810 × 10 –10 × 6 = 8 × 10 –10 J. 5. Plate area A = 25 cm 2 = 2.5 × 10 –3 m Separation d = 2 mm = 2 × 10 –3 m Potential v = 12 v (a) We know C = d A 0 ? = 3 3 12 10 2 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 11.06 ×10 –12 F C = v q ? 11.06 ×10 –12 = 12 q ? q 1 = 1.32 × 10 –10 C. (b) Then d = decreased to 1 mm ? d = 1 mm = 1 × 10 –3 m C = d A 0 ? = v q = 3 3 12 10 1 10 5 . 2 10 85 . 8 ? ? ? ? ? ? ? = 12 2 ? q 2 = 8.85 × 2.5 × 12 × 10 –12 = 2.65 × 10 –10 C. ? The extra charge given to plate = (2.65 – 1.32) × 10 –10 = 1.33 × 10 –10 C. 6. C 1 = 2 ?F, C 2 = 4 ?F , C 3 = 6 ?F V = 12 V cq = C 1 + C 2 + C 3 = 2 + 4 + 6 = 12 ?F = 12 × 10 –6 F q 1 = 12 × 2 = 24 ?C, q 2 = 12 × 4 = 48 ?C, q 3 = 12 × 6 = 72 ?C 5 cm 0.1 cm 1 mm C 1 V C 2 C 3 Capacitor 31.2 7. ? The equivalent capacity. C = 2 1 3 1 3 2 3 2 1 C C C C C C C C C ? ? = 30 20 40 20 40 30 40 30 20 ? ? ? ? ? ? ? = 2600 24000 = 9.23 ?F (a) Let Equivalent charge at the capacitor = q C = V q ? q = C × V = 9.23 × 12 = 110 ?C on each. As this is a series combination, the charge on each capacitor is same as the equivalent charge which is 110 ?C. ? (b) Let the work done by the battery = W ? V = q W ? W = Vq = 110 × 12 × 10 –6 = 1.33 × 10 –3 J. 8. C 1 = 8 ?F, C 2 = 4 ?F , C 3 = 4 ?F Ceq = 3 2 1 1 3 2 C C C C ) C C ( ? ? ? ? = 16 8 8 ? = 4 ?F Since B & C are parallel & are in series with A So, q 1 = 8 × 6 = 48 ?C q 2 = 4 × 6 = 24 ?C q3 = 4 × 6 = 24 ?C 9. (a) ? C 1 , C 1 are series & C 2 , C 2 are series as the V is same at p & q. So no current pass through p & q. 2 1 C 1 C 1 C 1 ? ? ? 2 1 C C 1 1 C 1 ? ? C p = 2 C 1 = 2 4 = 2 ?F And C q = 2 C 2 = 2 6 = 3 ?F ? C = C p + C q = 2 + 3 = 5 ?F (b) C 1 = 4 ?F, C 2 = 6 ?F, In case of p & q, q = 0 ? C p = 2 C 1 = 2 4 = 2 ?F C q = 2 C 2 = 2 6 = 3 ?F & C ? = 2 + 3 = 5 ?F C & C ? = 5 ?F ? The equation of capacitor C = C ? + C ?? = 5 + 5 = 10 ?F 12 V 8 ?F ? A 4 ?F ? B C 4 ?F ? A B C 1 C 2 C 1 C 2 C 1 = 4 C 2 = 6 A p B C 1 C 1 C 1 C 1 q R S C 2 C 2 C 2 C 2 30 ?F ? V = 12 V 40 ?F ? 20 ?F ? Capacitor 31.3 10. V = 10 v Ceq = C 1 +C 2 [ ? They are parallel] = 5 + 6 = 11 ?F q = CV = 11 × 10 110 ?C ? 11. The capacitance of the outer sphere = 2.2 ?F C = 2.2 ?F Potential, V = 10 v Let the charge given to individual cylinder = q. C = V q ? q = CV = 2.2 × 10 = 22 ?F ? The total charge given to the inner cylinder = 22 + 22 = 44 ?F 12. C = V q , Now V = R Kq So, C 1 = ? ? 1 R / Kq q = K R 1 = 4 ?? 0 R 1 Similarly c 2 = 4 ?? 0 R 2 The combination is necessarily parallel. Hence Ceq = 4 ?? 0 R 1 +4 ?? 0 R 2 = 4 ?? 0 (R 1 + R 2 ) 13. ?C = 2 ?F ? In this system the capacitance are arranged in series. Then the capacitance is parallel to each other. (a) ? The equation of capacitance in one row C = 3 C (b) and three capacitance of capacity 3 C are connected in parallel ? The equation of capacitance C = 3 C 3 C 3 C ? ? = C = 2 ?F As the volt capacitance on each row are same and the individual is = ce tan capaci of . No Total = 3 60 = 20 V 14. Let there are ‘x’ no of capacitors in series ie in a row So, x × 50 = 200 ? x = 4 capacitors. Effective capacitance in a row = 4 10 Now, let there are ‘y’ such rows, So, 4 10 × y = 10 ? y = 4 capacitor. So, the combinations of four rows each of 4 capacitors. 6 ?F ? B 5 ?F ? 10 V A A5 B6 C A B C C C C C C C C Capacitor 31.4 15. (a) Capacitor = 8 4 8 4 ? ? = 3 8 ?? and 3 6 3 6 ? ? = 2 ?F (i) The charge on the capacitance 3 8 ?F ? Q = 3 8 × 50 = 3 400 ? ? The potential at 4 ?F = 4 3 400 ? = 3 100 at 8 ?F = 8 3 400 ? = 6 100 The Potential difference = 6 100 3 100 ? = 3 50 ?V (ii) Hence the effective charge at 2 ?F = 50 × 2 = 100 ?F ? Potential at 3 ?F = 3 100 ; Potential at 6 ?F = 6 100 ?Difference = 6 100 3 100 ? = 3 50 ?V ? The potential at C & D is 3 50 ?V (b) ? S R q P ? = 2 1 = 2 1 = It is balanced. So from it is cleared that the wheat star bridge balanced. So the potential at the point C & D are same. So no current flow through the point C & D. So if we connect another capacitor at the point C & D the charge on the capacitor is zero. 16. Ceq between a & b = 2 1 2 1 3 2 1 2 1 C C C C C C C C C ? ? ? ? = 2 1 2 1 3 C C C C 2 C ? ? ( ?The three are parallel) 17. In the figure the three capacitors are arranged in parallel. All have same surface area = a = 3 A First capacitance C 1 = d 3 A 0 ? 2 nd capacitance C 2 = ) d b ( 3 A 0 ? ? 3 rd capacitance C 3 = ) d b 2 ( 3 A 0 ? ? Ceq = C 1 + C 2 +C 3 8 ?F ? B C A 4 ?F ? 3 ?F ? 6 ?F ? D 50 A C B D 6 ?F ? 8 ?F ? 3 ?F ? 4 ?F ? 8 ?F ? C D 50 4 ?F ? 3 ?F ? 6 ?F ? C 1 C 2 b a C 3 C 2 C 1 C 1C 2/C 1+C 2 b a C 3 C 1C 2/C 1+C 2 b C D B a A d a a b E Capacitor 31.5 = d 3 A 0 ? + ) d b ( 3 A 0 ? ? + ) d b 2 ( 3 A 0 ? ? = ? ? ? ? ? ? ? ? ? ? ? d b 2 1 d b 1 d 1 3 A 0 = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) d b 2 )( d b ( d d ) d b ( d ) d b 2 ( ) d b 2 )( d b ( 3 A 0 = ? ? ) d b 2 )( d b ( d 3 b 2 bd 6 d 3 A 2 2 0 ? ? ? ? ? 18. (a) C = ) R / R ( In L 2 1 2 0 ? = 2 In 10 10 85 . 8 14 . 3 e 1 2 ? ? ? ? ? ? [In2 = 0.6932] = 80.17 × 10 –13 ? 8 PF (b) Same as R 2 /R 1 will be same. 19. Given that C = 100 PF = 100 × 10 –12 F C cq = 20 PF = 20 × 10 –12 F V = 24 V q = 24 × 100 × 10 –12 = 24 × 10 –10 q 2 = ? Let q 1 = The new charge 100 PF V 1 = The Voltage. Let the new potential is V 1 After the flow of charge, potential is same in the two capacitor V 1 = 2 2 C q = 1 1 C q = 2 1 C q q ? = 1 1 C q = 12 1 10 10 24 q 10 24 ? ? ? ? ? = 12 1 10 100 q ? ? = 24 × 10 –10 – q 1 = 5 q 1 = 6q 1 = 120 × 10 –10 = q 1 = 6 120 ×10 –10 = 20 × 10 –10 ? V 1 = 1 1 C q = 12 10 10 100 10 20 ? ? ? ? = 20 V 20. Initially when ‘s’ is not connected, C eff = q 3 C 2 = 50 3 C 2 ? = 4 10 2 5 ? ? = 1.66 × 10 –4 C After the switch is made on, Then C eff = 2C = 10 –5 Q = 10 –5 × 50 = 5 × 10 –4 Now, the initial charge will remain stored in the stored in the short capacitor Hence net charge flowing = 5 × 10 –4 – 1.66 × 10 –4 = 3.3 × 10 –4 C. A S /Read More

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- HC Verma Solutions: Electric Current in Conductors (Chapter 32)
- HC Verma Solutions: Thermal and Chemical Effects of Current (Chapter 33)
- HC Verma Solutions: Magnetic Field (Chapter 34)
- HC Verma Solutions: Magnetic Field due to a Current (Chapter 35)
- HC Verma Solutions: Permanent Magnets (Chapter 36)
- HC Verma Solutions: Magnetic Properties of Matter (Chapter 37)