Two capacitors C1 = 7 µF and C2 = 14 µF are connected to a battery using the swi

Question

Two capacitors C1 = 7 µF and C2 = 14 µF are connected to a battery using the switch S. When the switch is in position A the capacitor C1 is connected to a VB = 12 Volt battery. When the switch is moved to position B the capacitor C1 is disconnected from the battery but remains with the same charge when it was connected to the battery. When the switch is moved to position C the charged capacitor C1 is connected to the uncharged capacitor C2.
(a) When the switch is in position B, what is the charge on the capacitor C1? 84 µC

(b) When the switch is in position C (after stabilizing), what is the charge on capacitor C1? µC

(c) When the switch is in position C (after stabilizing), what is the voltage across capacitor C1? V

(d) When the switch is in position C (after stabilizing), what is the charge on capacitor C2? µC

confused by B through D.. Thanks!

Explanation / Answer

C1 = 7F , C2 = 14F

When the switch is at position A

C1 connected to battery with V = 12V

so charge stored on C1 = 12*7 C = 84C

a)When switch at position B charge in C1 = 84C

When the switch is shifted to position C let the charge on C1 become q C then conservation of charges will imply charge on C2 will be (84-q)C.

But since they become in parallel so potential across them will be same so

q/C1 = (84-q)/C2

q/7 = (84-q)/14

on solving we get q = 28C

b) So charge on capacitor 1 after stabilization becomes 28C

c)Voltage across capacitor 1 becomes = 28/7 V = 4 V

d) Charge on capacitor 2 after stabilization = 84-28 C= 56C

e) Voltage across capacitor 2 becomes = 56/14 = 4V

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