Capacitors C110F and C220F are each charged to 20 V then di

Capacitors C1=10F and C2=20F are each charged to 20 V , then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of C1connected to the negative plate of C2 and vice versa.

What is the potential difference across each capacitor?

Solution

Charge on C1 after charging to 20 V,

Q1 = C1*V = 10*10^-6*20 = 2*10^-4 C

Similarly charge on C2,

Q2 = 20*10^-6*20 = 4*10^-4 C

Let after reconnecting the voltage on each capacitor be V\'

So, charges on both get canceled.

So, net charge on both = (4-2)*10^-4 = 2*10^-4 C

Now, let the charge on capacitor C1 after rearrangement be Q1\'

and on the capacitor C2 be Q2\'

So, Q1\' + Q2\' = 2*10^-4

Now, as they are connected in parallel

Q1\'/C1 = Q2\'/C2

So, (2*10^-4 - Q2\')/C1 = (Q2\')/C2

So, (2*10^4 - Q2\')/10 = Q2\'/20

So, Q2\' = 1.33*10^-4

So, potential difference on each = Q2\'/C2

= 1.33*10^-4/(20*10^-6)

= 6.65 V