A 50 mu F capacitor is charged to a potential difference of

A 5.0 mu F capacitor is charged to a potential difference of 70 V. The battery is then removed and replaced with an uncharged 2.0 mu F capacitor. Charge flows from the first capacitor to the second until they have the same potential difference, V_new. Determine V_new. 16V 22 V 37 V 50 V

Solution

Here,

C1 = 5 uF

V = 70 V

C2 = 2 uF

Now , let the new charge on the capacitors is Q1 and Q2

Q1 + Q2 = 70 * 5 ----(1)

Now , after attaching the capacitors

Q1/5 = Q2/2 -----(2)

solving 1 and 2

Q1 = 250 uC

Q2 = 100 uC

as Vnew = Q1/C1

Vnew = 250/5 = 50 V

the voltage Vnew is 50 V