A `muF` capacitor `C_1` is charged to `V_0 = 120 V`. The charging battery is then removed and the capacitor is connected in parallel to an uncharged `+ 4muF` capacitor `C_2`.

(a) what is the potential difference `V` across the combination?
(b) what is the stored energy before and after the switch `S` is closed?

Answer

C, D

Explanation

a. `V=q_(n et)/C_(n et)=(C_1V_0)/(C_1+C_2)=V_0/(1+C_2//C_1)`
`=120/(1+4//8)=80V`
b. `U_i=1/2C_1V_0^2=1/2xx8xx10^-6xx(120)^2`
`=5.76xx10^-2J`
`U_f=1/2(C_1+C_2)V^2`
`=1/2(C_1+C_2)V^2`
`=1/2xx12xx10^-6xx(80)^2`
`=3.81xx10^-2J`