Answer to Question #219167 in Electric Circuits for Solution

Question

A capacitor is charged with 9.6nC and has a 120v potential difference between it's terminals. Compute it's capacitance and energy stored in it

Answer 1

Gives

q=9.6nC

V=120V

$U=\frac{1}{2}qV$

Put value

$$U=\frac{1}{2}\times9.6\times10^{-9}\times120=5.76\times10^{-7}J$$