Question

How much energy in kJ will be stored by a capacitor of 470 $\mu F$ when charged by a battery of 20 V?

Answer 1

Energy stored in the capacitor is

$U=\frac{1}{2}C{V}^2$

$=\frac{1}{2}\times 470\times {10}^{-6}\times (20{)}^2$

$=0.094J=94mJ$