Question

A $10\mu F$ capacitor is connected to a $100V$ battery. What is the electrostatic energy stored?

Answer 1

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

$U=\frac{1}{2}\times 10\times {10}^{-4}\times {250}^2$

$U=5\times {10}^{-4}\times {250}^2$

$U=31.25J$