Question

You bought a parallel plate capacitor for the flash of your developed camera. It consists of two rectangular plates, each with area of 4.5 cm²_, that are separated by a 2.00-mm thick ordinary glass. You then connected it with a 12.0 V battery. How much energy is stored in the capacitor?

Answer 1

Given:

$d=0.002\:\rm m$

$A=4.5*10^{-4}\:\rm m^2$

$V=12\:\rm V$

$k=5$

The capacitance

$$C=\frac{k\epsilon_0A}{d}=\frac{5*8.85*10^{-12}*4.5*10^{-4}}{0.002}\\ =9.96*10^{-12}\:\rm F$$

The energy of electric field

$$W=\frac{CV^2}{2}=\frac{9.96*10^{-12}*12^2}{2}=7.2*10^{-10}\:\rm J$$