Question

A parallel plate capacitor consists of 2 plates of area 1.00 x 10^(-2) m^2 placed a distance 2.00 x 10^(-2) m apart in air. The capacitor is charged so that the potential difference between the plates is 1000V. Calculate;
(i) the electric field between the plates(neglect edge effects)
(ii)the capacitance of the capacitor and the energy stored in the capacitor.

Answer 1

(i)

$$E=\dfrac{V}{d}=\dfrac{1000\ V}{2.0\cdot10^{-2}\ m}=5\cdot10^4\ \dfrac{V}{m}.$$

(ii)

$$C=\epsilon_0\dfrac{A}{d},$$ $$C=8.854\cdot10^{-12}\ \dfrac{F}{m}\cdot\dfrac{1.0\cdot10^{-2}\ m^2}{2.0\cdot10^{-2}\ m}=4.43\cdot10^{-12}\ F.$$ $$E=\dfrac{1}{2}CV^2,$$ $$E=\dfrac{1}{2}\cdot4.43\cdot10^{-12}\ F\cdot(1000\ V)^2=2.2\cdot10^{-6}\ J.$$