Question

A parallel plate capacitor with a capacitance of 5µF is connected to a 500 V batter. Calculate the charge on the plates and the energy stored.

Answer 1

Explanation:

It is given that,

Capacitance, $C=5\ \mu F=5\times 10^{-6}\ F$

Voltage, V = 500 volts

Charge on the plates, $Q=C\times V$

$Q=5\times 10^{-6}\ F\times 500$

Q = 0.0025 C

Energy stored in the capacitor is given by :

$E=\dfrac{1}{2}CV^2$

$E=\dfrac{1}{2}\times 5\times 10^{-6}\ F\times (500\ V)^2$

E = 0.625 Joules

Hence, this is the required solution.