A parallel plate capacitor with air as the dielectric has a capacitance of 6∗10−4μF and is charged by a 100V battery. Calculate the charge and the energy stored in the capacitor.

Given,
Capacitance of the capacitor $(C) = 6 * 10^{-4}\; \mu F = 6 * 10^{-10}\;F$
Potential difference $(V) = 100\;V$
Charge $(q) = ?$
Energy Stored $(E) = ?$

We have,
Charge $(q) = C\;*\;V = 6 * 10^{-10}\;*\;100 = 6 * 10^{-8}\;C$

Again,
Energy Stored $ (E) = \frac{1}{2}\;C\;V^2 = 6 * 10^{-10}\;*\;100^2 = 3 * 10^{-6}\;J$

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