An electrostatic energy of $10 \mathrm{~nJ}$ is stored in an $8 \mathrm{~pF}$ capacitor. The voltage

Question

An electrostatic energy of $10 \mathrm{~nJ}$ is stored in an $8 \mathrm{~pF}$ capacitor. The voltage of the battery connected across it is

Professor PC Thomas Classes · Answer

(a) : Energy stored in a capacitor, $U = \frac{1}{2} C V^2$. We are given $U = 10 \mathrm{~nJ} = 10 	imes 10^{-9} \mathrm{~J}$ and $C = 8 \mathrm{~pF} = 8 	imes 10^{-12} \mathrm{~F}$. Substituting the values, we get $10 	imes 10^{-9} = \frac{1}{2} 	imes (8 	imes 10^{-12}) 	imes V^2$. Simplifying this gives $10 	imes 10^{-9} = 4 	imes 10^{-12} 	imes V^2$, which implies $V^2 = \frac{10 	imes 10^{-9}}{4 	imes 10^{-12}} = 2.5 	imes 10^3 = 2500$. Therefore, $V = 50 \mathrm{~V}$.