Question

A $1\mu F$ capacitor of a TV is subjected to a 4000V potential difference. The energy stored in the capacitor is
A) $8J$
B) $16J$
C) $4 \times {10^{ - 3}}J$
D) $2 \times {10^{ - 3}}J$

Answer 1

Use the formula for the energy stored in the capacitor. There are three formulae for energy stored in the capacitor, which relate the energy stored with Capacitance, Potential difference, and stored charge, but we have to use the one in which we can use the information provided in the question as the question has already given us the values of the capacitance of the TV and the applied potential difference and using them, you can reach at the answer directly.

Complete step by step answer:
Given,
Capacitance of the capacitor $C = 1\mu F$
Potential difference across the capacitor $V = 4000V$
We know that,
Energy stored in the capacitor $E = \dfrac{1}{2}C{V^2}$
$\Rightarrow E = \dfrac{1}{2}(1 \times {10^{ - 6}}){(4000)^2}$
$\Rightarrow E = 8J$

$\therefore$ The energy stored in the capacitor is 8J. Hence, option (A) is correct.

Additional information:
When a capacitor is being charged, there is initially no charge on either plate, so that the electric field intensity between the plates is zero. To move the first increment of charge from one plate to the other requires no work. Later increments of charge transferred from one plate to the other must-have work done upon them against the electric field of the charged plates. So, the total work done to transfer all the charges to other plates is being stored inside the capacitor in the form of an electric field. We can also say that the energy stored inside a capacitor is electric potential energy.

Note:
Normally, in real life, the capacitor is filled with the oil of some electrical permeability. The oil in the capacitor is used to remove any air voids and secondly it helps to cool or remove the heat away from the capacitor. The oil works to offset these higher temperatures so that the capacitor can remain working.