Question

$\huge\mathbb\pink{Question}$

Derive an expression for energy stored in a capacitor ?

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Answer 1

Answer: $U=\frac{1}{2}CV^2\;$  'or'  $\;U=\frac{1}{2}\frac{q^2}{C}$

Explanation:

Let that capacitance of a capacitor is 'C'.  Let at at time of charging the capacitor charge on the capacitor is q' and at that time  potential difference between its points is V',

Then

V' = q'/C

Let the work done in giving more very  small volume of charge dq is  dW,

So,

dW = V' × dq

Putting the value of V' = q'/C

dW = q'/C × dq

Now, If the work done in charging the capacitor from 0 to q  is W

Then,

$W=\int\limits^q_0{dW}\\\;\\W=\int\limits^q_0{\frac{q'}{C}.dW}\\\;\\W=\frac{1}{C}\int\limits^q_0{q'\;dq}\\\;\\W=\frac{1}{C}[\frac{{q'}^{2}}{2}]\limits^q_0\\\;\\W=\frac{1}{C}[\frac{q^2}{2}-0]\\\;\\W=\frac{1}{2}\frac{q^2}{C}$

This work has been stored in the capacitor as its electric potential energy, So we can denote it by 'U'. Putting W = U

$U=\frac{1}{2}\frac{q^2}{C}$

Also putting q = CV,

$U=\frac{1}{2}CV^2$

Answer 2

The energy stored on a capacitor can be expressed in terms of the work done by the battery. Voltage represents energy per unit charge, so the work to move a charge element dq from the negative plate to the positive plate is equal to V dq, where V is the voltage on the capacitor.

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