Energy Stored in a Capacitor

A capacitor is a device for storing energy. When we connect a battery across the two plates of a capacitor, it gets charged. The potential difference gradually increases across the two plates and the battery had to do more work to deliver the same amount of charge due to the continuous increase in potential difference.

One basic use is the use in fans. Think about the fans in our Indian railway system. If you don't know, let me tell you guys that, in Indian railways, most of the fans do not work on switching it, but we need to strike it with a stick.

But why?

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We shall know that the fans contain capacitors inside. Now, when we switch on the fan, what happens is that the fan is in a state of rest, so it needs some extra energy to go into rotational motion, and hence it uses the energy stored in the capacitors. And while the fan is running it again recharges itself.

In Indian railways, most of the fans do not work on switching it, The reason is that they don't contain capacitors which give them the extra energy to change their state of rest to motion. So we use external energy, which is by striking the stick. So this was a simple example

The work done to charge the capacitor is thus stored in the form of electric potential energy inside the capacitor.

Energy Stored In a Charged Capacitor

If the capacitance of a conductor is $C,$ it is uncharged initially and the potential difference between its plates is $V$ when connected to a battery. If $q$ is the charge on the plate at that time, then

$q = CV.$

We know that $W=Vq,$ i.e. work done is equivalent to the product of the potential and charge.

Hence, if the battery delivers the infinitesimally small amount of charge $dq$ to the capacitor at constant potential $V,$ then

$dW=Vdq=\frac{q}{C}dq.$

Total work done in delivering a charge of amount $q$ to the capacitor is given by

$W=\int_{0}^{q}\frac{q}{C}dq = \frac{1}{C}\left[\frac{q^2}{2}\right]_{0}^{q} = \frac{1}{2}\frac{q^2}{C}.$

Therefore, energy stored in a capacitor is

$U=\frac{1}{2}\frac{q^2}{C}.\qquad (1)$

Substituting $q=CV,$ we get

$U=\frac{1}{2}CV^2.\qquad (2)$

Substituting $C=\frac{q}{V},$ we get

$U=\frac{1}{2}qV.\qquad (3)$

If the capacitance of a capacitor is 100 F charged to a potential of 100 V, Calculate the energy stored in it.

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We have C = 100 F and V = 100 V.

Then we have $U = \frac{1}{2}CV^2=\frac{1}{2}\cdot 100\cdot 100^2=500000\text{ J}.$