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 dont know , let me tell you guys that, In Indian Railways, most of the fans do not work on switching it, we need to strike it with a stick.

But WHY?

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

Now in Indian Railways, most of the fans do not work on switching it, The reason is because they dont contain capacitors which give them the extra energy to change their state of rest to motion.So we use external energy, that is by striking the stick. So this was an simple example

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

If the capacitance of a conductor is C , it is uncharged initially & the potential difference between it's 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 & charge.

Hence if if the battery delivers 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 :

\(W=\int_{0}^{q}\frac{q}{C}dq = \frac{1}{C}[\frac{q^2}{2}]_{0}^{q} = \frac{1}{2}\frac{q^2}{C}\)

Therefore energy stored in a capacitor is

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

Substituting q=CV we get,

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

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

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

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

Solution : C = 100 F , V = 100 V

We have U = \(\frac{1}{2}CV^2=\frac{1}{2}.100.100=5000 J\)