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, the current charges the capacitor, leading to an accumulation of charges on opposite plates of the capacitor. As charges accumulate, the potential difference gradually increases across the two plates. While discharging, this potential difference can drive a current in the opposite direction.
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.
We have C = 100 F and V = 100 V.
Then we have \(U = \frac{1}{2}CV^2=\frac{1}{2}\cdot 100\cdot 100^2=500000\text{ J}.\)