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The electrostatic energy stored in electric field is given by \[\mathcal E = \frac{\epsilon_0}{2}\int |\vec E|^2 d^3r.\] In this article a few selected applications of the above expression for electrostatic energy to the problems involving capacitors are discussed.
What is a capacitors?
A capacitor is a device which can store charge. Wikipedia defines it as follows.
A capacitor is a device that stores electrical energy in an electric field by virtue of accumulating electric charges on two close surfaces insulated from each other. It is a passive electronic component with two terminals.
Parallel plate capacitor
In the following discussion about parallel plate capacitors, it will be assumed that the linear dimensions of the plates are much larger than the separation. This will allow the edge effects to be neglected and the known results, exact for infinite plates, will be used.
Capacitance
The capacitance of a parallel plate capacitor is given by \begin{equation} C = \frac{\epsilon_0 A}{d} \end{equation} where \(A\) is the area of plates and \(d\) is the separation between them.
Energy stored
We now consider the case when the plates of a parallel plate capacitor are connected to a battery.
Let the charge, voltage difference and capacitance of the capacitor be \(Q, V\) and \(C\) respectively. When the charge on the plates is increased from \(q\) to \(q+dq\), the potential between the plates increases by an amount \(\Delta V\). The electrical work is done in this process is \(V\Delta q\), where \(V=q/C\). The total work done in charging the capacitor is \begin{equation} W = \int_0^Q V dq = \int_0^Q \frac{q}{C} dq = \frac{1}{2} \frac{Q^2}{C}. \end{equation} This expression coincides with the field energy stored in the electric field between the plates. \begin{equation}\label{ESEnergy} \text{Energy stored in field} = U \frac{\epsilon_0}{2} \int |\vec E|^2 d^3 r, \end{equation} where the integral is taken over all space. For a parallel plate condenser, the electric field outside volume enclosed between the plates will be neglected. Taking the field to be uniform it is given by \(E =V/d\) and is perpendicular to the plates. The expression \eqref{ESEnergy} then gives the answer\begin{eqnarray} U=\frac{\epsilon_0}{2} \int |\vec E|^2 d^3 r &=& \frac{\epsilon_0}{2}|\vec E|^2 (A \times d)\qquad \text{Why (A d) factor?} \\ &=& \frac{\epsilon_0 }{2} \Big(\frac{V}{d}\Big)^2\,(A\, d)\\ &=& \frac{1}{2}\, \Big(\frac{\epsilon_0 A}{d}\Big) V^2 = \frac{1}{2} C V^2 = \frac{Q^2}{2C}. \end{eqnarray}Thus we see that the electrical work done is stored as the energy, \(U\), of the field between the plates.
Force between the plates-I
The electrostatic energy expression for a charged capacitor can be used to obtain an expression for force between the charged capacitor. if \(F\) is the force between the plates, an increase in the separation between the plates by a small amount \(\Delta x\) would requires mechanical work \(\Delta W\) \begin{equation} \label{EQ07} (\Delta W)_\text{mech} = F \Delta x. \end{equation} Equating this work to change in energy, \(U\), of the capacitor we get
\begin{eqnarray} F \Delta x &=& \Delta U\\ \text{or}\qquad F \Delta x &=&\Delta \Big(\frac{Q^2}{2C}\Big). \end{eqnarray}
If the charge on the capacitor remains constant, we have \begin{equation} F \Delta x = \frac{Q^2}{2}\Delta \Big(\frac{1}{C}\Big) = \frac{Q^2}{2}\frac{-\Delta C}{C^2} \end{equation} For a parallel plate capacitor we have \(C=\epsilon_0 A/x\). Therefore, taking derivative of \(\log C\), we get \begin{equation} \frac{\Delta C}{C} = -\frac{\Delta x}{x} \end{equation} and the force is given by \begin{equation}\label{EQ11} F = \frac{Q^2}{2 C x} = \frac{Q^2}{2 \epsilon_0 A}. \end{equation}
Warning: The charge \(Q\) on a plate of a capacitor is assumed to be spread uniformly over the area. The electric field between the plates of the capacitor is \(E\). One may, therefore, (wrongly) conclude that the force on each plate should be
\begin{equation}
|E|Q = \frac{V}{d}Q = \frac{Q^2}{\epsilon_0 A}
\end{equation}
This is incorrect!
Force between the plates-II
In the above derivation it has been assumed that the charge on the capacitor remains constant. If the capacitor is connected to a battery, the charge will not remain constant is the plates are displaced by a small amount. The voltage will remain constant, but in this case computing change in energy by means of
\begin{equation}
\Delta U = \Delta \Big(\frac{1}{2} C V^2\Big) = \frac{1}{2}V^2 \Delta C
\end{equation}
would give a repulsive force. This is a wrong result because the force will be attractive whether the battery is connected or not. To resolve this problem, one must take into account of the work done by the battery. In the process of keeping the voltage constant, the battery supplies charge \(\Delta Q = V (\Delta C)\) and the electrical work done by the battery is given by
\begin{equation}\label{EQ13}
(\Delta W)_\text{elec} -V \Delta Q =V^2 \Delta C = \frac{Q^2}{C^2}\Delta C \end{equation}
Thus the total work done in changing the separation by a small amount is the sum of the expressions \eqref{EQ07} and \eqref{EQ13}. Equating the total, electrical and mechanical, work to change in the energy of the capacitor gives
\begin{equation}
F \Delta x + V^2 \Delta C = \Delta U
\end{equation}
and it is easy to see that we get the same expression, \eqref{EQ11}, for the force as before.
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