2.15 ENERFGY STORED IN A CAPACITOR
A capacitor, as we have seen above, is a system of two conductors with charge \( Q \) and \( -Q \). To determine the energy stored in this configuration, consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by bit, so that at the end, conductor 1 gets charge \( Q \). By charge conservation, conductor 2 has charge \( -Q \) at the end (Fig 2.30 ).
(a) Work done in a small step of building charge on conductor 1 from \( Q^\prime \) to \( Q^\prime+\delta Q^\prime \).
(b) Total work done in charging the capacitor may be viewed as stored in the energy of electric field between the plates.
In transferring positive charge from conductor 2 to conductor 1, work will be done externally, since at any stage conductor 1 is at a higher potential than conductor 2. To calculate the total work done, we first calculate the work done in a small step involving transfer of an infinitesimal (i.e., vanishingly small) amount of charge. Consider the intermediate situation when the conductors 1 and 2 have charges \( Q^\prime \) and \( -Q^\prime \) respectively. At this stage, the potential difference \( V^\prime \) between conductors 1 to 2 is \( Q^\prime/C \), where \( C \) is the capacitance of the system. Next imagine that a small charge \( \delta Q^\prime \) is transferred from conductor 2 to 1. Work done in this step (\( \delta Q^\prime \)), resulting in charge \( Q^\prime \) on conductor 1 increasing to \( Q^\prime+\delta Q^\prime \), is given by
\[ \delta W=V^{\prime} \delta Q^{\prime}=\frac{Q^{\prime}}{C} \delta Q^{\prime} \] (2.68)
Since \( \delta Q^\prime \) can be made as small as we like, Eq. (2.68) can be written as
\[ \delta W=\frac{1}{2 C}\left[\left(Q^{\prime}+\delta Q^{\prime}\right)^{2}-Q^{\prime 2}\right] \](2.69)
Equations (2.68) and (2.69) are identical because the term of second order in \( \delta Q^\prime \), i.e., \( \delta {Q^\prime}^2/2C \), is negligible, since \( \delta Q^\prime \) is arbitrarily small. The total work done (W) is the sum of the small work \( \delta W\) over the very large number of steps involved in building the charge \( Q^\prime \) from zero to \( Q \).
\[ \begin{aligned} W &=\sum_{\text {sum over all steps }} \delta W \\ &=\sum_{\text {sum over all steps }} \frac{1}{2 C}\left[\left(Q^{\prime}+\delta Q^{\prime}\right)^{2}-Q^{\prime 2}\right] \end{aligned} \](2.70)
\( \begin{array}{l} =\frac{1}{2 C}\left[\left\{\delta^{\prime 2}-0\right\}+\left\{\left(2 \delta Q^{\prime}\right)^{2}-\delta Q^{\prime 2}\right\} \right. \\ \left.+\left\{\left(3 \delta Q^{\prime}\right)^{2}-\left(2 \delta Q^{\prime}\right)^{2}\right\}+\ldots+\left\{Q^{2}-(Q-\delta Q)^{2}\right\}\right] \end{array} \) (2.71)
\[ =\frac{1}{2 C}\left[Q^{2}-0\right]=\frac{Q^{2}}{2 C} \](2.72)
The same result can be obtained directly from Eq. (2.68) by integration \[ W=\int_{0}^{8} \frac{Q^{\prime}}{C} \delta Q^{\prime}=\left.\frac{1}{C} \frac{Q^{\prime 2}}{2}\right|_{0} ^{8}=\frac{Q^{2}}{2 C} \] This is not surprising since integration is nothing but summation of a large number of small terms.
We can write the final result, Eq. (2.72) in different ways
\[ W=\frac{Q^{2}}{2 C}=\frac{1}{2} C V^{2}=\frac{1}{2} Q V \] (2.73)
Since electrostatic force is conservative, this work is stored in the form of potential energy of the system. For the same reason, the final result for potential energy [Eq. (2.73)] is independent of the manner in which the charge configuration of the capacitor is built up. When the capacitor discharges, this stored-up energy is released. It is possible to view the potential energy of the capacitor as ‘stored’ in the electric field between the plates. To see this, consider for simplicity, a parallel plate capacitor [of area \( A \) (of each plate) and separation \( d \) between the plates].
Energy stored in the capacitor \[ =\frac{1}{2} \frac{Q^{2}}{C}=\frac{(A \sigma)^{2}}{2} \times \frac{d}{\varepsilon_{0} A} \] (2.74)
The surface charge density \( \sigma \) is related to the electric field \( E \) between the plates,
\[ E= \frac{\sigma}{\varepsilon_0} \](2.75)
From Eqs. (2.74) and (2.75) , we get
Energy stored in the capacitor \[ U=(1 / 2) \varepsilon_{0} E^{2} \times A d \] (2.76)
Note that \( Ad \) is the volume of the region between the plates (where electric field alone exists). If we define energy density as energy stored per unit volume of space, Eq (2.76) shows that
Energy density of electric field, \[ u=(1 / 2) \varepsilon_{0} E^{2} \] (2.77)
Example 2.10
(a) A \( 900 pF \) capacitor is charged by \( 100 V \) battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another \( 900 pF \) capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system?
(a)The charge on the capacitor is \[ Q=C V=900 \times 10^{-12} \mathrm{~F} \times 100 \mathrm{~V}\] \[=9 \times 10^{-8} \mathrm{C} \]
The energy stored by the capacitor is \[ \begin{array}{l} =(1 / 2) C V^{2}=(1 / 2) Q V \\ =(1 / 2) \times 9 \times 10^{-8} \mathrm{C} \times 100 \mathrm{~V} \end{array} \] \[ =4.5 \times 10^{-6} \mathrm{~J} \]
(b)In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be \( V^\prime \). The charge on each capacitor is then \( Q^\prime=C V^\prime \). By charge conservation, \( Q^\prime=Q/2 \). This implies \( V^\prime=V/2 \). The total energy of the system is \[ =2 \times \frac{1}{2} Q^{\prime} V^{\prime}=\frac{1}{4} Q V=2.25 \times 10^{-6} \mathrm{~J} \]
Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone?
There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.