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 1).
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| Fig.1:(a) Work done in a small step of building charge on conductor 1 from $Q′$ to $Q′ + \delta Q′$. (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 $Q′$ and $–Q′$ respectively. At this stage, the potential difference V′ between conductors 1 to 2 is $Q′/C$, where $C$ is the capacitance of the system. Next imagine that a small charge $\delta Q′$ is transferred from conductor 2 to 1. Work done in this step ($\delta W$), resulting in charge $Q′$ on conductor 1 increasing to $Q′+ \delta Q′$, is given by
$\delta W=V'.\delta Q=\frac{Q'}{C}\delta Q'$ (1)
Since $\delta Q'$ can be made as small as we like, Eq. (1) can be written as
$\delta W=\frac{1}{2C}[(Q'+\delta Q')^2 -Q'^2]$ (2)
Equations (1) and (2) are identical because the term of second order in $\delta Q′$, i.e., $\delta Q′/2C$, is negligible, since $\delta Q′$ 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′ from zero to Q.
$W=\Sigma_{sum \ over \ all \ steps} \delta W$
$= \Sigma_{sum \ over \ all \ steps} \frac{1}{2C}[(Q'+\delta Q')^2 -Q'^2]$ (3)
$=\frac{1}{2C}[(\delta Q'^2 -0)+(2\delta Q')^2-\delta Q'^2+(3\delta Q')^2-(2\delta Q')^2+...$
$+Q^2-(Q-\delta Q')^2]$
$W=\frac{1}{2C}[(\delta Q'^2 -0)=\frac{Q^2}{2C}$ (4)
The same result can be obtained directly from Eq. (1) by integration
$W=\int_{0}^{Q}\frac{Q'}{C}\delta Q'$
$W=\frac{1}{C}\frac{Q'^2}{2}|\begin{matrix}Q \\ 0\end{matrix}=\frac{Q^2}{2C}$
This is not surprising since integration is nothing but summation of a large number of small terms.
We can write the final result, Eq. (4) in different ways
$W=\frac{Q^2}{2C}=\frac{1}{2}CV^2=\frac{1}{2}QV$ (5)
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. (5)] 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
$W=\frac{Q^2}{2C}=\frac{(A\sigma)^2}{2}\times \frac{d}{\epsilon_0A}$ (6)
The surface charge density σ is related to the electric field E between the plates,
$E=\frac{\sigma}{\epsilon_0}$ (7)
From Eqs. (2.74) and (2.75) , we get
Energy stored in the capacitor
$u=\frac{1}{2}\epsilon_0E^2 \times Ad$ (8)
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,
$u=\frac{1}{2}\epsilon_0E^2$ (9)
Though we derived Eq. (9) for the case of a parallel plate capacitor, the result on energy density of an electric field is, in fact, very general and holds true for electric field due to any configuration of charges.
Example 1
(a) What is the magnitude of the electric field required to store 1.00 J of electric potential energy in a volume of $1.00 \ m^3$ in vacuum? (b) If the field magnitude is 10 times larger than that, how much energy is stored per cubic meter?
Solution:
(a) The desired energy density is $u=1.00 \ J/m^3$. Then from Eq. (24.11),
$E=\sqrt{\frac{2u}{\epsilon_0}}$
$E=\sqrt{\frac{2(1.00 \ J/m^3)}{8.85 \times 10^{-12} \ C^2/N.m^2}}$
$E=4.75 \times 10^5 \ N/C$
(b) Equation (9) shows that u is proportional to $E^2$. If increases by a factor of 10, u increases by a factor of $10^2 =100$, so the energy density becomes $u=100 \ J/m^3$.
Example 2
We connect a capacitor $C_1=8.0 \ \mu F$ to a power supply, charge it to a potential difference $V_0=120 \ V$, and disconnect the power supply (Fig. 2). Switch is open. (a) What is the charge $Q_0$ on $C_1$? (b) What is the energy stored in $C_1$? (c) Capacitor $C_2=4.0 \ \mu F$ is initially uncharged. We close switch S. After charge no longer flows, what is the potential difference across each capacitor, and what is the charge on each capacitor? (d) What is the final energy of the system?
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| Fig.2: When the switch is closed, the charged capacitor is connected to an uncharged capacitor The center part of the switch is an insulating handle; charge can flow only between the two upper terminals and between the two lower terminals. |
Solution
In parts (a) and (b) we find the charge $Q_0$ and stored energy $U_{initial}$ for the single charged capacitor $C_1$ using Eqs. $C=\frac{Q}{V}$ and (24.9), respectively. After we close switch S, one wire connects the upper plates of the two capacitors and another wire connects the lower plates; the capacitors are now connected in parallel. In part (c) we use the character of the parallel connection to determine how $Q_0$ is shared between the two capacitors. In part (d) we again use Eq. (5) to find the energy stored in capacitors $C_1$ and $C_2$ the energy of the system is the sum of these values.
(a) The initial charge $Q_0$ on $C_1$ is
$Q_0=C_1V_0=(8.0 \ \mu F)(120 \ V)=960 \ \mu C$
(b) The energy initially stored in $C_1$ is
$U_{initial}=\frac{1}{2}Q_0V_0$
$U_{initial}=\frac{1}{2}(960 \times 10^{-6} \ C)(120 \ V)=0.058 \ J$
(c) When we close the switch, the positive charge $Q_0$ is distributed over the upper plates of both capacitors and the negative charge $-Q_0$ is distributed over the lower plates. Let $Q_1$ and $Q_2$ be the magnitudes of the final charges on the capacitors. Conservation of charge requires that $Q_1+Q_2=Q_0$. The potential difference between the plates is the same for both capacitors because they are connected in parallel, so the charges are $Q_1=C_1V$ and $Q_2=C_2V$. We now have three independent equations relating the three unknowns $Q_1$, $Q_2$, and V. Solving these, we find
$V=\frac{Q_0}{C_1+C_2}=\frac{960 \ \mu C}{8.0 \ \mu F+4.0 \ \mu F}=80 \ V$
$Q_1=640 \ \mu C$; $Q_2=320 \ \mu C$
(d) The final energy of the system is
$u_{final}=\frac{1}{2}Q_1V+\frac{1}{2}Q_2V=\frac{1}{2}Q_0V$
$u_{final}=\frac{1}{2}(960 \times 10^{-6} \ C)(80 \ V)=0.038 \ J$
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