The energy of a condenser. Forces on charged conductors
We consider now the energy required to charge a condenser. If the charge $Q$ has been taken from one of the conductors of a condenser and placed on the other, the potential difference between them is \begin{equation} \label{Eq:II:8:8} V=\frac{Q}{C}, \end{equation}  where $C$ is the capacity of the condenser. How much work is done in charging the condenser? Proceeding as for the sphere, we imagine that the condenser has been charged by transferring charge from one plate to the other in small increments $dQ$. The work required to transfer the charge $dQ$ is \begin{equation*} dU=VdQ. \end{equation*}  Taking $V$ from Eq. (8.8), we write \begin{equation*} dU=\frac{Q\,dQ}{C}. \end{equation*}  Or integrating from zero charge to the final charge $Q$, we have \begin{equation} \label{Eq:II:8:9} U=\frac{1}{2}\,\frac{Q^2}{C}. \end{equation}  This energy can also be written as \begin{equation} \label{Eq:II:8:10} U=\tfrac{1}{2}CV^2. \end{equation}  Recalling that the capacity of a conducting sphere (relative to infinity) is \begin{equation*} C_{\text{sphere}}=4\pi\epsO a, \end{equation*}  we can immediately get from Eq. (8.9) the energy of a charged sphere, \begin{equation} \label{Eq:II:8:11} U=\frac{1}{2}\,\frac{Q^2}{4\pi\epsO a}. \end{equation}  This, of course, is also the energy of a thin spherical shell of total charge $Q$ and is just $5/6$ of the energy of a uniformly charged sphere, Eq. (8.7). We now consider applications of the idea of electrostatic energy. Consider the following questions: What is the force between the plates of a condenser? Or what is the torque about some axis of a charged conductor in the presence of another with opposite charge? Such questions are easily answered by using our result Eq. (8.9) for electrostatic energy of a condenser, together with the principle of virtual work (Chapters 4, 13, and 14 of Vol. I). Let’s use this method for determining the force between the plates of a parallel-plate condenser. If we imagine that the spacing of the plates is increased by the small amount $\Delta z$, then the mechanical work done from the outside in moving the plates would be \begin{equation} \label{Eq:II:8:12} \Delta W=F\,\Delta z, \end{equation}  where $F$ is the force between the plates. This work must be equal to the change in the electrostatic energy of the condenser. By Eq. (8.9), the energy of the condenser was originally \begin{equation*} U=\frac{1}{2}\,\frac{Q^2}{C}. \end{equation*}  The change in energy (if we do not let the charge change) is \begin{equation} \label{Eq:II:8:13} \Delta U=\frac{1}{2}\,Q^2\,\Delta\biggl(\frac{1}{C}\biggr). \end{equation}  Equating (8.12) and (8.13), we have \begin{equation} \label{Eq:II:8:14} F\,\Delta z=\frac{Q^2}{2}\,\Delta\biggl(\frac{1}{C}\biggr). \end{equation}  This can also be written as \begin{equation} \label{Eq:II:8:15} F\,\Delta z=-\frac{Q^2}{2C^2}\,\Delta C. \end{equation}  The force, of course, results from the attraction of the charges on the plates, but we see that we do not have to worry in detail about how they are distributed; everything we need is taken care of in the capacity $C$. It is easy to see how the idea is extended to conductors of any shape, and for other components of the force. In Eq. (8.14), we replace $F$ by the component we are looking for, and we replace $\Delta z$ by a small displacement in the corresponding direction. Or if we have an electrode with a pivot and we want to know the torque $\tau$, we write the virtual work as \begin{equation*} \Delta W=\tau\,\Delta\theta, \end{equation*}  where $\Delta\theta$ is a small angular displacement. Of course, $\Delta(1/ C)$ must be the change in $1/C$ which corresponds to $\Delta\theta$. We could, in this way, find the torque on the movable plates in a variable condenser of the type shown in Fig. 8–3. Returning to the special case of a parallel-plate condenser, we can use the formula we derived in Chapter 6 for the capacity: \begin{equation} \label{Eq:II:8:16} \frac{1}{C}=\frac{d}{\epsO A}, \end{equation}  where $A$ is the area of each plate. If we increase the separation by $\Delta z$, \begin{equation*} \Delta\biggl(\frac{1}{C}\biggr)=\frac{\Delta z}{\epsO A}. \end{equation*}  From Eq. (8.14) we get that the force between the plates is \begin{equation} \label{Eq:II:8:17} F=\frac{Q^2}{2\epsO A}. \end{equation}  Let’s look at Eq. (8.17) a little more closely and see if we can tell how the force arises. If for the charge on one plate we write \begin{equation*} Q=\sigma A, \end{equation*}  Eq. (8.17) can be rewritten as \begin{equation*} F=\frac{1}{2}\,Q\,\frac{\sigma}{\epsO}. \end{equation*}  Or, since the electric field between the plates is \begin{equation} E_0=\frac{\sigma}{\epsO},\notag \end{equation}  then \begin{equation} \label{Eq:II:8:18} F=\tfrac{1}{2}QE_0. \end{equation}  One would immediately guess that the force acting on one plate is the charge $Q$ on the plate times the field acting on the charge. But we have a surprising factor of one-half. The reason is that $E_0$ is not the field at the charges. If we imagine that the charge at the surface of the plate occupies a thin layer, as indicated in Fig. 8–4, the field will vary from zero at the inner boundary of the layer to $E_0$ in the space outside of the plate. The average field acting on the surface charges is $E_0/2$. That is why the factor one-half is in Eq. (8.18). You should notice that in computing the virtual work we have assumed that the charge on the condenser was constant—that it was not electrically connected to other objects, and so the total charge could not change. Suppose we had imagined that the condenser was held at a constant potential difference as we made the virtual displacement. Then we should have taken \begin{equation*} U=\tfrac{1}{2}CV^2 \end{equation*}  and in place of Eq. (8.15) we would have had \begin{equation*} F\Delta z=\tfrac{1}{2}V^2\,\Delta C, \end{equation*}  which gives a force equal in magnitude to the one in Eq. (8.15) (because $V=Q/C$), but with the opposite sign! Surely the force between the condenser plates doesn’t reverse in sign as we disconnect it from its charging source. Also, we know that two plates with opposite electrical charges must attract. The principle of virtual work has been incorrectly applied in the second case—we have not taken into account the virtual work done on the charging source. That is, to keep the potential constant at $V$ as the capacity changes, a charge $V\,\Delta C$ must be supplied by a source of charge. But this charge is supplied at a potential $V$, so the work done by the electrical system which keeps the potential constant is $V^2\,\Delta C$. The mechanical work $F\,\Delta z$ plus this electrical work $V^2\,\Delta C$ together make up the change in the total energy $\tfrac{1}{2}V^2\,\Delta C$ of the condenser. Therefore $F\,\Delta z$ is $-\tfrac{1}{2}V^2\,\Delta C$, as before.