In Section 6.5, it was mentioned that the capacitance C=Q/V of a capacitor can be found by either assuming Q and finding V, as in Section 6.5, or by assuming V and finding Q, as in Example 6.10. Use the latter method to derive eq. (6.22).
C=\frac{Q}{V}=\frac{\varepsilon S}{d}
Assume that the parallel plates in Figure 6.13 are maintained at a potential difference V_{o} so that V(x=0) and V(x=d)=V_{o}. This necessitates solving a one-dimensional boundary-value problem; that is, we solve Laplace’s equation
\nabla^2 V=\frac{d^{2}V}{dx^{2}}=0
Integrating twice gives
V=Ax+B
where A and B are integration constants to be determined from the boundary conditions. At x=0, V=0\rightarrow 0=0+B or B=0
and at x=d, V=V_{o}\rightarrow V_{o}=Ad+0 or A=V_{o}/d.
Hence
V=\frac{V_{o}}{d}x
Notice that this solution satisfies Laplace’s equation and the boundary conditions.
We have assumed the potential difference between the plates to be V_{o}. Our goal is to find the charge Q on either plate so that we can eventually find the capacitance C=Q/V_{o}. The charge on either plate is
Q=\int_{S}\rho_{S}dS
But \rho_{S}=D\cdot a_{n}=\varepsilon E\cdot a_{n}, where
E=-\nabla V=-\frac{dV}{dx}a_{x}=-Aa_{x}=-\frac{V_{o}}{d}a_{x}
On the lower plate, a_{n}=a_{x}, so
\rho_{S}=-\frac{\varepsilon V_{o}}{d} and Q=-\frac{\varepsilon V_{o}S}{d}
On the upper plate, a_{n}=-a_{x}, so
\rho_{S}=\frac{\varepsilon V_{o}}{d} and Q=\frac{\varepsilon V_{o}S}{d}
As expected, Q is equal but opposite on each plate. Thus
C=\frac{|Q|}{V_{o}}=\frac{\varepsilon S}{d}
which is in agreement with eq. (6.22).
