Page 181 - Engineering Electromagnetics, 8th Edition

CHAPTER 6 Capacitance 163

and the partial derivative may be replaced by an ordinary derivative, since V is not a
function of y or z,
2
d V = 0
dx 2
We integrate twice, obtaining
dV = A
dx
and
V = Ax + B (31)
where A and B are constants of integration. Equation (31) contains two such constants,
as we would expect for a second-order differential equation. These constants can be
determined only from the boundary conditions.
Since the field varies only with x and is not a function of y and z, then V is a
constant if x is a constant or, in other words, the equipotential surfaces are parallel
planes normal to the x axis. The field is thus that of a parallel-plate capacitor, and as
soon as we specify the potential on any two planes, we may evaluate our constants of
integration.

EXAMPLE 6.2
Start with the potential function, Eq. (31), and find the capacitance of a parallel-plate
capacitor of plate area S, plate separation d, and potential difference V 0 between
plates.
Solution. Take V = 0at x = 0 and V = V 0 at x = d. Then from Eq. (31),
V 0
A = B = 0
d
and

V 0 x
V = (32)
d
We still need the total charge on either plate before the capacitance can be found.
We should remember that when we first solved this capacitor problem, the sheet of
charge provided our starting point. We did not have to work very hard to find the
charge, for all the fields were expressed in terms of it. The work then was spent in
finding potential difference. Now the problem is reversed (and simplified).
The necessary steps are these, after the choice of boundary conditions has been
made:
1. Given V, use E =−∇V to find E.
2. Use D = E to find D.
3. Evaluate D at either capacitor plate, D = D S = D N a N .
4. Recognize that ρ S = D N .

5. Find Q by a surface integration over the capacitor plate, Q = S ρ S dS.