Page 167 - Engineering Electromagnetics, 8th Edition

CHAPTER 6 Capacitance 149

where C 1 = 1 S/d 1 and C 2 = 2 S/d 2 . This is the correct result, but we can obtain it
using less intuition and a more basic approach.
Because the capacitance definition, C = Q/V ,involves a charge and a voltage,
we may assume either and then find the other in terms of it. The capacitance is not a
function of either, but only of the dielectrics and the geometry. Suppose we assume
a potential difference V 0 between the plates. The electric field intensities in the two
regions, E 2 and E 1 , are both uniform, and V 0 = E 1 d 1 + E 2 d 2 .At the dielectric
interface, E is normal, and our boundary condition, Eq. (35) Chapter 5, tells us that
D N1 = D N2 ,or 1 E 1 = 2 E 2 . This assumes (correctly) that there is no surface charge
at the interface. Eliminating E 2 in our V 0 relation, we have

V 0
E 1 =
d 1 + d 2 ( 1 / 2 )
and the surface charge density on the lower plate therefore has the magnitude
V 0
ρ S1 = D 1 = 1 E 1 =
d 1 d 2
+
1 2
Because D 1 = D 2 , the magnitude of the surface charge is the same on each plate.
The capacitance is then
Q ρ S S 1 1
C = = = = 1 1
V 0 V 0 d 1 d 2
1 S + 2 S C 1 + C 2
As an alternate (and slightly simpler) solution, we might assume a charge Q on
one plate, leading to a charge density Q/S and a value of D that is also Q/S. This is
true in both regions, as D N1 = D N2 and D is normal. Then E 1 = D/ 1 = Q/( 1 S),
E 2 = D/ 2 = Q/( 2 S), and the potential differences across the regions are V 1 =
E 1 d 1 = Qd 1 /( 1 S), and V 2 = E 2 d 2 = Qd 2 /( 2 S). The capacitance is

Q Q 1
C = = = (9)
V V 1 + V 2 d 1 d 2
1 S + 2 S
How would the method of solution or the answer change if there were a third
conducting plane along the interface? We would now expect to find surface charge on
each side of this conductor, and the magnitudes of these charges should be equal. In
other words, we think of the electric lines not as passing directly from one outer plate
to the other, but as terminating on one side of this interior plane and then continuing
on the other side. The capacitance is unchanged, provided, of course, that the added
conductor is of negligible thickness. The addition of a thick conducting plate will
increase the capacitance if the separation of the outer plates is kept constant, and this
is an example of a more general theorem which states that the replacement of any
portion of the dielectric by a conducting body will cause an increase in the capacitance.
If the dielectric boundary were placed normal to the two conducting plates and
the dielectrics occupied areas of S 1 and S 2 , then an assumed potential difference V 0
would produce field strengths E 1 = E 2 = V 0 /d. These are tangential fields at the