Resistance vs. Capacitance

a) Consider two conductors of some shape. Use them in two alternative ways, as a capacitor and as a resistor as shown in (a) and (b) of Figure 1.1, respectively. In case (a), the space between the conductors is filled with a homogeneous material of permittivity ε, while in case (b), it is filled with a homogeneous material of finite conductivity σ. By considering separately these two cases, prove the relation

Figure 1.1

between the capacitance C in case (a) and the resistance R in case

(b). If you cannot give a general proof, try at least some special example, e.g., parallel plates.

b) Two conducting spheres have their centers a distance apart. Their respective radii are a and b. Show that when c >> a, b the capacitance of this system will be given approximately by

c) Two small, spherical, perfectly conducting electrodes of radii a and b are embedded in an infinite medium of conductivity σ. Their centers are separated by a distance c >> a, b Find the resistance between them without using (a) and (b).

Hint: If two electrodes at potentials V₁ and V₂ are embedded in a medium of finite conductivity, the currents I₁ and I₂ leaving each of them are related to the potentials by the formulae V₁ = R₁₁ I₁ + R₁₂ I₂, V₂ = R₂₁ I₁ + R₂₂ I_2. Determine the coefficients R_ij by considering cases with I₂ = 0 and I₁ = 0.

d) Check the results of (b) and (c) by using (a).

SOLUTION

a) Enclose one of the conductors in a surface and use Maxwell’s equation (see Figures 1.2a and 1.2b)

So

(1)

Take the volume integral of (1) and transform into a surface integral:

(2)

Figures 1.2a, b

Using this result and the definition of C, we find

which yields

(3)

Now treat the resistor problem by starting with J = σE. Finding the current flux through the same surface,

So

(4)

Equating (3) and (4), we find

and finally

(5)

The parallel plate capacitor has the following capacitance and resistance:

Thus, we confirm the general result for RC.

b) Find the potential at the surface of each conductor:

(6)

So

(7)

c) Following the hint, consider the potential drop from each conductor to infinity when first I₂ = 0, and then I₁ = 0:

so

In the same fashion

Now, V₁ = (R₁₁ – R₁₂)I and V₂ = (R₂₁ – R₂₂)I, where I = I₁ = -I₂. So we obtain for R

(8)

d) Multiplying (7) by (8), we find the result in (5) with ε = 1