Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 3.4.2

Half-filled spherical capacitor

A conducting sphere of radius $a$ is held at the potential $V_{0}$ . It is surrounded by a concentric, grounded conducting shell of radius $2 a$ . One half of the space between them is air ( $ε_{0}$ ), the other half is filled with a dielectric of permittivity $3 ε_{0}$ , and the plane interface passes through the centre.

Find $V$ and $E$ between the conductors.
Find the surface charge densities at the points A (air side) and B (dielectric side) of the inner sphere.
Find the charge of the inner sphere and the capacitance.

Adapted from: B. M. Notaroš, Electromagnetics, Example 2.20.

Inner sphere of radius a at V0 inside a grounded shell of radius 2a, upper half air, lower half dielectric 3 epsilon0, points A and B

Try it yourself first.

Strategy

Guess that the potential is still spherically symmetric, solve Laplace's equation for $V (r)$ , and then check the boundary conditions on the flat interface. If they hold, the uniqueness theorem makes the guess the solution. The charges then follow from $ρ_{S} = D_{n}$ , which is different in the two halves.

(a)

In each half the medium is simple and charge-free, so $V$ satisfies Laplace's equation. Both conductors are spheres centred at O and the interface is a plane through O, so we try a potential that depends on $r$ only. In spherical coordinates (Section 2.3) Laplace's equation then reads

\frac{1}{r ^{2}} \frac{d}{d r} (r^{2} \frac{d V}{d r}) = 0 ⟹ r^{2} \frac{d V}{d r} = - A ⟹ V (r) = \frac{A}{r} + B .

The two conductor potentials give

V (a) = \frac{A}{a} + B = V_{0}, V (2 a) = \frac{A}{2 a} + B = 0 ⟹ A = 2 a V_{0}, B = - V_{0} .

Answer

V (r) E = \frac{2 a V _{0}}{r} - V_{0}, = - \nabla V = \frac{2 a V _{0}}{r ^{2}} a_{r} [V/m] (a < r < 2 a)

Interface conditions

This single function must also satisfy the boundary conditions on the plane interface. $E$ is radial, hence tangential to the interface, and has the same value on both sides: $E_{1 τ} = E_{2 τ}$ $✓$ . The normal components of $D = ε_{0} E$ and $3 ε_{0} E$ are both zero: $D_{1 n} = D_{2 n}$ $✓$ . All conditions are met, so by the uniqueness theorem this is the solution, in both halves.

(b)

On the inner sphere the normal out of the conductor is $a_{r}$ , and at $r = a$ the field is $E = 2 V_{0} / a$ . With $ρ_{S} = D \cdot a_{r}$ :

Answer

ρ_{S} (A) ρ_{S} (B) = ε_{0} \frac{2 V _{0}}{a} = \frac{2 ε _{0} V _{0}}{a}, = 3 ε_{0} \frac{2 V _{0}}{a} = \frac{6 ε _{0} V _{0}}{a} [C/ m^{2}]

(c)

Each hemisphere has the area $2 π a^{2}$ and a uniform charge density:

Q = 2 π a^{2} \frac{2 ε _{0} V _{0}}{a} + 2 π a^{2} \frac{6 ε _{0} V _{0}}{a} = 4 π ε_{0} a V_{0} + 12 π ε_{0} a V_{0} = 16 π ε_{0} a V_{0},

Answer

Q C = 16 π ε_{0} a V_{0} [C], = \frac{Q}{V _{0}} = 16 π ε_{0} a [F]

Check

If both halves were air, $C = 4 π ε_{0} ab / (b - a)$ with $b = 2 a$ gives $8 π ε_{0} a$ , and if both were the dielectric, $24 π ε_{0} a$ . Our $16 π ε_{0} a$ is exactly the average, as expected for two halves in parallel.

Remark

The field is the same in both halves, but $D$ and the free charge are three times larger on the dielectric side. The bound charge on the dielectric face at $r = a$ is $P \cdot (- a_{r}) = - 2 ε_{0} E (a) = - 4 ε_{0} V_{0} / a$ , so the total charge there, $6 ε_{0} V_{0} / a - 4 ε_{0} V_{0} / a = 2 ε_{0} V_{0} / a$ , equals that on the air side, which is why $E$ is spherically symmetric. Since $E$ is tangential to the interface, the two halves are two hemispherical capacitors in parallel: a full spherical capacitor with $b = 2 a$ has $C = 4 π ε ab / (b - a) = 8 π ε a$ , and $\frac{1}{2} (8 π ε_{0} a) + \frac{1}{2} (8 π \cdot 3 ε_{0} a) = 16 π ε_{0} a$ $✓$ . The outer shell carries $ρ_{S} = - ε_{0} V_{0} / (2 a)$ and $- 3 ε_{0} V_{0} / (2 a)$ on its two halves, in total $- Q$ .

Home exercise 3.4.2: Half-filled spherical capacitor