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

Problem 3.3.3

Laplace's equation in $r$ only

Suppose that there is no source between two concentric spherical conductors of radii $a$ and $b$ . The conductors at $r = a$ and $r = b$ have potentials of $V_{a}$ and $V_{b}$ , respectively. Determine the electric field between these spherical conductors. Find the total charge on the outer sphere.

Try it yourself first.

Strategy

Spherical symmetry: $V = V (r)$ , so only the radial part of the spherical Laplacian (3.3.17) survives. Integrate twice, impose $V (a) = V_{a}$ and $V (b) = V_{b}$ , then use $E = - \nabla V$ and $ρ_{S} = D_{n}$ .

Symmetry and the equation

Laplace's equation in spherical coordinates (3.3.17) is

Δ V = \frac{1}{r ^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial V}{\partial r}) + \frac{1}{r ^{2} sin θ} \frac{\partial}{\partial θ} (sin θ \frac{\partial V}{\partial θ}) + \frac{1}{r ^{2} sin ^{2} θ} \frac{\partial ^{2} V}{\partial ϕ ^{2}} = 0.

The geometry and the boundary values are spherically symmetric, so $V = V (r)$ and

\frac{1}{r ^{2}} \frac{d}{d r} (r^{2} \frac{d V}{d r}) = 0 \Rightarrow r^{2} \frac{d V}{d r} = C \Rightarrow \frac{d V}{d r} = \frac{C}{r ^{2}} \Rightarrow V (r) = - \frac{C}{r} + D .

Boundary conditions

$V (a) = V_{a}$ and $V (b) = V_{b}$ give

- \frac{C}{a} + D = V_{a}, - \frac{C}{b} + D = V_{b} .

Subtracting the first from the second, $C (\frac{1}{a} - \frac{1}{b}) = C \frac{b - a}{ab} = V_{b} - V_{a}$ , and then $D = V_{a} + C / a$ :

C = \frac{ab ( V _{b} - V _{a} )}{b - a}, D = \frac{b V _{b} - a V _{a}}{b - a} .

Hence

V (r) = \frac{ab ( V _{a} - V _{b} )}{b - a} \frac{1}{r} + \frac{b V _{b} - a V _{a}}{b - a} [V] .

Field

Answer

E = - \nabla V = - \frac{d V}{d r} a_{r} = \frac{ab ( V _{a} - V _{b} )}{b - a} \frac{1}{r ^{2}} a_{r} [V/m]

Charge on the outer sphere

On the inner face of the outer sphere ( $r = b$ , area $4 π b^{2}$ ) the normal pointing into the gap is $- a_{r}$ :

ρ_{S} = D \cdot (- a_{r})_{r = b} = - ε_{0} E_{r} (b) = - ε_{0} \frac{a ( V _{a} - V _{b} )}{b ( b - a )},

Answer

Q_{outer} = 4 π b^{2} ρ_{S} = - 4 π ε_{0} \frac{ab ( V _{a} - V _{b} )}{b - a} [C]

Problem 3.3.3 (drawn for V_a>V_b): equipotential spheres r=\text{const} and the radial field between the conductors. — **Figure 3.3.24.** Problem 3.3.3 (drawn for $V_{a} > V_{b}$ ): equipotential spheres $r = const$ and the radial field between the conductors.

Right: V varies as 1/r (plus a constant), E_r as 1/r^2. — Right: $V$ varies as $1/ r$ (plus a constant), $E_{r}$ as $1/ r^{2}$ .

Check

At $r = a$ ,

\frac{ab ( V _{a} - V _{b} )}{( b - a ) a} + \frac{b V _{b} - a V _{a}}{b - a} = \frac{b V _{a} - a V _{a}}{b - a} = V_{a},

and similarly $V (b) = V_{b}$ . For $V_{a} > V_{b}$ the field points outwards and the outer sphere carries negative charge, and $4 π ε_{0} ab V / (b - a)$ has the unit $(F/m) m V = C$ .

Reading the answer

$V \propto 1/ r$ and $E \propto 1/ r^{2}$ : between the spheres the field is that of a point charge $Q_{a} = 4 π ε_{0} ab (V_{a} - V_{b}) / (b - a)$ at the centre, as Gauss's law demands (Section 3.2), and $Q_{outer} = - Q_{a}$ (Figure 3.3.24). Letting $b \to \infty$ with $V_{b} = 0$ gives $V = V_{a} a / r$ , the potential of an isolated charged sphere. The same $1/ r$ solution describes the current field around the earth electrode of Application 3.3.2. The ratio $Q_{a} / (V_{a} - V_{b}) = 4 π ε_{0} ab / (b - a)$ is the capacitance of the spherical capacitor (Home exercise 3.3.9).

Problem 3.3.3: Laplace's equation in r only