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

Home exercise 3.4.3

Interactive: Radial Field Profiles

Home exercise 3.4.3

Two-layer spherical capacitor

A spherical capacitor has an inner conductor of radius $a$ at the potential $V_{a}$ and an outer conductor of radius $b$ at the potential $V_{b}$ . The region $a < r < d$ is filled with a dielectric of permittivity $ε_{1}$ , the region $d < r < b$ with $ε_{2}$ .

Find $V$ and $E$ in both regions.
Find $ρ_{S}$ on the outer conductor and its charge $Q_{b}$ .
Find the capacitance $C$ and the stored energy $W$ .

Hint: solve Laplace's equation in each layer. At $r = d$ , $V$ and $D_{r}$ are continuous.

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Problem P.3-37.

Spherical capacitor with inner radius a, interface radius d and outer radius b, permittivities epsilon1 and epsilon2

Try it yourself first.

Strategy

This is the "solve in each region, then glue" method. Laplace's equation gives $A / r + B$ in each layer: four constants. Two conductor potentials and two interface conditions (continuity of $V$ and of $D_{r}$ ) fix them.

Laplace's equation holds in each layer and, by spherical symmetry, $V = V (r)$ :

V_{1} (r) V_{2} (r) = \frac{A _{1}}{r} + B_{1} (a < r < d), = \frac{A _{2}}{r} + B_{2} (d < r < b) .

Four conditions are needed to fix the four constants. Two are the conductor potentials,

V_{1} (a) = V_{a}, V_{2} (b) = V_{b},

and two are the boundary conditions at the interface $r = d$ , which carries no free charge: $V$ is continuous (first condition) and so is $D_{r} = - ε d V / d r$ (second condition),

V_{1} (d) = V_{2} (d), ε_{1} \frac{A _{1}}{d ^{2}} = ε_{2} \frac{A _{2}}{d ^{2}} \Rightarrow ε_{1} A_{1} = ε_{2} A_{2} .

The last condition says that $D_{r} = ε_{1} A_{1} / r^{2} = ε_{2} A_{2} / r^{2}$ is the same function in both layers, as Gauss's law requires. Writing $V_{a} - V_{b} = \int_{a}^{b} E_{r} d r$ with $E_{r} = A_{1} / r^{2}$ in layer 1 and $E_{r} = A_{2} / r^{2} = (ε_{1} / ε_{2}) A_{1} / r^{2}$ in layer 2,

V_{a} - V_{b} = A_{1} [(\frac{1}{a} - \frac{1}{d}) + \frac{ε _{1}}{ε _{2}} (\frac{1}{d} - \frac{1}{b})] = A_{1} \frac{ε _{2} b ( d - a ) + ε _{1} a ( b - d )}{ε _{2} ab d} .

With the shorthand $N = ε_{2} b (d - a) + ε_{1} a (b - d)$ ,

A_{1} A_{2} = \frac{ε _{2} ab d ( V _{a} - V _{b} )}{N}, = \frac{ε _{1} ab d ( V _{a} - V _{b} )}{N} .

(a)

Then $B_{1} = V_{a} - A_{1} / a$ and $B_{2} = V_{b} - A_{2} / b$ give $V_{1} = V_{a} - A_{1} (r - a) / (a r)$ and $V_{2} = V_{b} + A_{2} (b - r) / (b r)$ :

Answer

V_{1} (r) V_{2} (r) = V_{a} - \frac{ε _{2} b d ( V _{a} - V _{b} )}{N} \frac{r - a}{r}, = V_{b} + \frac{ε _{1} a d ( V _{a} - V _{b} )}{N} \frac{b - r}{r}

Answer

E_{1} E_{2} = \frac{ε _{2} ab d ( V _{a} - V _{b} )}{N} \frac{a _{r}}{r ^{2}}, = \frac{ε _{1} ab d ( V _{a} - V _{b} )}{N} \frac{a _{r}}{r ^{2}} [V/m]

(b)

On the inner face of the outer conductor the normal pointing out of the conductor, into the gap, is $- a_{r}$ :

ρ_{S} (b) = D_{2} \cdot (- a_{r})_{r = b} = - ε_{2} \frac{A _{2}}{b ^{2}} = - \frac{ε _{1} ε _{2} a d ( V _{a} - V _{b} )}{b N},

Answer

ρ_{S} (b) Q_{b} = \frac{ε _{1} ε _{2} a d ( V _{b} - V _{a} )}{b N}, = 4 π b^{2} ρ_{S} (b) = \frac{4 π ε _{1} ε _{2} ab d ( V _{b} - V _{a} )}{N} [C]

(c)

The inner conductor carries $Q_{a} = - Q_{b}$ (check: $4 π a^{2} ε_{1} A_{1} / a^{2} = 4 π ε_{1} A_{1} = - Q_{b}$ ), and

Answer

C = \frac{Q _{b}}{V _{b} - V _{a}} = \frac{4 π ε _{1} ε _{2} ab d}{ε _{2} b ( d - a ) + ε _{1} a ( b - d )} [F]

The energy follows from $w = \frac{1}{2} ε E^{2}$ integrated over the two spherical shells, with the volume element $d v = 4 π r^{2} d r$ and each layer's own permittivity:

W = \frac{1}{2} \int_{a}^{d} ε_{1} \frac{A _{1}^{2}}{r ^{4}} 4 π r^{2} d r + \frac{1}{2} \int_{d}^{b} ε_{2} \frac{A _{2}^{2}}{r ^{4}} 4 π r^{2} d r = 2 π ε_{1} A_{1}^{2} \frac{d - a}{a d} + 2 π ε_{2} A_{2}^{2} \frac{b - d}{b d},

which, with $A_{1}$ and $A_{2}$ inserted, reduces to

Answer

W = \frac{1}{2} C (V_{b} - V_{a})^{2} = \frac{Q _{b}^{2}}{2 C} [J]

Check

$ε_{1} = ε_{2} = ε$ gives $N = ε d (b - a)$ and $C = 4 π ε ab / (b - a)$ , the single-dielectric result.

Remark

\frac{1}{C} = \frac{1}{4 π ε _{1}} (\frac{1}{a} - \frac{1}{d}) + \frac{1}{4 π ε _{2}} (\frac{1}{d} - \frac{1}{b}) :

two spherical capacitors in series (Home exercise 3.4.1), since the interface $r = d$ is an equipotential. Because $E \propto 1/ (ε r^{2})$ , a layer of high permittivity next to the inner conductor (where $r$ is small) evens out the field. This is the principle of graded insulation in high-voltage cables (Exercise 3.4.3).

Move the interface radius $d$ and change $ε_{1}$ and $ε_{2}$ in the tool below: $D_{r}$ stays continuous at $r = d$ while $E_{r}$ jumps by the factor $ε_{1} / ε_{2}$ , and putting the higher permittivity inside evens out the field across the gap.

InteractiveLoading simulation…

Home exercise 3.4.3: Two-layer spherical capacitor