Exercise 3.4.2: Point charge inside a dielectric shell

Exercise 3.4.2

Interactive: Radial Field Profiles

About the further exercises of Section 3.4

The ten exercises below are in the spirit of Problem 3.4.1, the home exercises and the worked examples. They use only $D = ε_{0} E + P = ε E$ , Gauss's law $\oint_{S} D \cdot d S = Q_{free}$ , the boundary conditions, the bound charges, $C = Q / V$ , $W = \frac{1}{2} \int D \cdot E d v$ and the image principle.

Exercise 3.4.2

Point charge inside a dielectric shell

cf. the polarisation picture

A point charge $Q = 10$ nC sits at the centre of a spherical dielectric shell, $a = 5$ cm $< r < b = 10$ cm, $ε_{r} = 5$ , and everywhere else is air. Find (a) $D$ , $E$ and $P$ in all regions, (b) the bound charges, and (c) the potential at $r = a$ , compared with the value without the shell.

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 3-12.

Point charge Q at the centre of a dielectric shell a less than r less than b, with negative bound charge on the inner face and positive on the outer face

Try it yourself first.

Strategy

With spherical symmetry, Gauss's law for $D$ gives $D$ everywhere at once, because it sees only the free charge $Q$ and does not care about the shell. Then $E = D / ε$ region by region, and $P = D - ε_{0} E$ .

(a)

By symmetry $D = D_{r} (r) a_{r}$ . Gauss's law on a sphere of radius $r$ encloses only the free charge $Q$ , whatever the medium:

D_{r} 4 π r^{2} = Q ⟹ D = \frac{Q}{4 π r ^{2}} a_{r} for all r .

Then $E = D / ε$ and $P = D - ε_{0} E$ :

Answer

E = ⎩ ⎨ ⎧ \frac{Q}{4 π ε _{0} r ^{2}} a_{r} \frac{Q}{4 π ε _{r} ε _{0} r ^{2}} a_{r} r < a, r > b a < r < b

P = ⎩ ⎨ ⎧ 0 (1 - \frac{1}{ε _{r}}) \frac{Q}{4 π r ^{2}} a_{r} r < a, r > b a < r < b

$E$ jumps from $36.0$ kV/m to $7.20$ kV/m at $r = a$ and from $1.80$ kV/m to $9.00$ kV/m at $r = b$ , while $D_{r}$ is continuous at both surfaces (Figure 3.4.31).

Exercise 3.4.2: the field E (orange) and D/\varepsilon_0 (blue, dashed) of a 10 nC point charge at the centre of a dielectric shell. D ignores the shell, while E drops by the factor \varepsilon_r=5 inside it and returns to the vacuum value outside. — **Figure 3.4.31.** Exercise 3.4.2: the field $E$ (orange) and $D / ε_{0}$ (blue, dashed) of a $10$ nC point charge at the centre of a dielectric shell. $D$ ignores the shell, while $E$ drops by the factor $ε_{r} = 5$ inside it and returns to the vacuum value outside.

(b)

Inside the shell $\nabla \cdot P = \frac{1}{r ^{2}} \frac{d}{d r} (r^{2} P_{r}) = 0$ , because $r^{2} P_{r}$ is constant, so $ρ_{b} = 0$ . On the faces, with the outward normals $- a_{r}$ at $r = a$ and $+ a_{r}$ at $r = b$ :

ρ_{S b} (a) ρ_{S b} (b) = - (1 - \frac{1}{ε _{r}}) \frac{Q}{4 π a ^{2}} = - 255 nC/ m^{2}, = + (1 - \frac{1}{ε _{r}}) \frac{Q}{4 π b ^{2}} = + 63.7 nC/ m^{2} .

Multiplying by the areas $4 π a^{2}$ and $4 π b^{2}$ :

Answer

Q_{b} (a) Q_{b} (b) = - (1 - \frac{1}{ε _{r}}) Q = - 8 nC, = + 8 nC

(c)

Integrating $E$ inwards from infinity:

V (b) V (a) = \frac{Q}{4 π ε _{0} b} = 900 V, = V (b) + \int_{a}^{b} \frac{Q d r}{4 π ε _{r} ε _{0} r ^{2}} = \frac{Q}{4 π ε _{0} b} + \frac{Q}{4 π ε _{r} ε _{0}} (\frac{1}{a} - \frac{1}{b}) = 1080 V,

against $Q / (4 π ε_{0} a) = 1800$ V without the shell.

Check

The total bound charge is $- 8 + 8 = 0$ nC, as for any neutral body.

Remark

Inside the shell the field is that of the charge $Q - 8 nC = Q / ε_{r}$ , i.e. $Q$ screened by the bound charge on the inner face. Outside, the two bound layers cancel and the field is that of $Q$ alone: a dielectric shell cannot shield a charge. The total bound charge is zero, as it must be.

Change $ε_{r}$ and the radii of the shell in the tool below: $D_{r}$ ignores the shell, while $E_{r}$ drops by the factor $ε_{r}$ inside it and returns to the vacuum curve outside.

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