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

Problem 3.2.5

Interactive: Radial Field Profiles

Problem 3.2.5

Spherical shell with volume charge

By using Gauss's law, find the electric field in the regions $r < a$ , $a < r < b$ and $r > b$ regarding the spherical shell with an inner and outer radii of $a$ and $b$ , respectively. Note that the spherical shell carries a uniform volume charge density of $ρ_{v}$ .

Try it yourself first.

Strategy

Spherical symmetry: the field is radial and depends on $r$ only, so a concentric sphere is a Gaussian surface. One sphere, drawn with three different radii, serves all three regions, and only the enclosed charge (a volume of the shell times $ρ_{v}$ ) changes.

Symmetry

The charge distribution is unchanged by every rotation about the centre. The field must therefore be radial (any tangential component would single out a direction on the sphere) and its magnitude can depend on $r$ only:

E = E_{r} (r) a_{r} .

Gaussian surface and left-hand side

We take a sphere of radius $r$ concentric with the shell (Figure 3.2.21). On it

d S = r^{2} sin θ d θ d ϕ a_{r}, E \cdot d S = E_{r} r^{2} sin θ d θ d ϕ,

with $E_{r}$ constant on $S$ , to give

ε_{0} \oint_{S} E \cdot d S = ε_{0} \int_{θ = 0}^{π} \int_{ϕ = 0}^{2 π} E_{r} r^{2} sin θ d ϕ d θ = 4 π r^{2} ε_{0} E_{r} .

The same Gaussian sphere, drawn with three different radii, serves all three regions. Only the enclosed charge changes.

Region

r < a

The Gaussian sphere lies in the empty cavity,

total charge inside S = 0,

which gives

E = 0 for r < a .

Region

a < r < b

The Gaussian sphere encloses the part of the shell between radii $a$ and $r$ , whose volume is $\frac{4 π}{3} r^{3} - \frac{4 π}{3} a^{3} = \frac{4 π}{3} (r^{3} - a^{3})$ :

total charge inside S = ρ_{v} \frac{4 π}{3} (r^{3} - a^{3}) .

Equating both sides,

\Rightarrow 4 π r^{2} ε_{0} E_{r} = ρ_{v} \frac{4 π}{3} (r^{3} - a^{3}) E_{r} = \frac{ρ _{v}}{3 ε _{0}} \frac{r ^{3} - a ^{3}}{r ^{2}} .

Region

r > b

The Gaussian sphere encloses the whole shell,

total charge inside S = ρ_{v} \frac{4 π}{3} (b^{3} - a^{3}) = Q,

and equating both sides,

\Rightarrow 4 π r^{2} ε_{0} E_{r} = ρ_{v} \frac{4 π}{3} (b^{3} - a^{3}) E_{r} = \frac{ρ _{v}}{3 ε _{0}} \frac{b ^{3} - a ^{3}}{r ^{2}} .

So we find

Answer

E = ⎩ ⎨ ⎧ 0, \frac{ρ _{v}}{3 ε _{0}} \frac{r ^{3} - a ^{3}}{r ^{2}} a_{r}, \frac{ρ _{v}}{3 ε _{0}} \frac{b ^{3} - a ^{3}}{r ^{2}} a_{r}, r < a a < r < b r > b

where, for $r > b$ ,

E = \frac{ρ _{v}}{3 ε _{0}} \frac{b ^{3} - a ^{3}}{r ^{2}} a_{r} = \frac{Q}{4 π ε _{0} r ^{2}} a_{r} [V/m]

Problem 3.2.5. E_r for a shell with b=2a, in units of \rho_va/(3\varepsilon_0): zero in the cavity, rising through the shell to its maximum \tfrac74 at r=b, then the 1/r^2 field of the total charge Q=\frac43\pi\rho_v(b^3-a^3) placed at the centre (dashed continuation). — **Figure 3.2.22.** Problem 3.2.5. $E_{r}$ for a shell with $b = 2 a$ , in units of $ρ_{v} a / (3 ε_{0})$ : zero in the cavity, rising through the shell to its maximum $\frac{7}{4}$ at $r = b$ , then the $1/ r^{2}$ field of the total charge $Q = \frac{4}{3} π ρ_{v} (b^{3} - a^{3})$ placed at the centre (dashed continuation).

Change the radii and the charge density below and watch the field profile: zero in the cavity, rising through the shell, and the $1/ r^{2}$ field of the total charge outside.

InteractiveLoading simulation…

Check

$E_{r}$ is continuous at $r = a$ (both expressions give $0$ ) and at $r = b$ (both give $ρ_{v} (b^{3} - a^{3}) / (3 ε_{0} b^{2})$ ): there is no surface charge, so no jump. The field rises through the shell, peaks at $r = b$ and outside equals the field of a point charge $Q$ at the centre (Figure 3.2.22): a spherically symmetric charge distribution looks, from outside, like a point charge. The field in the cavity is zero whatever the thickness of the shell. With $a \to 0$ the result becomes the uniformly charged solid sphere of Home exercise 3.2.1.

Problem 3.2.5: Spherical shell with volume charge