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

Home exercise 3.2.2

Interactive: Radial Field Profiles

Home exercise 3.2.2

Nonuniformly charged sphere

A sphere of radius $b$ carries the volume charge density

ρ_{v} = k r^{2} (r < b),

where $k$ is a constant. (a) Find the total charge of the sphere. (b) By using Gauss's law, find the electric field in the regions $r < b$ and $r > b$ .

Try it yourself first.

Strategy

The density depends on $r$ only, so the symmetry is still spherical and the concentric Gaussian sphere works. The only new element is that the enclosed charge must be found by integrating $ρ_{v}$ , because with a density that varies with $r$ we cannot simply multiply by a volume.

The total charge is the volume integral of the density over the sphere, with $d V = r^{2} sin θ d r d θ d ϕ$ :

Q = \int_{r = 0}^{b} \int_{θ = 0}^{π} \int_{ϕ = 0}^{2 π} k r^{2} r^{2} sin θ d ϕ d θ d r = k ϕ (2 π) θ (2) \int_{0}^{b} r^{4} d r = 4 π k \frac{b ^{5}}{5},

Answer

Q = \frac{4 π}{5} k b^{5} [C]

The distribution is spherically symmetric and $E = E_{r} (r) a_{r}$ . For the left-hand side of Eq. (3.2.13), exactly as in Home exercise 3.2.1,

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} .

For the right-hand side, we can write for $r < b$ (primed variables for the integration inside the Gaussian sphere)

total charge inside S = \int_{r^{'} = 0}^{r} \int_{θ = 0}^{π} \int_{ϕ = 0}^{2 π} k (r^{'})^{4} sin θ d ϕ d θ d r^{'} = \frac{4 π}{5} k r^{5},

and equating both sides, $4 π r^{2} ε_{0} E_{r} = \frac{4 π}{5} k r^{5}$ , which gives

Answer

E = \frac{k r ^{3}}{5 ε _{0}} a_{r} for r < b

For $r > b$ the Gaussian sphere encloses the whole charge,

total charge inside S = Q = \frac{4 π}{5} k b^{5},

which gives

Answer

E = \frac{k b ^{5}}{5 ε _{0} r ^{2}} a_{r} for r > b

Home exercise 3.2.2. With the charge concentrated towards the surface (\rho_v=kr^2), the field rises as r^3 inside, more slowly at first than the linear rise of a uniform sphere with the same total charge (dashed), and reaches the same value kb^3/(5\varepsilon_0) at the surface. — **Figure 3.2.24.** Home exercise 3.2.2. With the charge concentrated towards the surface ( $ρ_{v} = k r^{2}$ ), the field rises as $r^{3}$ inside, more slowly at first than the linear rise of a uniform sphere with the same total charge (dashed), and reaches the same value $k b^{3} / (5 ε_{0})$ at the surface.

Change $k$ and the radius below and watch the $r^{3}$ rise inside and the $1/ r^{2}$ fall outside.

InteractiveLoading simulation…

Check

Both expressions give $k b^{3} / (5 ε_{0})$ at $r = b$ : the field is continuous, as it must be without surface charge. Units: $k$ is in $C/ m^{5}$ , so $k r^{3} / ε_{0}$ is in $(C/ m^{2}) / (F/m) = V/m$ . Because the charge is concentrated towards the surface, little charge is enclosed at small $r$ and the field rises as $r^{3}$ (Figure 3.2.24). A check with the differential form of Gauss's law in spherical coordinates:

\nabla \cdot D = \frac{1}{r ^{2}} \frac{d}{d r} (r^{2} ε_{0} \frac{k r ^{3}}{5 ε _{0}}) = \frac{1}{r ^{2}} \frac{d}{d r} (\frac{k r ^{5}}{5}) = k r^{2} = ρ_{v} .

Home exercise 3.2.2: Nonuniformly charged sphere