Question 5 of 10: Field and potential of a radially graded charge cloud

Question 5: Field and potential of a radially graded charge cloud

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A spherically symmetric cloud of charge in free space whose density falls linearly from 1 C/m$^3$ at the centre to zero at $r = 2$ m, with nothing outside.

Find. $\vec{E}$ inside and outside the cloud, and the potential at an exterior point referred to infinity.

Approach. Spherical symmetry makes Gauss' law one-dimensional: integrate the density over a concentric sphere to get the enclosed charge, divide by $4\pi\epsilon_0 r^2$, then integrate the exterior field radially inward from infinity for the potential.

1. Exploit the symmetry. Because $\rho_v$ depends only on $r$, the field can only be radial and can only depend on $r$: $\vec{E} = E_r(r)\hat{a}_r$. Gauss' law over a concentric sphere of radius $r$ then reduces to one scalar equation, $$\oint \vec{D}\cdot d\vec{S} = 4\pi r^2 D_r = Q_{enc}(r) \;\Rightarrow\; E_r(r) = \frac{Q_{enc}(r)}{4\pi\epsilon_0 r^2}$$
2. Integrate the density to get the enclosed charge. Using the thin-spherical-layer element $dv = 4\pi r'^2\,dr'$, $$Q_{enc}(r) = \int_0^r \left(1 - \frac{r'}{2}\right)4\pi r'^2\,dr' = 4\pi\left[\frac{r^3}{3} - \frac{r^4}{8}\right]\ \text{C}$$
3. (a)(i) Field inside the cloud, $r \le 2$ m. Substituting $Q_{enc}$ into the Gauss result, the $4\pi$ and one power of $r^2$ cancel: $$\boxed{\vec{E} = \frac{1}{\epsilon_0}\left(\frac{r}{3} - \frac{r^2}{8}\right)\hat{a}_r\ \text{V/m}, \qquad 0 \le r \le 2\ \text{m}}$$ This is not monotonic: differentiating, $E_r$ peaks where $1/3 = r/4$, i.e. at $r = 4/3$ m, where $E_r = 2.510\times10^{10}$ V/m. Inside that radius the enclosed charge grows faster than $r^2$; beyond it the thinning outer layers can no longer keep up. At $r = 1$ m, $E_r = 2.353\times10^{10}$ V/m.
4. (a)(ii) Field outside the cloud, $r \gt 2$ m. All the charge is now enclosed: $$Q_{tot} = 4\pi\left[\frac{8}{3} - 2\right] = \frac{8\pi}{3} = 8.378\ \text{C}$$ so the cloud looks like a point charge at the origin, $$\boxed{\vec{E} = \frac{Q_{tot}}{4\pi\epsilon_0 r^2}\hat{a}_r = \frac{2/3}{\epsilon_0 r^2}\hat{a}_r = \frac{7.530\times10^{10}}{r^2}\hat{a}_r\ \text{V/m}, \qquad r \gt 2\ \text{m}}$$ At $r = 2$ m this returns $1.882\times10^{10}$ V/m, exactly matching the interior expression there — the continuity check that the boundary carries no surface charge.
5. (b) Potential at an exterior point. With the reference at infinity, $V(r) = -\int_\infty^r \vec{E}\cdot d\vec{l} = \int_r^\infty E_r\,dr'$, and for $r \gt 2$ m the field is the inverse-square form throughout the path, so $$V(r) = \int_r^\infty \frac{Q_{tot}}{4\pi\epsilon_0 r'^2}\,dr' = \frac{Q_{tot}}{4\pi\epsilon_0 r}$$ $$\boxed{V(r,\theta,\phi) = \frac{2/3}{\epsilon_0 r} = \frac{7.530\times10^{10}}{r}\ \text{V}, \qquad r \gt 2\ \text{m}}$$ The potential is independent of $\theta$ and $\phi$, as spherical symmetry requires. Sample values: $3.764\times10^{10}$ V at the cloud surface and $1.506\times10^{10}$ V at $r = 5$ m.
6. Confirm the potential numerically. Integrating the exterior field from $r = 5$ m out to 2000 m and comparing with $V(5) - V(2000)$ from the closed form agrees to six figures, so the boxed potential is consistent with the boxed field.