Question 7 of 7: Charged conducting sphere

Question 7: Charged conducting sphere (20 marks)

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 uniformly charged hollow conducting sphere centred on the origin in free space, and a small test charge moved radially inward along the $x$ axis.

Find. The total charge; the exterior field by Gauss' law; the work done moving the test charge inward; and the potential difference between the two points.

Approach. Multiply the surface density by the spherical surface area, exploit spherical symmetry in Gauss' law to extract E, integrate E to get the potential, and evaluate the work as $q\Delta V$.

1. Total charge on the sphere. The charge is spread uniformly over the full spherical surface: $$Q = \rho_s\,(4\pi a^{2}) = (20\times 10^{-6})\,4\pi(0.01)^{2} = \boxed{25.13\ \text{nC}}.$$
2. Choose the symmetry and the Gaussian surface. The charge distribution is invariant under any rotation about the origin, so in spherical coordinates E can depend only on $r$ and can only point radially: $\mathbf{E} = E_r(r)\,\hat{\mathbf{a}}_r$. Choose as the Gaussian surface a concentric sphere of radius $r \gt a$; on it $E_r$ is constant and $d\mathbf{S} = dS\,\hat{\mathbf{a}}_r$ is everywhere parallel to E.
3. Apply Gauss' law and solve for the field. $$\oint_S \mathbf{D}\cdot d\mathbf{S} = Q_{\text{enc}} \;\Longrightarrow\; \varepsilon_0 E_r\,(4\pi r^{2}) = Q \;\Longrightarrow\; E_r = \frac{Q}{4\pi\varepsilon_0 r^{2}} = \frac{\rho_s a^{2}}{\varepsilon_0 r^{2}}.$$ Numerically $Q/(4\pi\varepsilon_0) = 225.9\ \text{V}\cdot\text{m}$, so $$\boxed{\mathbf{E}(r) = \frac{225.9}{r^{2}}\, \hat{\mathbf{a}}_r\ \text{V/m}\quad (r \gt 1\ \text{cm}, \ r\ \text{in metres})}.$$ Outside the conductor the field is indistinguishable from that of a point charge $Q$ at the origin. As a check, at the surface $E(a) = \rho_s/\varepsilon_0 = 2.259\ \text{MV/m}$, which is the standard conductor boundary condition.
4. Build the potential function. Integrating the radial field inward from infinity (the usual reference), $$V(r) = -\int_{\infty}^{r} E_r\,dr' = \frac{Q}{4\pi\varepsilon_0 r} = \frac{225.9}{r}\ \text{V},$$ so $$V(r_A) = \frac{225.9}{0.05} = 4517.7\ \text{V}, \qquad V(r_B) = \frac{225.9}{0.03} = 7529.5\ \text{V}.$$
5. Work done moving the test charge. The work an external agent must do against the field, moving the charge quasi-statically from A to B, is $$W = -q\int_{A}^{B}\mathbf{E}\cdot d\mathbf{l} = q\left[V(r_B) - V(r_A)\right] = (1\times 10^{-9})(7529.5 - 4517.7)$$ $$\Longrightarrow\; \boxed{W = 3.01\ \mu\text{J}}.$$ The sign is positive and must be: both charges are positive, so pushing the test charge closer costs energy. (The field itself does $-3.01\ \mu$J of work, and that energy is stored in the configuration.) Because the electrostatic field is conservative, the path along the $x$ axis is irrelevant — only the two radii matter. A direct numerical evaluation of $-q\int_{0.05}^{0.03}(225.9/r^{2})\,dr$ returns the same 3.012 $\mu$J.
6. Potential difference between the two points. $$V_{BA} = V(r_B) - V(r_A) = 7529.5 - 4517.7 = \boxed{3.01\ \text{kV}}$$ with B (the inner point, 3 cm) at the higher potential. This is just $W/q$, so parts (c) and (d) are the same physical statement expressed per-coulomb rather than in joules — a useful self-check that the two answers are consistent.