Question 2 of 7: Charge Distribution and Field of a Charged Conducting Spherical Shell
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2019 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only, one two-sided aid sheet permitted). Constants given: $\varepsilon_0=8.854\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.602\times10^{-19}$ C. Format: seven questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All seven are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — Coulomb's law and superposition of point charges, Gauss's law and conductors in electrostatic equilibrium, Ampère's law for cylindrical and sheet current distributions, the force between parallel currents, Faraday's law (rotational and transformer EMF), parallel-plate capacitance and the force between charged plates, and magnetic-circuit (reluctance) analysis of a gapped core.
Question 2: Charge Distribution and Field of a Charged Conducting Spherical Shell (20 marks)
Net charge on the conducting shell, $q_{\text{shell}}$
$10$ nC
Field points $P_1$, $P_2$
$0.95$ m, $1.05$ m from centre
Gaussian sphere radius
$2$ m
Find.(a) charge on the shell's inner surface. (b) charge density on the outer surface. (c) $E$ at $P_1$ ($0.95$ m) and $P_2$ ($1.05$ m). (d) flux through a radius-2 m Gaussian sphere.
Cross-section: point charge at the centre of the cavity; the grey annulus is the conducting shell (0.90m to 1.00m radius). P1 lies inside the conductor itself; P2 lies outside the shell.
Approach. Inside a conductor in electrostatic equilibrium $E=0$, which by Gauss's law forces the inner-surface charge to exactly cancel the enclosed point charge; the outer surface then carries whatever remains of the shell's own net charge. Outside the shell, all enclosed charge (point charge plus shell) acts as if concentrated at the centre.
Part (a) — inner-surface charge. A Gaussian surface drawn inside the conductor material must enclose zero net charge ($E=0$ there), so the inner surface carries charge exactly cancelling the enclosed point charge:
$$\boxed{q_{\text{inner}}=-40\ \text{nC}}$$
Part (b) — outer-surface charge density. The shell's total net charge splits between its two surfaces: $q_{\text{inner}}+q_{\text{outer}}=q_{\text{shell}}$, so
$$q_{\text{outer}}=q_{\text{shell}}-q_{\text{inner}}=10-(-40)=50\ \text{nC}$$
Spread over the outer surface area $4\pi R_{\text{out}}^2$:
$$\sigma_{\text{outer}}=\frac{q_{\text{outer}}}{4\pi R_{\text{out}}^2}=\frac{50\times10^{-9}}{4\pi(1.00)^2}$$
$$\boxed{\sigma_{\text{outer}}=3.979\times10^{-9}\ \text{C/m}^2=3.979\ \text{nC/m}^2}$$
Part (c) — field at $P_1$ and $P_2$. $P_1=0.95$ m lies INSIDE the conductor material itself ($R_{\text{in}}=0.90 \lt 0.95 \lt R_{\text{out}}=1.00$), so
$$\boxed{E(P_1)=0}$$
$P_2=1.05$ m lies outside the whole assembly; by Gauss's law the enclosed charge is $q+q_{\text{shell}}=40+10=50$ nC, acting as a point charge at the centre:
$$E(P_2)=\frac{k(q+q_{\text{shell}})}{P_2^2}=\frac{(8.988\times10^9)(50\times10^{-9})}{(1.05)^2}$$
$$\boxed{E(P_2)=407.6\ \text{V/m, radially outward}}$$
Part (d) — flux through the $r=2$ m Gaussian sphere. By Gauss's law, flux depends only on the total enclosed charge (the same $50$ nC found above, since the $r=2$m sphere still encloses everything):
$$\Phi=\frac{Q_{\text{enc}}}{\varepsilon_0}=\frac{50\times10^{-9}}{8.854\times10^{-12}}$$
$$\boxed{\Phi=5647\ \text{N}\cdot\text{m}^2/\text{C}}$$
