Point Charges Enclosed by a Charged Conducting Shell: 04-BS-9, Undated paper, Q2

Question 2: Point Charges Enclosed by a Charged Conducting Shell (15 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.

Quantity	Value
Pyramid base charges	4 × $+10$ nC (square base, 10 cm edge)
Pyramid apex charge	$-30$ nC
Shell inner / outer radius	$0.90$ m / $1.00$ m (10 cm thick)
Shell net charge	$+10$ nC
$P_1$, $P_2$ radii	$0.95$ m (in shell), $1.05$ m (outside)

Find. (a) inner-surface charge. (b) outer-surface charge density. (c) $E$ at $P_1$ and $P_2$. (d) flux through a 10 m Gaussian sphere.

The pyramid's five point charges sit deep inside the conductor's hollow cavity (edge 10 cm, well inside the 90 cm inner radius); the shell's inner and outer surfaces carry induced/free charge in electrostatic equilibrium.

Approach. Total the pyramid's enclosed charge, then apply Gauss's law at three nested surfaces: just inside the conductor (forces the inner-surface charge), the conductor's own net charge (fixes the outer surface by charge conservation), and two spheres outside the shell (one very close, one very far) to get $E$ and the total flux.

Enclosed pyramid charge. $q_{\text{pyr}}=4(+10)+(-30)=+10$ nC (net).
Part (a) — inner-surface charge. In electrostatic equilibrium $\vec E=0$ everywhere inside the conductor material, so a Gaussian surface drawn just inside the inner radius must enclose zero net charge. The pyramid's $+10$ nC forces an equal and opposite induced charge onto the inner surface: $$\boxed{q_{\text{inner}}=-q_{\text{pyr}}=-10\ \text{nC}}$$
Part (b) — outer-surface charge density. The conductor's total charge (inner $+$ outer surfaces) equals its stated net charge of $+10$ nC: $$q_{\text{outer}}=q_{\text{net}}-q_{\text{inner}}=10-(-10)=20\ \text{nC}$$ Spread uniformly over the outer sphere ($R=1.00$ m): $$\sigma_{\text{outer}}=\frac{q_{\text{outer}}}{4\pi R^2}=\frac{20\times10^{-9}}{4\pi(1.00)^2}$$ $$\boxed{\sigma_{\text{outer}}=1.592\ \text{nC/m}^2}$$
Part (c) — field at $P_1$ (95 cm) and $P_2$ (105 cm). $P_1$ lies inside the conductor material ($0.90<0.95<1.00$ m), so: $$\boxed{E(P_1)=0}$$ $P_2$ lies outside the shell; a Gaussian sphere there encloses the pyramid and the whole shell ($q_{\text{pyr}}+q_{\text{net}}=10+10=20$ nC): $$E(P_2)=\frac{k(20\times10^{-9})}{(1.05)^2}$$ $$\boxed{E(P_2)=163.0\ \text{V/m, radially outward}}$$
Part (d) — flux through a 10 m Gaussian sphere. At 10 m the Gaussian surface encloses the entire system (pyramid $+$ shell $=20$ nC); by Gauss's law the flux depends only on that total, not on how far outside it the surface sits: $$\Phi=\frac{Q_{\text{enc}}}{\varepsilon_0}=\frac{20\times10^{-9}}{8.854\times10^{-12}}$$ $$\boxed{\Phi=2259\ \text{V}\cdot\text{m}\ (\text{N}\cdot\text{m}^2/\text{C})}$$

Quantity	Result
Inner-surface charge	$-10$ nC
Outer-surface density	$1.592$ nC/m$^2$
$E(P_1{=}95\text{cm})$	$0$
$E(P_2{=}105\text{cm})$	$163.0$ V/m
Flux, 10 m sphere	$2259$ V·m

Concept & Theory

Topic. Gauss's law with a conductor in electrostatic equilibrium.

Key relations. $\oint\vec E\cdot d\vec A=Q_{\text{enc}}/\varepsilon_0$; inside a conductor $\vec E=0$; conservation of the conductor's own net charge across its two surfaces.

Why this works. A conductor's free charges rearrange until the field inside the material vanishes; the only way to enforce that around an embedded charge is an equal-and-opposite induced layer on the cavity wall. Whatever charge remains (the conductor's stated net charge minus that induced layer) must appear on the outer surface, because the conductor as a whole has no other charge to place. Once a Gaussian surface encloses everything, the flux is fixed by total enclosed charge alone — the shell's exact geometry becomes irrelevant.

Common pitfall. Forgetting the sign convention (the inner surface charge is opposite to the enclosed charge, not equal to it), and mis-reading a Gaussian radius given in metres as centimetres (or vice versa) — here 10 m, not 10 cm, which changes which charges the surface encloses.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (Gauss's law, conductors in electrostatic equilibrium).

Question 2 of 7: Point Charges Enclosed by a Charged Conducting Shell

Question 2: Point Charges Enclosed by a Charged Conducting Shell (15 marks)