Potential Inside a Point-Charge-Plus-Layer Distribution: 04-BS-9, December 2013, Q1

Question 1: Potential Inside a Point-Charge-Plus-Layer Distribution (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.

Quantity	Value
Point charge at centre	$q=2e=3.2\times10^{-19}$ C
Spherical charge layer	$Q=-e=-1.6\times10^{-19}$ C at radius $R=0.5\times10^{-10}$ m
Field point distance from centre	$r=0.25\times10^{-10}$ m (inside the layer, since $r<R$)

Find. The electric potential $V(r)$ relative to infinity at the field point.

Point charge +2e at the centre; charged layer −e at R = 0.5×10⁻¹⁰ m. Field point P at r = 0.25×10⁻¹⁰ m lies inside the layer.

Approach. Superpose the potential of the central point charge (valid at every radius) with the potential of the uniformly charged spherical layer, which is constant and equal to its surface value for any point inside the layer ($r<R$).

Potential of the point charge at $r$. $$V_{\text{pt}}=\frac{q}{4\pi\varepsilon_0 r}=\frac{2e}{4\pi\varepsilon_0(0.25\times10^{-10})}$$ $$V_{\text{pt}}=\boxed{115.1\ \text{V}}$$
Potential of the layer, evaluated inside it. Because the layer is a thin uniform spherical charge layer, $E=0$ inside it and $V$ is constant there, equal to the value at the surface $r=R$: $$V_{\text{layer}}=\frac{Q}{4\pi\varepsilon_0 R}=\frac{-e}{4\pi\varepsilon_0(0.5\times10^{-10})}$$ $$V_{\text{layer}}=\boxed{-28.77\ \text{V}}$$
Superpose. Potentials add algebraically (superposition holds for $V$, not just $E$): $$V(r)=V_{\text{pt}}+V_{\text{layer}}=115.1-28.77$$ $$V=\boxed{86.3\ \text{V}}$$

Quantity	Result
$V_{\text{pt}}$ (point charge)	$115.1$ V
$V_{\text{layer}}$ (charge layer, inside)	$-28.77$ V
$V(r=0.25\times10^{-10}\,\text{m})$	$86.3$ V

Concept & Theory

Topic. Superposition of potential for a point charge plus a uniformly charged spherical layer.

Key relations. Point-charge potential $V=q/4\pi\varepsilon_0 r$; for a uniform spherical charge layer of total charge $Q$ and radius $R$, $V(r)=Q/4\pi\varepsilon_0 R$ for $r\le R$ (constant) and $V(r)=Q/4\pi\varepsilon_0 r$ for $r\ge R$.

Why this works. Gauss's law shows the field inside a uniform spherical charge layer is exactly zero, so no work is done moving a test charge from $r=0$ out to $r=R$ within the layer — the potential is therefore flat there and pinned to the surface value. The point charge, being a true point, keeps its ordinary $1/r$ potential at every radius, so the two contributions simply add.

Common pitfall. Using $V_{\text{layer}}=Q/4\pi\varepsilon_0 r$ (the outside formula) for a field point that is actually inside the layer — this overstates $|V_{\text{layer}}|$ because $r<R$. Also forgetting the sign of $Q=-e$ when superposing.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (Electric Flux Density, Gauss's Law, Electric Potential for spherical charge distributions).

Question 1 of 8: Potential Inside a Point-Charge-Plus-Layer Distribution

Question 1: Potential Inside a Point-Charge-Plus-Layer Distribution (20 marks)