Field Jump Across a Charged Surface Layer Covering a Uniformly Charged Sphere: 04-BS-9, May 2017, Q1

Question 1: Field Jump Across a Charged Surface Layer Covering a Uniformly Charged 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.

Quantity	Value
Point charge at centre	$+3e$
Volume charge (uniform, filling the sphere)	$-e$, radius $R=5\times10^{-11}$ m
Thin surface-layer charge (adheres to the sphere's surface, same radius $R$)	$-e$

Find. $E(R^-)-E(R^+)$, the jump in radial field magnitude across the thin surface layer at $r=R$.

Point charge $+3e$ at the centre of a uniformly charged sphere ($-e$ total); a thin surface layer (also $-e$) coats the sphere's outer surface at the same radius $R$.

Approach. Apply Gauss's law with a spherical Gaussian surface just inside ($r=R^-$) and just outside ($r=R^+$) the surface layer. Just inside, the layer contributes nothing (not yet enclosed); just outside, it is fully enclosed. Subtract the two fields — this is exactly the standard boundary-condition jump $\Delta E=\sigma/\varepsilon_0$ for the layer's own surface charge.

Enclosed charge just inside the surface layer ($r=R^-$). Only the point charge and the FULL volume charge are enclosed (the surface layer sits exactly at $R$, not yet included): $$Q_{\text{enc}}(R^-)=3e+(-e)=2e$$ $$E(R^-)=\frac{kQ_{\text{enc}}(R^-)}{R^2}=\frac{k(2e)}{R^2}=\boxed{1.1505\times10^{12}\ \text{V/m}}$$
Enclosed charge just outside the surface layer ($r=R^+$). The layer's $-e$ is now fully enclosed as well: $$Q_{\text{enc}}(R^+)=3e-e-e=e$$ $$E(R^+)=\frac{kQ_{\text{enc}}(R^+)}{R^2}=\frac{k(e)}{R^2}=\boxed{5.7523\times10^{11}\ \text{V/m}}$$
Field difference. $$\Delta E=E(R^-)-E(R^+)=\frac{k(2e)}{R^2}-\frac{k(e)}{R^2}=\frac{ke}{R^2}$$ $$\Delta E=\boxed{5.752\times10^{11}\ \text{V/m}}$$

Quantity	Result
$E$ just inside the surface layer, $E(R^-)$	$1.1505\times10^{12}$ V/m
$E$ just outside the surface layer, $E(R^+)$	$5.7523\times10^{11}$ V/m
Field jump $\Delta E$	$5.752\times10^{11}$ V/m (field drops on crossing outward)

Concept & Theory

Topic. Gauss's law for a point charge plus a uniformly charged sphere plus a thin coincident surface charge layer; the field discontinuity across a surface charge.

Key relations. $E(r)=kQ_{\text{enc}}(r)/r^2$ by spherical symmetry; a thin surface charge layer of density $\sigma$ produces a radial field jump $\Delta E=E(r^-)-E(r^+)=\sigma/\varepsilon_0$ (equivalently $ke\Delta n/r^2$ here, since the enclosed point charge count changes by exactly the layer's charge as $r$ crosses $R$).

Why this works. Because the sphere and the surface layer sit at the SAME radius $R$, a Gaussian surface at $r=R^-$ excludes the layer entirely while one at $r=R^+$ includes it entirely — the only thing that changes between the two calculations is whether the layer's $-e$ is counted, so the difference is a clean, sign-correct application of the field-jump boundary condition.

Common pitfall. Including or excluding the surface layer inconsistently (e.g. treating it as always enclosed), or forgetting that the field difference has the SAME sign as $-Q_{\text{layer}}$ acting alone (a negative layer charge makes the field weaker, not stronger, just outside it).

Source. Sadiku, Elements of Electromagnetics, Ch. 3–4 (Gauss's law, field discontinuity across a surface charge).

Question 1 of 8: Field Jump Across a Charged Surface Layer Covering a Uniformly Charged Sphere

Question 1: Field Jump Across a Charged Surface Layer Covering a Uniformly Charged Sphere (20 marks)