Electric Field Between Two Nested Uniformly-Charged Spheres: 04-BS-9, December 2015, Q2

Question 2: Electric Field Between Two Nested Uniformly-Charged Spheres (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
Inner sphere radius $a$, charge (uniform, own full volume)	$0.5\times10^{-10}$ m, $-e$
Outer sphere radius $b$, charge (uniform, own full volume)	$1\times10^{-10}$ m, $-e$
Point charge at common centre	$+e$
Observation radius $r$	$0.75\times10^{-10}$ m (i.e. $a<r<b$)

Find. The magnitude and direction of $\vec E$ at radius $r$ from the common centre.

Two independent uniform spherical charge distributions (radii $a$, $b$, each $-e$) plus a central point charge $+e$; the dashed Gaussian sphere of radius $r$ lies between $a$ and $b$.

Approach. Apply Gauss's law with the enclosed charge built from three independent contributions: the always-fully-enclosed point charge, the inner sphere (fully enclosed since $r>a$), and the outer sphere (only its volume fraction $(r/b)^3$ is enclosed, since $r<b$).

Enclosed charge from the point charge and the inner sphere. Both lie entirely within radius $r$: $$Q_{\text{point}}+Q_{a}= e+(-e)=0$$
Enclosed charge from the outer sphere (partial, since $r<b$). A uniform sphere of total charge $Q$ and radius $b$ encloses a charge fraction equal to its volume fraction, $(r/b)^3$: $$Q_{b,\text{enc}}=(-e)\left(\frac{r}{b}\right)^3=(-e)\left(\frac{0.75}{1.0}\right)^3=(-e)(0.4219)$$ $$Q_{b,\text{enc}}=\boxed{-0.4219\,e}$$
Total enclosed charge and Gauss's law. $$Q_{\text{enc}}=0+Q_{b,\text{enc}}=-0.4219\,e$$ $$E=\frac{k\,|Q_{\text{enc}}|}{r^2}=\frac{(8.9918\times10^9)(0.4219)(1.6\times10^{-19})}{(0.75\times10^{-10})^2}$$ $$E=\boxed{1.079\times10^{11}\ \text{V/m, radially inward (toward the centre)}}$$

Quantity	Result
$Q_{\text{enc}}$ within $r$	$-0.4219\,e$
$E$ at $r=0.75\times10^{-10}$ m	$1.079\times10^{11}$ V/m, inward

Concept & Theory

Topic. Gauss's law with several superposed, independently uniform, spherical charge distributions.

Key relations. For a uniform sphere of total charge $Q$ and radius $R$, the charge enclosed within $r<R$ is $Q(r/R)^3$; field $E=kQ_{\text{enc}}/r^2$.

Why this works. Gauss's law depends only on the total charge inside the Gaussian surface, not on how many separate physical distributions contribute to it, so each sphere (and the point charge) can be evaluated independently — using its full charge if the surface lies outside it, or the volume-scaled fraction if the surface lies inside it — and the results simply add.

Common pitfall. Reading "concentric sphere" as a thin spherical surface layer (all-or-nothing enclosure) rather than a solid uniform sphere occupying its own full volume, which instead requires the $(r/b)^3$ volume-fraction rule when the observation point lies inside it; also forgetting the always-enclosed point charge in the running total.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (Gauss's law, spherically symmetric charge distributions).

Question 2 of 8: Electric Field Between Two Nested Uniformly-Charged Spheres

Question 2: Electric Field Between Two Nested Uniformly-Charged Spheres (20 marks)