Field Inside a Point Charge Embedded in a Uniformly Charged Sphere: 04-BS-9, December 2016, Q1

Question 1: Field Inside a Point Charge Embedded in 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	$+2e$
Sphere's total charge (uniform)	$-e$, radius $R=0.5$ Å $=5\times10^{-11}$ m
Field point radius	$r=0.25$ Å $=2.5\times10^{-11}$ m (inside the sphere, $r<R$)

Find. The electric field intensity $E$ at $r=0.25$ Å from the point charge.

Point charge $+2e$ at the centre of a uniformly charged sphere ($-e$ total, radius $R$); the field point sits inside the sphere, at half the sphere's radius.

Approach. Apply Gauss's law with a spherical Gaussian surface of radius $r<R$: the enclosed charge is the point charge plus the FRACTION of the sphere's charge inside radius $r$ (the enclosed volume scales as $(r/R)^3$ for a uniform density), then $E=kQ_{\text{enc}}/r^2$.

Enclosed charge at $r=0.25$Å. The uniform sphere's charge inside radius $r$ scales with volume fraction $(r/R)^3$: $$Q_{\text{enc}}=2e+(-e)\left(\frac{r}{R}\right)^3=2e-e\left(\frac{0.25}{0.5}\right)^3=2e-e(0.125)$$ $$Q_{\text{enc}}=\boxed{1.875\,e=3.000\times10^{-19}\ \text{C}}$$
Field from the enclosed charge (Gauss's law, spherical symmetry). $$E=\frac{kQ_{\text{enc}}}{r^2}=\frac{(8.992\times10^9)(3.000\times10^{-19})}{(2.5\times10^{-11})^2}$$ $$E=\boxed{4.316\times10^{12}\ \text{V/m, radially outward from the centre}}$$

Quantity	Result
Enclosed charge $Q_{\text{enc}}(r=0.25\text{\AA})$	$1.875\,e = 3.000\times10^{-19}$ C
Electric field $E$	$4.316\times10^{12}$ V/m, radially outward

Concept & Theory

Topic. Gauss's law for a point charge embedded at the centre of a uniformly charged sphere — superposing a point-charge field with a uniform-volume-charge field inside the sphere.

Key relations. $Q_{\text{enc}}(r)=Q_{\text{point}}+Q_{\text{sphere}}(r/R)^3$ for $r<R$ (uniform density, charge scales as volume); $E=kQ_{\text{enc}}/r^2$ by spherical symmetry, exactly as if $Q_{\text{enc}}$ were a point charge at the centre.

Why this works. By Gauss's law, only the charge enclosed by the Gaussian sphere of radius $r$ contributes to the field at $r$; the sphere's own charge outside $r$ produces zero net field inside it (the standard uniform-spherical-layer result), so the point charge and the enclosed fraction of the sphere's charge simply add algebraically before applying the point-charge formula.

Common pitfall. Using the FULL sphere charge $-e$ instead of the enclosed fraction $-e(r/R)^3$, or forgetting that a positive net $Q_{\text{enc}}$ (here $1.875e>0$, since the point charge dominates at this radius) means the field still points outward despite the sphere's charge being negative.

Source. Sadiku, Elements of Electromagnetics, Ch. 3–4 (Gauss's law, spherically symmetric charge distributions).

Question 1 of 8: Field Inside a Point Charge Embedded in a Uniformly Charged Sphere

Question 1: Field Inside a Point Charge Embedded in a Uniformly Charged Sphere (20 marks)