Field Inside a Point Charge Surrounded by a Uniform Charge Sphere: 04-BS-9, December 2014, Q2

Question 2: Field Inside a Point Charge Surrounded by a Uniform Charge 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	$+e=1.6\times10^{-19}$ C
Surrounding sphere (uniform density)	radius $R=10^{-10}$ m, total charge $-e$
Field point radius $r$	$0.5\times10^{-10}$ m (inside the sphere, $r=R/2$)

Find. The magnitude and direction of $\vec{E}$ at $r=0.5\times10^{-10}$ m from the centre.

Point charge +e at centre, surrounded by a uniformly charged sphere of total charge −e, radius R. Field point lies at r = R/2, inside the sphere.

Approach. By spherical symmetry, apply Gauss's law with a Gaussian sphere of radius $r$: the enclosed charge is the point charge $+e$ plus the fraction of the sphere's uniformly-distributed $-e$ that lies inside radius $r$, which scales as $(r/R)^3$ for a uniform volume density.

Charge enclosed within radius $r$. The sphere's charge enclosed scales with volume fraction: $$q_{\text{sphere}}(r)=-e\left(\frac{r}{R}\right)^3=-e\left(\frac12\right)^3=-\frac{e}{8}$$ $$Q_{\text{enc}}=e+q_{\text{sphere}}(r)=e-\frac{e}{8}=\frac{7e}{8}$$ $$Q_{\text{enc}}=\boxed{1.400\times10^{-19}\ \text{C}}$$
Apply Gauss's law. $$E=\frac{kQ_{\text{enc}}}{r^2}=\frac{(8.992\times10^9)(1.400\times10^{-19})}{(0.5\times10^{-10})^2}$$ $$E=\boxed{5.04\times10^{11}\ \text{V/m, radially outward}}$$

Quantity	Result
$Q_{\text{enc}}(r=R/2)$	$7e/8=1.400\times10^{-19}$ C
$E(r=R/2)$	$5.04\times10^{11}$ V/m, radially outward

Concept & Theory

Topic. Gauss's law for a point charge embedded in a uniform spherical volume-charge background (a classic "Thomson atom" style field problem).

Key relations. $\oint \vec E\cdot d\vec A=Q_{\text{enc}}/\varepsilon_0$; for a uniform sphere, the fraction of charge inside radius $r<R$ is $(r/R)^3$.

Why this works. Spherical symmetry guarantees $\vec E$ is radial and depends only on $r$, so the enclosed-charge argument works exactly as for a simple point charge, provided the charge that has been "counted in" is only what lies inside the Gaussian surface — here the full point charge plus a fraction of the background.

Common pitfall. Using the sphere's total charge $-e$ instead of the fraction enclosed within $r$, which would badly overstate the cancellation and give the wrong sign or magnitude for $E$.

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

Question 2 of 8: Field Inside a Point Charge Surrounded by a Uniform Charge Sphere

Question 2: Field Inside a Point Charge Surrounded by a Uniform Charge Sphere (20 marks)