Electric Field and Voltage of a Radially-Varying Spherical Charge: 17-Phys-A3 Electromagnetics, Undated paper, Q5

Question 5: Electric Field and Voltage of a Radially-Varying Spherical Charge (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. $\rho_v(r)=(1-r/2)$ C/m³ for $0\le r\le2$ m (spherically symmetric), $\rho_v=0$ for $r>2$; free space.

The given charge density $\rho_v(r)$: linear, positive throughout $0\le r\le2$ m, dropping from $1$ C/m³ at the origin to $0$ at $r=2$ m, and zero beyond.

Find. $\mathbf E(r)$ for $r\le2$ and $r>2$; $V(r)$ for $r>2$ referenced to infinity.

Approach. Spherical symmetry means $\mathbf D=D(r)\hat a_r$ everywhere, so Gauss's law reduces to $D(r)(4\pi r^2)=Q_{enc}(r)$ with $Q_{enc}(r)=\int_0^r\rho_v(r')4\pi r'^2\,dr'$; then $V(r)=\int_r^\infty E\,dr'$.

Part (a)(i) — Enclosed charge and field for $r\le2$. $$Q_{enc}(r)=\int_0^r\left(1-\frac{r'}{2}\right)4\pi r'^2\,dr'=4\pi\left[\frac{r^3}{3}-\frac{r^4}{8}\right]=\frac{\pi r^3(8-3r)}{6}.$$ $$\mathbf E(r)=\frac{Q_{enc}(r)}{4\pi\varepsilon_0r^2}\hat a_r \Rightarrow \boxed{\mathbf E(r)=\frac{r(8-3r)}{24\varepsilon_0}\,\hat a_r\ \text{V/m},\qquad 0\le r\le2\ \text{m}.}$$
Part (a)(ii) — Total charge and field for $r>2$. Beyond $r=2$ m there is no more charge, so $Q_{enc}$ freezes at its value at $r=2$: $$Q_{total}=Q_{enc}(2)=\frac{\pi(2)^3(8-6)}{6}=\frac{8\pi}{3}\ \text{C}.$$ $$\boxed{\mathbf E(r)=\frac{Q_{total}}{4\pi\varepsilon_0r^2}\hat a_r=\frac{2}{3\varepsilon_0r^2}\,\hat a_r\ \text{V/m},\qquad r>2\ \text{m}.}$$ Check: both expressions give $E(2)=1/(6\varepsilon_0)$ — the field is continuous across $r=2$, as it must be for a bounded (non-singular) charge density.
Part (b) — Voltage for $r>2$, reference at infinity. $$V(r)=\int_r^\infty E(r')\,dr'=\int_r^\infty\frac{2}{3\varepsilon_0r'^2}\,dr'=\boxed{\frac{2}{3\varepsilon_0r}\ \text{V},\qquad r>2\ \text{m}.}$$

Final results
Quantity	Value
$\mathbf E(r)$, $r\le2$ m	$\dfrac{r(8-3r)}{24\varepsilon_0}\hat a_r$ V/m
$\mathbf E(r)$, $r>2$ m	$\dfrac{2}{3\varepsilon_0r^2}\hat a_r$ V/m
Total enclosed charge	$8\pi/3$ C
$V(r)$, $r>2$ m (ref. at $\infty$)	$\dfrac{2}{3\varepsilon_0r}$ V

Topic: Gauss's law for a spherically symmetric, radially-varying volume charge.

Key relations: $Q_{enc}(r)=\int_0^r\rho_v4\pi r'^2dr'$; $E(r)=Q_{enc}(r)/(4\pi\varepsilon_0r^2)$; $V(r)=\int_r^\infty E\,dr'$.

Why this works: Spherical symmetry of $\rho_v(r)$ forces $\mathbf D$ to be purely radial and constant on any sphere, collapsing the surface integral in Gauss's law to a single scalar equation; outside the charged region, the field looks exactly like that of a point charge equal to the total enclosed charge.

Common pitfall: Forgetting the $r'^2$ in the volume-charge integral (it is NOT $\int\rho_v\,dr'$), and forgetting that $Q_{enc}$ stops growing once $r$ exceeds the charge's outer radius (2 m here) even though the field expression for $r>2$ keeps the OLD frozen charge, not a re-evaluated integral.

Source: Sadiku, Elements of Electromagnetics, Ch. 4 (Gauss's law, spherical symmetry).

Question 5 of 10: Electric Field and Voltage of a Radially-Varying Spherical Charge

Question 5: Electric Field and Voltage of a Radially-Varying Spherical Charge (20 marks)