Question 7 of 8: Charge Distribution Producing a Linear-in- r Field That Vanishes Outside a Sphere

Question 7: Charge Distribution Producing a Linear-in-r Field That Vanishes Outside a 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.

Find. The full charge distribution (volume density inside, plus any surface charge needed) that produces this field.

Approach. Inside $R$, apply the differential (point) form of Gauss's law $\nabla\cdot\vec E=\rho_v/\varepsilon_0$ directly to the given field to get the volume density. Since the field is required to be EXACTLY zero outside $R$ (not the usual $1/r^2$ falloff of an isolated charged sphere), the total charge enclosed for any $r>R$ must be zero — this forces a compensating surface charge layer exactly at $r=R$.

1. Volume charge density from the divergence of $\vec E$ inside the sphere. With $\vec E=(Q/\varepsilon_0)\vec r$ and $\nabla\cdot\vec r=3$: $$\nabla\cdot\vec E=\frac{Q}{\varepsilon_0}\nabla\cdot\vec r=\frac{3Q}{\varepsilon_0}=\frac{\rho_v}{\varepsilon_0}$$ $$\rho_v=3Q=3(10^{-6})=\boxed{3\times10^{-6}\ \text{C/m}^3,\quad r\le R}$$
2. Total volume charge enclosed. $$Q_{\text{vol}}=\rho_v\cdot\frac{4}{3}\pi R^3=(3\times10^{-6})\cdot\frac{4}{3}\pi(10^{-3})^3=\boxed{1.257\times10^{-14}\ \text{C}}$$
3. Surface charge needed to cancel the field outside. For $E=0$ at $r>R$, the total enclosed charge (volume $+$ surface) beyond $R$ must be zero, so $Q_{\text{surf}}=-Q_{\text{vol}}$: $$\sigma=\frac{Q_{\text{surf}}}{4\pi R^2}=\frac{-Q_{\text{vol}}}{4\pi R^2}=-QR=-(10^{-6})(10^{-3})$$ $$\boxed{\sigma=-1\times10^{-9}\ \text{C/m}^2\ \text{(at }r=R\text{)}}$$