# Homework Help: Electrostatics: spherical shell

1. Oct 1, 2009

### Bapelsin

Electrostatics: spherical shell [SOLVED]

1. The problem statement, all variables and given/known data

A point charge $$Q_1$$ is located in the centre of a spherical conducting shell with inner radius $$a$$ and outer radius $$b$$. The shell has total charge $$Q_2$$. Determine the electrostatic field $$\vec{E}$$ and the potential $$\phi$$ everywhere in space, and determine how much charge is on the inner and outer surfaces of the shell after electrostatic equilibrium has been reached.

2. Relevant equations

Gauss law: $$\oint_S\vec{E}\cdot d\vec{a} =\frac{ Q_{encl}}{\epsilon_0}$$
Area of a sphere: $$A=4\pi r^2$$
Surface charge density (homogenous charge distribution): $$\sigma = Q/A$$
Electric field from a point charge: $$\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}$$
Electric field from a surface charge: $$\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma da}{r^2}\hat{r}$$

3. The attempt at a solution

I call the charges on the inner and outer surfaces $$Q_a$$ and $$Q_b$$ respectively. The electric field inside a conductor is always zero, so Gauss law on a surface $$a<r<b$$ gives

$$0=\oint_S\vec{E}\cdot d\vec{a} =\frac{ Q_1+Q_a}{\epsilon_0}$$
$$\Rightarrow Q_a=-Q_1$$

The total charge of the conductor:

$$Q_2=Q_a+Q_b=Q_b-Q_1$$
$$\Rightarrow Q_b=Q_1+Q_2$$

The total electric field could be calculated by taking the sum of the electric field from each charge.

$$\vec{E}_{tot}(\vec{r})=\vec{E}(\vec{r})_1+\vec{E}(\vec{r})_a+\vec{E}(\vec{r})_b$$

E-field from the point charge:

$$\vec{E}_1(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{Q_1}{r}\hat{r}$$

Introduce the surface charge density $$\sigma_{a,b}=Q_{a,b}/A_{sphere}$$

\begin{align*}\vec{E}_a(\vec{r})=&\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma da}{r^2}\hat{r} \\ &=\frac{1}{4\pi\epsilon_0}\int_0^r\frac{Q_a da}{4\pi a^2 r^2}\hat{r} \\ &= \frac{Q_a}{16\pi^2\epsilon_0a^2}\int_S\frac{rdrd\theta}{r^2}\hat{r} \\ & =\frac{Q_a}{8\pi\epsilon_0a^2}\int_0^r\frac{1}{r}\hat{r} \\ & = \frac{Q_a}{8\pi\epsilon_0a^2}\left[ln|r|\right]_0^r \end{align*}

which does not make sense. Similar problems arise for E-field from $$\sigma_b$$. Can anyone help me out, please?

Last edited: Oct 1, 2009
2. Oct 1, 2009

### ehild

Use the spherical symmetry of the problem: the magnitude of E depends only on r, and it is parallel with the radius at any point. As $$d\vec{a}$$ is normal to the surface of the sphere and so is $$\vec{E}$$

$$\oint_S\vec{E} \cdot d\vec{a} =\oint_SEda$$

where E is the magnitude of the electric field at distance r from the centre of the sphere, and it is the same at each point of the surface, therefore

$$\oint_S\vec{E} \cdot d\vec{a} =\oint_SEda = 4\pi r^2 E=\frac{ Q_{encl}}{\epsilon_0}$$

You know well that the electric field is zero inside the conducting shell, you know that the enclosed charge is Q1 for r<a and Q1+Q2 for r>b...

ehild

3. Oct 1, 2009

### Bapelsin

Thank you! That was really helpful.