Electrostatics: spherical shell

[Bapelsin]

Electrostatics: spherical shell [SOLVED]

Homework Statement

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.

Homework 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}

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})=&amp;\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma da}{r^2}\hat{r} \\<br /> &amp;=\frac{1}{4\pi\epsilon_0}\int_0^r\frac{Q_a da}{4\pi a^2 r^2}\hat{r} \\<br /> &amp;= \frac{Q_a}{16\pi^2\epsilon_0a^2}\int_S\frac{rdrd\theta}{r^2}\hat{r} \\<br /> &amp; =\frac{Q_a}{8\pi\epsilon_0a^2}\int_0^r\frac{1}{r}\hat{r} \\<br /> &amp; = \frac{Q_a}{8\pi\epsilon_0a^2}\left[ln|r|\right]_0^r<br /> \end{align*}

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

Thanks in advance.

 

[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}

<br /> \oint_S\vec{E} \cdot d\vec{a} =\oint_SEda<br />

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

<br /> \oint_S\vec{E} \cdot d\vec{a} =\oint_SEda<br /> = 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

 

[Bapelsin]

Thank you! That was really helpful.