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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 4.36

Recall Gauss's law for the electric displacement $\mathbf{D}$:

$$\oint \mathbf{D}\cdot d\mathbf{a}=Q_{f_{\text{enc}}}$$

We have spherical symmetry, so we can break down the integral.

$$\begin{array}{rl}D\cdot (\text{area})& =Q_{f_{\text{enc}}}\\ D\cdot (4\pi r^2)& =q\\ D& =\frac{q}{4\pi r^2}\\ \mathbf{D}& =\frac{q}{4\pi r^2}\hat{\mathbf{r}}\end{array}$$

The electric field will then be:

$$\begin{array}{rl}\text{(Recall }\mathbf{D}=ϵ\mathbf{E}\text{)}& \mathbf{E}& =\frac{1}{ϵ}\mathbf{D}\text{(As found above)}& & =\frac{1}{ϵ}\frac{q}{4\pi r^2}\hat{\mathbf{r}}& =\frac{1}{{ϵ}_0(1+{\chi }_e)}\frac{q}{4\pi r^2}\hat{\mathbf{r}}\\ & =\frac{1}{4\pi {ϵ}_0}\frac{q}{(1+{\chi }_e)r^2}\hat{\mathbf{r}}\end{array}$$

The polarization follows from $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$:

$$\mathbf{P}=\frac{{\chi }_e}{4\pi }\frac{q}{(1+{\chi }_e)r^2}\hat{\mathbf{r}}$$

Now that we know the polarization, we can compute the bound charges. For volume:

$$\begin{array}{rl}{\rho }_b& =-\mathrm{\nabla }\cdot \mathbf{P}\\ & =-\frac{{\chi }_e}{4\pi }\frac{q}{(1+{\chi }_e)}\mathrm{\nabla }\cdot \frac{\hat{\mathbf{r}}}{r^2}\\ & =-\frac{{\chi }_e}{4\pi }\frac{q}{(1+{\chi }_e)}4\pi {\delta }^3(\mathbf{r})\\ & =-\frac{{\chi }_{e}q}{(1+{\chi }_e)}{\delta }^3(\mathbf{r})\end{array}$$

And for surface:

$$\begin{array}{rl}{\sigma }_b& =\mathbf{P}\cdot \hat{\mathbf{r}}\\ & =\frac{{\chi }_e}{4\pi }\frac{q}{(1+{\chi }_e)R^2}\end{array}$$

The total charge on the surface is:

$$\begin{array}{rl}{Q}_{\text{surface}}& =\int \sigma bda\\ & ={\sigma }_b(4\pi R^2)\\ & =\frac{{\chi }_e}{4\pi }\frac{q}{(1+{\chi }_e)R^2}(4\pi R^2)\\ & =\frac{q{\chi }_e}{1+{\chi }_e}\end{array}$$

Total bound charge should be zero, so the rest is in the center as indicated by the dirac delta:

$$\begin{array}{rl}{Q}_{\text{center}}& =\int {\rho }_{b}d\tau \\ & =\int -\frac{{\chi }_{e}q}{(1+{\chi }_e)}{\delta }^3(\mathbf{r})d\tau \\ & =-\frac{{\chi }_{e}q}{(1+{\chi }_e)}\int {\delta }^3(\mathbf{r})d\tau \\ & =-\frac{{\chi }_{e}q}{(1+{\chi }_e)}\end{array}$$

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