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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 4.35

To find the bound charges (${\rho }_b$ and ${\sigma }_b$) we need to know what $\mathbf{P}$ is. To find $\mathbf{P}$, we can use the fact that

$$\begin{array}{}\text{(Eq. 4.30, page 186)}& \mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}\end{array}$$

So we need to find $\mathbf{E}$. To do so, we use $\mathbf{D}=ϵ\mathbf{E}$ (Eq. 4.32). We don't have the permittivity, but we know about ${ϵ}_r$.

${ϵ}_r$, the dielectric constant (also known as the relative permittivity, see page 187), is defined as

$$\begin{array}{}\text{(4.34, page 187)}& {ϵ}_r=1+{\chi }_e=\frac{ϵ}{{ϵ}_0}\end{array}$$

Since ${ϵ}_r(0)=1$ and ${ϵ}_r(d)=2$, it follows that

$$\begin{array}{rl}{\chi }_e(0)& =0\\ {\chi }_e(d)& =1\end{array}$$

The linear function that satisfies this conditions is

$${\chi }_e(x)=\frac{x}{d}$$

Using Eq. 4.34 again, it follows that

$$\begin{array}{rl}\text{Permittivity}ϵ& ={ϵ}_0(1+{\chi }_e)\\ & ={ϵ}_0(1+\frac{x}{d})\end{array}$$

Using this in Eq. 4.32 for $\mathbf{E}$, we see that

$$\mathbf{E}=\frac{1}{ϵ}\mathbf{D}=\frac{1}{{ϵ}_0(1+\frac{x}{d})}\mathbf{D}$$

It is clear that $\mathbf{D}={\sigma }_f\hat{\mathbf{x}}$, so we conclude that

$$\mathbf{E}=\frac{1}{{ϵ}_0(1+\frac{x}{d})}{\sigma }_f\hat{\mathbf{x}}$$

We can relate ${\sigma }_f$ to $V$ as follows:

$$\begin{array}{rl}V& =-{\int }_d^0\mathbf{E}\cdot d\mathbf{l}\\ & =\frac{{\sigma }_f}{{ϵ}_0}{\int }_0^d\frac{1}{1+\frac{x}{d}}dx\\ & =\frac{{\sigma }_f}{{ϵ}_0}d\ln (1+(x/d)){|}_0^d\\ & =\frac{{\sigma }_f}{{ϵ}_0}d\ln (2)-\frac{{\sigma }_f}{{ϵ}_0}d\ln (1)\\ & =\frac{{\sigma }_f}{{ϵ}_0}d\ln (2)\end{array}$$

So

$${\sigma }_f=\frac{{ϵ}_0}{d\ln (2)}V$$

This implies that

$$\mathbf{E}=\frac{V}{d\ln (2)}\frac{1}{1+\frac{x}{d}}\hat{\mathbf{x}}$$

Now we can go back to Eq. 4.30 and see that

$$\begin{array}{rl}\mathbf{P}& ={ϵ}_0{\chi }_e\mathbf{E}\\ & ={ϵ}_0{\chi }_e\frac{V}{d\ln (2)}\frac{1}{1+\frac{x}{d}}\hat{\mathbf{x}}\\ & ={ϵ}_0\frac{x}{d}\frac{V}{d\ln (2)}\frac{1}{1+\frac{x}{d}}\hat{\mathbf{x}}\\ & ={ϵ}_0\frac{V}{d^2\ln (2)}\frac{x}{1+\frac{x}{d}}\hat{\mathbf{x}}\end{array}$$

To find the bound charges, we recall:

$$\begin{array}{}\text{(Eq. 4.12, page 173)}& {\rho }_{\text{b}}& =-\mathrm{\nabla }\cdot \mathbf{P}\text{(Eq. 4.11, page 173)}& {\sigma }_{\text{b}}& =\mathbf{P}\cdot \hat{\mathbf{n}}\end{array}$$

So:

$$\begin{array}{rl}{\rho }_{\text{b}}& =-\frac{V{ϵ}_0}{(d+x{)}^2\ln (2)}\\ {\sigma }_{\text{b}}& =\{\begin{array}{l}\begin{array}{cc}0& x=0\\ \frac{{ϵ}_0}{2d\ln 2}V& x=d\end{array}\end{array}\end{array}$$

Finally, we verify that the total bound charge is zero (where $S$ denotes surface area):

$$\begin{array}{rl}{Q}_{\text{b}}& =\int {\rho }_{b}d\tau +\int {\sigma }_{b}da\\ & ={\int }_0^d-\frac{V{ϵ}_0}{(d+x{)}^2\ln (2)}Sdx+\int \frac{{ϵ}_0}{2d\ln 2}VdA\\ & =-\frac{V{ϵ}_0}{\ln 2}S{\int }_0^d\frac{1}{(d+x{)}^2}dx+\frac{{ϵ}_0}{2d\ln 2}VS\\ & =-\frac{V{ϵ}_0}{\ln 2}S\cdot \frac{1}{2d}+\frac{{ϵ}_0}{2d\ln 2}VS\\ & =0\end{array}$$

As expected.

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