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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 4.41

In this scenario, the external field is that of the uniform line charge. The electric field of such line charge points in its cylindrical radial direction:

$${\mathbf{E}}_{\text{ext}}=\frac{\lambda }{2\pi {ϵ}_0}\frac{\hat{\mathbf{s}}}{s}$$

In this solution I will assume that this field is roughly constant around the tiny sphere, so for practical purposes we can consider the scenario presented in Example 4.7 (page 193). In that example, after using boundary conditions to solve Laplace's equation, we find that the field inside the sphere is

$${\mathbf{E}}_{\text{in}}=\frac{3}{{ϵ}_r+2}{\mathbf{E}}_0$$

Where ${\mathbf{E}}_0$ denotes the constant field outside. Now that we know the $\mathbf{E}$ field, we can compute $\mathbf{P}$:

$$\begin{array}{rl}\text{(Eq. 4.30, page 186)}& \mathbf{P}& ={ϵ}_0{\chi }_e{\mathbf{E}}_0& ={ϵ}_0{\chi }_e\frac{3}{{ϵ}_r+2}{\mathbf{E}}_0\\ & ={ϵ}_0{\chi }_e\frac{3}{1+{\chi }_e+2}{\mathbf{E}}_0\\ & ={ϵ}_0{\chi }_e\frac{3}{3+{\chi }_e}{\mathbf{E}}_0\\ & ={ϵ}_0{\chi }_e\frac{1}{1+\frac{{\chi }_e}{3}}{\mathbf{E}}_0\end{array}$$

In the above, we will replace ${\mathbf{E}}_0$ by ${\mathbf{E}}_{\text{ext}}$. Both represent the same thing, but it's important to keep notation consistent!

$$\mathbf{P}={ϵ}_0{\chi }_e\frac{1}{1+\frac{{\chi }_e}{3}}{\mathbf{E}}_{\text{ext}}$$

We are now ready to use Equation 4.69:

$$\begin{array}{rl}\mathbf{F}& =\int (\mathbf{P}\cdot \mathrm{\nabla }){\mathbf{E}}_{\text{ext}}d\tau \\ & =\int ({ϵ}_0{\chi }_e\frac{1}{1+\frac{{\chi }_e}{3}}{\mathbf{E}}_{\text{ext}}\cdot \mathrm{\nabla }){\mathbf{E}}_{\text{ext}}d\tau \\ & =\left({ϵ}_0{\chi }_e\frac{1}{1+\frac{{\chi }_e}{3}}\right)\frac{\lambda }{2\pi {ϵ}_{0}s}\frac{\partial }{\partial s}\frac{\lambda }{2\pi {ϵ}_0}\frac{\hat{\mathbf{s}}}{s}\int d\tau \\ & =-\frac{{\chi }_e}{1+\frac{{\chi }_e}{3}}\frac{{\lambda }^2}{4{\pi }^2{ϵ}_0}\frac{1}{s^2}\hat{\mathbf{s}}\int d\tau \\ & =-\frac{{\chi }_e}{1+\frac{{\chi }_e}{3}}\frac{{\lambda }^2}{4{\pi }^2{ϵ}_0}\frac{1}{s^2}\left(\frac{4}{3}\pi R^3\right)\hat{\mathbf{s}}\end{array}$$

Where I took $s$'s out of the integral since we assume that the sphere is tiny, so $s$ won't vary much as we integrate over the sphere. Further, the integral $\int d\tau$ was performed over the tiny sphere of radius $R$.

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