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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 2.43

An important keyword here is "A metal sphere," meaning that we are dealing with a conductor. As such, it follows that

$$\begin{array}{rl}{\mathbf{E}}_{\text{Inside}}& =\mathbf{0}\hat{\mathbf{r}}\\ {\mathbf{E}}_{\text{Outside}}& =\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}\hat{\mathbf{r}}\end{array}$$

Where the first comes from the fact that it is a conductor, and ideal conductors (see page 95) are such that $\mathbf{E}=\mathbf{0}$ inside (property 1 of conductors). Furthermore, since conductors hide the internal charge configuration, we can simply treat this sphere as a uniformly charged sphere. The field of such a sphere is the one given above. (For a derivation of this result, see Example 2.3, page 69.)

Recall the force per unit area equation:

$$\begin{array}{}\text{(Eq. 2.50, page 100)}& \begin{array}{rl}\mathbf{f}& =\sigma {\mathbf{E}}_{\text{avg}}\\ & =\frac{1}{2}\sigma ({\mathbf{E}}_{\text{above}}+{\mathbf{E}}_{\text{below}})\end{array}\end{array}$$

$\sigma$ will be:

$$\begin{array}{rl}\sigma & =\frac{\text{Total charge inside conductor}}{\text{Conductor's surface area}}\\ & =\frac{Q}{4\pi R^2}\end{array}$$

(A result that's familiar to us after doing problem 2.40!) And because ${\mathbf{E}}_{\text{below}}=\mathbf{0}$, we see that:

$$\begin{array}{rl}\mathbf{f}& =\sigma \frac{1}{2}{\mathbf{E}}_{\text{above}}\\ & =\frac{1}{2}\frac{Q}{4\pi R^2}{\mathbf{E}}_{\text{Outside}}{|}_{r=R}\\ & =\frac{1}{2}\frac{Q}{4\pi R^2}\frac{1}{4\pi {ϵ}_0}\frac{Q}{R^2}\hat{\mathbf{r}}\end{array}$$

To get the force of repulsion between the "northern" hemisphere and the "southern" hemisphere, we need only the vertical component of the force per unit unit area:

$$f_z=|\mathbf{f}|\cos \left(\theta \right)$$

We finish by integrating this over the northern hemisphere:

$$\begin{array}{rl}{F}_z& =\int f_{z}da\\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi /2}\left[\frac{1}{2}\frac{Q}{4\pi R^2}\frac{1}{4\pi {ϵ}_0}\frac{Q}{R^2}\right]\cos \left(\theta \right)(R^2\sin \left(\theta \right)d\theta d\varphi )\\ & =\frac{1}{2}\frac{Q}{4\pi R^2}\frac{1}{4\pi {ϵ}_0}\frac{Q}{R^2}{R}^2(2\pi ){\int }_{\theta =0}^{\pi /2}\cos \left(\theta \right)\sin \left(\theta \right)d\theta \\ & =\frac{Q}{4\pi {ϵ}_0}\frac{Q}{4R^2}{\left(\frac{1}{2}{u}^2\right)}_0^1\\ & =\frac{Q^2}{32\pi R^2{ϵ}_0}\end{array}$$

Where $u$-substitution was used as follows: Let $u=\sin \left(\theta \right)$, then $du=\cos \left(\theta \right)d\theta$, so $d\theta =\frac{du}{\cos \left(\theta \right)}$ and the integral simplifies to ${\int }_0^{\pi /2}\sin \left(\theta \right)\cos \left(\theta \right)d\theta ={\int }_0^{1}du$.

We see than that the force of repulsion between the "northern" hemisphere and the "southern" hemisphere is

$$F_z=\frac{Q^2}{32\pi R^2{ϵ}_0}$$

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