- #1

Tony Hau

- 101

- 30

- Homework Statement
- Find the electric field produced by a uniformly polarized sphere of radius R.

- Relevant Equations
- Potential of a polarized object: ##V(\mathbf r) = \frac{1}{4 \pi \epsilon_o} \oint_{S} \frac{\sigma_b}{r}{da} + \frac{1}{4\pi \epsilon_o} \int_V \frac{\rho_b}{r} d{\tau} ##

Charged density of bound charge: ##\rho_b \equiv - \nabla \cdot \mathbf P##

Surface charge density of bound charge: ##\sigma_b \equiv \mathbf P \cdot \hat {\mathbf n} ##

This is an example of Griffith's book on bound charge, and the following is the solution to this example.

We choose the z-axis to conincide with the direction of polarization.

By $$\sigma_b \equiv \mathbf P \cdot \hat {\mathbf n} $$ and $$\rho_b \equiv - \nabla \cdot \mathbf P$$ we can see that ##\rho_b## is 0 and ##\sigma_b = Pcos\theta##.

The book then gives the solution to the potential without steps, which is: $$V(r,\theta) =

\begin{cases}

\frac{P}{3 \epsilon_o} rcos\theta & \text{for} \text{ }r \leq R \\

\frac{P}{3\epsilon_o}\frac{R^3}{r^2}cos\theta & \text{for } \text{ }r \geq R

\end{cases} $$

This is a figure from the book:

My attempt to the solution is like this:

$$V(\mathbf r) = \frac{1}{4 \pi \epsilon_o} \oint_{S} \frac{\sigma_b}{r}{da} $$, because ##\rho_b = 0##.

Then, ##\oint_{S} \frac{\sigma_b}{r}da= \int_0^ {\pi}\int_0^{2\pi} \frac{Pcos\theta}{R} R^{2}sin\theta d\theta d\phi##

I think the variable r is a constant because it is integrating over a surface; the r is fixed. However, it is not the case for the solution. Can anyone help?

We choose the z-axis to conincide with the direction of polarization.

By $$\sigma_b \equiv \mathbf P \cdot \hat {\mathbf n} $$ and $$\rho_b \equiv - \nabla \cdot \mathbf P$$ we can see that ##\rho_b## is 0 and ##\sigma_b = Pcos\theta##.

The book then gives the solution to the potential without steps, which is: $$V(r,\theta) =

\begin{cases}

\frac{P}{3 \epsilon_o} rcos\theta & \text{for} \text{ }r \leq R \\

\frac{P}{3\epsilon_o}\frac{R^3}{r^2}cos\theta & \text{for } \text{ }r \geq R

\end{cases} $$

This is a figure from the book:

My attempt to the solution is like this:

$$V(\mathbf r) = \frac{1}{4 \pi \epsilon_o} \oint_{S} \frac{\sigma_b}{r}{da} $$, because ##\rho_b = 0##.

Then, ##\oint_{S} \frac{\sigma_b}{r}da= \int_0^ {\pi}\int_0^{2\pi} \frac{Pcos\theta}{R} R^{2}sin\theta d\theta d\phi##

I think the variable r is a constant because it is integrating over a surface; the r is fixed. However, it is not the case for the solution. Can anyone help?

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