 #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 zaxis 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 zaxis 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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