# Solve Bound Charge Question: Griffith's Book Example

• Tony Hau
In summary: So in your case we want to impose \left(\frac{\partial V_{out}}{\partial r} - \frac{\partial V_{in}}{\partial r}\right) = \frac{\sigma}{\epsilon_0} as you had.
Tony Hau
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?

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I don't think the integral formulation will necessarily help you; I believe it should be $$V(\mathbf{r}) = \int_S \frac{\sigma_b}{\|\mathbf{r} - \mathbf{r}'\|}\,dS'$$ where the integration is with respect to the primed variable.

I would instead note that since $\rho = 0$ the potential satisfies $$\nabla^2 V = 0$$ in $r < R$ and $r > R$.

I don't know enough EM to say what boundary condition it must satisfy at $r = R$ due to the charged sphere (presumably Griffiths will tell you), but I suspect that it will end up being a condition of the form $\frac{\partial V}{\partial r} = A\cos\theta$ for some constant $A$. That would suggest looking for solutions of the form $$V(r) = f(r) \cos \theta.$$

Tony Hau
pasmith said:
I don't think the integral formulation will necessarily help you; I believe it should be $$V(\mathbf{r}) = \int_S \frac{\sigma_b}{\|\mathbf{r} - \mathbf{r}'\|}\,dS'$$ where the integration is with respect to the primed variable.

I would instead note that since $\rho = 0$ the potential satisfies $$\nabla^2 V = 0$$ in $r < R$ and $r > R$.

I don't know enough EM to say what boundary condition it must satisfy at $r = R$ due to the charged sphere (presumably Griffiths will tell you), but I suspect that it will end up being a condition of the form $\frac{\partial V}{\partial r} = A\cos\theta$ for some constant $A$. That would suggest looking for solutions of the form $$V(r) = f(r) \cos \theta.$$
The Laplace's equation ##\nabla ^{2} V =0##, has the following solution in spherical coordinates: $$V = \sum_{l=0}^{\infty} (A_l r^l + \frac{B_l}{r^{l+1})P_{l}(cos\theta)$$.

The potential boundary condition can be calculated by letting ##V_{in} = V_{out}##. But I don't know as well how you can satisfy the boundary condition for this case. The book hasn't mentioned it.

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pasmith said:
I don't think the integral formulation will necessarily help you; I believe it should be $$V(\mathbf{r}) = \int_S \frac{\sigma_b}{\|\mathbf{r} - \mathbf{r}'\|}\,dS'$$ where the integration is with respect to the primed variable.

I would instead note that since $\rho = 0$ the potential satisfies $$\nabla^2 V = 0$$ in $r < R$ and $r > R$.

I don't know enough EM to say what boundary condition it must satisfy at $r = R$ due to the charged sphere (presumably Griffiths will tell you), but I suspect that it will end up being a condition of the form $\frac{\partial V}{\partial r} = A\cos\theta$ for some constant $A$. That would suggest looking for solutions of the form $$V(r) = f(r) \cos \theta.$$
You are totally correct and smart! Thanks.
Here is the solution I have worked out:

The potetial for the interior of the sphere in spherical coordinates is: $$V_{in} = \sum_{l=0}^\infty A_l r^l P_l(cos\theta)$$

By the boundary condition that: $$V_{in} = V_{out}$$ and $$\frac{\partial V_{out}}{\partial r} - \frac{\partial V_{in}}{\partial r} = -\frac{\sigma}{\epsilon_o}$$ at ## r=R##, we obtain $$\sum_{l=0}^{\infty}(2l+1)A_l R^{l-1} P_l (cos\theta) = \frac {\sigma}{\epsilon_o}$$ $$= \frac{Pcos\theta}{\epsilon_o}$$

Obviously only the term for ##l=1## survives, and ##A_1 = \frac{P}{3\epsilon_o}##.

Subsituting ##A_1 = \frac{P}{3\epsilon_o}## into ##V_{in} = \sum_{l=0}^\infty A_l r^l P_l(cos\theta)##, we get $$V_{in}=\frac{P}{3\epsilon_o}r cos\theta$$

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Having now checked my copy of Grant & Phillips, it appears that if $\mathbf{D}_1$ is the field inside the sphere and $\mathbf{D}_2$ the field outside then at $r = R$ the condition is $$(\mathbf{D}_2 - \mathbf{D}_1) \cdot \mathbf{e}_r = \sigma$$ with $V$ continuous. In the absence of volume charge density we have $\mathbf{D} = \epsilon_0\mathbf{E} = -\epsilon_0\nabla V$.

Tony Hau

## 1. What is a bound charge?

A bound charge is a type of electric charge that is associated with the alignment of polar molecules in a material. It is not free to move like a free charge, but rather remains bound to the molecules.

## 2. How is bound charge different from free charge?

Bound charge is different from free charge in that it is not able to move freely within a material. Free charge, on the other hand, is able to move and flow through a material.

## 3. How does Griffith's book example illustrate bound charge?

In Griffith's book example, a dielectric material is placed between two parallel metal plates. The polar molecules in the dielectric material become aligned in the presence of an external electric field, creating bound charges on the surface of the material. This results in an overall decrease in the electric field between the plates.

## 4. How is bound charge calculated in Griffith's book example?

In Griffith's book example, the bound charge is calculated by taking the difference between the total charge on the plates and the free charge on the plates. This difference is equal to the bound charge on the surface of the dielectric material.

## 5. Why is understanding bound charge important in electromagnetism?

Understanding bound charge is important in electromagnetism because it helps to explain the behavior of materials in the presence of an electric field. It also plays a role in the formation of electric dipoles and the overall behavior of capacitors and other electronic devices.

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