# Spherical capacitor with dielectrics

#### Karl86

1. The problem statement, all variables and given/known data
Consider the following system:

which consists of a conducting sphere with free charge , a dielectric shell with permittivity $\epsilon_1$, another dielectric shell with permittivity $\epsilon_2$ and finally a conducting spherical shell with no free charge.
2. Relevant equations
$D = \epsilon_0 E+P$
$\sigma_b=P \cdot \hat{n}$
3. The attempt at a solution
After finding $D=\frac{Q}{4\pi r^2}\hat{r}$, I calculated the polarization vector $P_1,P_2$ on each region to be
$$P_1=\left(1-\frac{\epsilon_0}{\epsilon_1}\right)\frac{Q}{4\pi r^2}\hat{r} \\ P_2=\left(1-\frac{\epsilon_0}{\epsilon_2}\right)\frac{Q}{4\pi r^2}\hat{r}$$
and the surface charge due to bound charges to be
$$\sigma_{b_1}=\begin{cases}\frac{\epsilon_0 \chi_1 Q}{4\pi \epsilon_1 R_2^2} & \text{on outer surface} \\ -\frac{\epsilon_0 \chi_1 Q}{4\pi \epsilon_1 R_1^2} & \text{on inner surface} \end{cases}$$
$$\sigma_{b_2}=\begin{cases}\frac{\epsilon_0 \chi_2 Q}{4\pi \epsilon_2 R_3^2}\hat{r} & \text{on outer surface} \\ -\frac{\epsilon_0 \chi_2 Q}{4\pi \epsilon_2 R_2^2}\hat{r} & \text{on inner surface} \end{cases}$$
$\rho_b=0$ for both dielectrics because the divergence of $P_i$ is zero. Now I want to see that the overall induced charge is zero. I try to do this by integrating the surface distributions on each sphere and I get that the total charge induced by the dielectric 1 is $\frac{\epsilon_0 \chi_1 Q}{\epsilon_1}$ at $R_2$, $-\frac{\epsilon_0 \chi_1 Q}{\epsilon_1}$ at $R_1$ and for the dielectric 2 it's $\frac{\epsilon_0 \chi_2 Q}{\epsilon_2}$ at $R_3$, $-\frac{\epsilon_0 \chi_2 Q}{\epsilon_2}$ at $R_2$. This seems to add up to zero, but I wonder whether these are the only induced charges in this setup or I'm forgetting something. Also: is integrating like that just the right way to compute the induced charge?
Thanks if you'll find time to help me somehow.

#### Attachments

• 28.3 KB Views: 104
Related Advanced Physics Homework News on Phys.org

#### kuruman

Homework Helper
Gold Member
This seems to add up to zero, but I wonder whether these are the only induced charges in this setup or I'm forgetting something.
See if there are induced free charges on the surfaces of the outer conducting shell at $r=R_3$ and $r=R_4$. Use the boundary condition on the normal component of $\vec D$ to find what they are.
Also: is integrating like that just the right way to compute the induced charge?
Yes except surface charge density is a scalar quantity. You have a unit vector $\hat r$ attached to $\sigma_{b_2}$ which does not belong. I'm sure it's a typo.

#### Karl86

See if there are induced free charges on the surfaces of the outer conducting shell at $r=R_3$ and $r=R_4$. Use the boundary condition on the normal component of $\vec D$ to find what they are.

Yes except surface charge density is a scalar quantity. You have a unit vector $\hat r$ attached to $\sigma_{b_2}$ which does not belong. I'm sure it's a typo.
Thanks a lot for your reply. For sure that's scalar and that was a typo. About your first remark: I was inclined to think that the total induced charge at $R_3$ would be equal in value to the one induced by the dielectric at $R_3$ and of opposite sign, because there is indeed some free charge due to the fact that the last spherical shell is a conductor. Similarly at $R_4$. Is this wrong?

#### kuruman

Homework Helper
Gold Member
Thanks a lot for your reply. For sure that's scalar and that was a typo. About your first remark: I was inclined to think that the total induced charge at $R_3$ would be equal in value to the one induced by the dielectric at $R_3$ and of opposite sign, because there is indeed some free charge due to the fact that the last spherical shell is a conductor. Similarly at $R_4$. Is this wrong?
It is correct. You might wish to calculate the free charge distributions and see what they are. Do you know how to do that?

#### Karl86

It is correct. You might wish to calculate the free charge distributions and see what they are. Do you know how to do that?
Hmm. I don't really know how to calculate them. I thought it was just a consequence of saying that it had to be the opposite of the total charge induced by the dielectric and that would be it.

Homework Helper
Gold Member

#### Karl86

The free surface charge density at the boundary between regions 1 and 2 is given by $\sigma=(\vec D_2-\vec D_1)\cdot \hat n$ where $\hat n$ is the normal to the boundary and is directed from 1 into 2. See https://en.wikipedia.org/wiki/Interface_conditions_for_electromagnetic_fields
In my case $D=\frac{Q}{4\pi r^2}\hat{r}$ in 2 and $0$ in 1 and $\hat{r} \cdot \hat{n} = 1$ so $(\vec D_2-\vec D_1)\cdot \hat n = \frac{Q}{2\pi R_1^2}$, right? Is it correct to pick the value of $D$ right inside the boundary between the regions?

#### kuruman

Homework Helper
Gold Member
In my case $D=\frac{Q}{4\pi r^2}\hat{r}$ in 2 and $0$ in 1 and $\hat{r} \cdot \hat{n} = 1$ so $(\vec D_2-\vec D_1)\cdot \hat n = \frac{Q}{2\pi R_1^2}$, right? Is it correct to pick the value of $D$ right inside the boundary between the regions?
It is correct to do that, but there is another typo, the $2$ in the denominator should be a $4$. What is the total charge on the surface at $R_1$? Repeat for the interfaces at $R_3$ and $R_4$.

#### Karl86

It is correct to do that, but there is another typo, the $2$ in the denominator should be a $4$. What is the total charge on the surface at $R_1$? Repeat for the interfaces at $R_3$ and $R_4$.
$Q$, $-Q$,$Q$. Ok, I think I got the gist of it, at least it feels much clearer now thanks to you. One quick question that I think is not worth opening a thread about. Is the normal component of $P$ discontinuous at the interface of the two dielectrics?

Last edited:

#### kuruman

Homework Helper
Gold Member
$Q$, $-Q$,$Q$. Ok, I think I got the gist of it, at least it feels much clearer now thanks to you. One quick question that I think is not worth opening a thread about. Is the normal component of $P$ discontinuous at the interface of the two dielectrics?
You can answer that by yourself. Look at your expressions for $\vec P_1$ and $\vec P_2$ in post #1. Evaluate each expression separately at $r=R_2$. If the two expressions are not equal then the normal component of $P$ is discontinuous; if they are equal then it is continuous.

#### Karl86

You can answer that by yourself. Look at your expressions for $\vec P_1$ and $\vec P_2$ in post #1. Evaluate each expression separately at $r=R_2$. If the two expressions are not equal then the normal component of $P$ is discontinuous; if they are equal then it is continuous.
Yeah, I was confused because it seemed impossible for $P$ to be discontinuous but not $D$, but now I realised that the normal component of $E$ is discontinuous too, so that fixes things. Thanks

### Want to reply to this thread?

"Spherical capacitor with dielectrics"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving