Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to freshen up my memory about electrical fields and I came across this exercise from school.

A sphere with a constantly distributed charge is located in between two different dielectrics (see picture at the bottom) and the task is to calculate the electrical field. I have a sample solution that makes the following assumptions:

  1. The electrical field at the boundary between the two dielectrics has to be constant. (I understand this)
  2. The electrical field has only a radial component. (I accept this as well)
  3. The electrical field has spherical symmetry i.e. it has the same magnitude regardless if it is in the upper or lower dielectric. I am sort of stumbling over this assumption, because I feel that the setup is not symmetric.

So, how come the electrical field is radial symmetric? This doesn't seem intuitive to me. Or is the sample solution incorrect.

enter image description here

EDIT: Okay, I think I figured it out. The sphere is actually a conducting sphere. Now since the whole sphere has the same potential and lets assume that at infinity the potential is zero. The static electrical field has to be symmetrical around the sphere, since the integral from infinity over the electrical field onto the sphere has to be consistent from all directions. This also means, that the charges inside the sphere will shift accordingly. Can somebody confirm this?

share|cite|improve this question
@Lipschitz Your edit is too radical. If you want to ask that, why not make a new question of your own? – Emilio Pisanty Jul 27 '13 at 11:11

The total charge density, which is the sum of free charge density and bound charge density is constant all over the sphere, but the free and bound charge densities differ for upper and lower half of the sphere.

Your question about symmetry of total charge density can be answered easily, assuming that you know the symmetry of electric field. (your reasoning for it's spherical symmetry is correct):

($E_1$ means electric field in the first region and $E_2$ the field of the second region)

$$\mathbf{E}_1=\mathbf{E}_2 =\mathbf{E}_{out} \tag{as you stated}$$ $$\mathbf{E}_{out} . \hat r-\mathbf{E}_{in}.\hat r=\frac{\sigma}{\epsilon}_0 \,\,\,,\,\,\,\mathbf{E}_{in}=0 \to \sigma={\epsilon}_0 \mathbf{E}_{out} . \hat r$$

You can arrive at this result by calculating free and bound charge densities directly by solving for potential using Laplace equation too:

The problem is clearly azimuthal symmetric, and also according to your argument, it isn't a function of $\theta$ neither; i.e., the problem is 1-D and $V$ is a function of $r$.

Using the fact that potential is continuous at the boundary of the two regions and also vanishes at $\infty$, the only remaining term of the potential expansion in spherical coordinates will be:

$$V=\frac{Cq}{r}$$ $$ \mathbf{D}=-\epsilon \nabla V \to \cases{\mathbf{D}_1=\frac{\epsilon_1Cq}{r^2}\hat r \\ \mathbf{D}_2=\frac{\epsilon_2Cq}{r^2}\hat r}$$

Now, assuming a spherical surface surrounding the sphere and applying Gauss law for $\mathbf{D}$, we will find $C$:

$$\oint{\mathbf{D}.d\mathbf{S}}=Q_{free}\to C=\frac{1}{2\pi (\epsilon_1 + \epsilon_2)}$$

$$V=\frac{1}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r}$$

Now we find $\mathbf{D}_1$ , $\mathbf{D}_2$ and $\mathbf{E}$ :

$$\mathbf{E}=\frac{1}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r$$

$$\mathbf{D}_1=\frac{\epsilon_1}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r$$

$$\mathbf{D}_2=\frac{\epsilon_2}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r$$

using the relation $\sigma_f=\mathbf{D}.\hat r $ we will have:

$$\to \cases{\sigma_{1f}=\frac{q\epsilon_1}{2\pi (\epsilon_1 + \epsilon_2)} \frac{1}{R^2} \\ \sigma_{2f}=\frac{q\epsilon_2}{2\pi (\epsilon_1 + \epsilon_2)} \frac{1}{R^2}}$$

From these two relations for $\sigma_{f}$s it is evident that the total free charge is equal to $q$, as expected: $$\sigma_{1f}\times 2\pi R^2+\sigma_{2f}\times 2\pi R^2=q$$

Now we find bound charge densities. First we should find polarization vectors $\mathbf{P}_1$ and $\mathbf{P}_2$:

$$\mathbf{P}=\mathbf{D}-\epsilon_0 \mathbf{E} \to \cases{\mathbf{P}_1= \frac{\epsilon_1- \epsilon_0}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r \\ \mathbf{P}_2=\frac{\epsilon_2- \epsilon_0}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{r^2}\hat r }$$ and

$$\sigma_b =\mathbf{P}.\hat r\to \cases{\sigma_{b1}=-\frac{\epsilon_1- \epsilon_0}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{R^2} \\ \sigma_{b2}=-\frac{\epsilon_2- \epsilon_0}{2\pi (\epsilon_1 + \epsilon_2)} \frac{q}{R^2} }$$

Now you can see that on each half sphere the sum of bound and free charge densities is equal, as expected: $$\sigma_{b1}+\sigma_{f1}=\sigma_{b2}+\sigma_{f2}$$

share|cite|improve this answer

I'm not good with equations, therefore let's try an intuitive approach.

You have two contradictory statements in your question: you wrote that the distribution of charge is uniform on the sphere, but in the EDIT section you state that the sphere is conducting. Decide what exactly the conditions of your problem are. I'll try to address the both.

First of all, what happens in dielectric when it is placed inside an electric field? Well, the dipoles inside the material are being affected by the field - they experience a torque and rotate slightly in the direction dictated by the field. This creates a polarization inside the dielectric. The field associated with this polarization is opposite (in direction) to the externally applied field, therefore the net (effective) field inside the dielectric is reduced (this, for example, the reason for higher value dielectric capacitors to have larger capacitance).

Non-conducting sphere having uniform charge distribution:

Since the sphere is non-conducting, the uniform charge distribution is doomed to persist.

We already know that the effective electric field will be stronger in a low value dielectric. Can it be the case that the field is spherically symmetric in each half of the space, just being stronger in one of the halves? The answer is NO!

I'm sure that there are many ways to show that the above scenario is impossible (including solving the exact equations). I'll reason my answer using a continuity argument: the electric field must be continuous!

Let's assume that the field is indeed spherically symmetric in each half of the space, but have different magnitudes. Well, this means that the field lines in the two halves of the space are parallel at the boundary. This means that an infinitesimal step across the boundary will change the radial component of the electric field by a non-infinitesimal amount, which is discontinuity. A contradiction.

So, what the field will look like? Away from the boundary it will resemble a spherically symmetric field. However, the closer to the boundary the more interaction there will be between differently polarized dipoles. My intuition says that the field will still be radial, but the net effect of this interaction will be to make the magnitude of the field at the boundary an average of the maximal magnitudes in each half of the space, but it's just an intuition.

Conducting sphere:

This case is very different from the previous one - now, the charge can move freely on the sphere.

The intuition screams that in this case we will get perfectly spherically symmetric field in all the space. Why? Again, there are many ways to show that. This time I'm choosing a "hand-waving" story!

Lets see what happens at time zero: uniform charge on the sphere causes non uniform electric field around it (the same as in the non-conductive case). This field changes more in magnitude closer to the boundary (its gradient grows). Well, this means that there is some tangential component in the gradient of the field, which means that there is some tangential component in the field itself. Hey, remember what tangential component of the electric field does to charges on the sphere? Correct, it moves them! Until when? Until there are no more tangential component to the field. What does it mean? Correct, it means that the field is spherically symmetric on the surface of the sphere, which in turn implies spherical symmetry throughout the whole space.


When I come to think about it now, I myself made a contradictory statements:

  1. The field is radial in all directions in the non-conducting case
  2. There is a tangential component of the field in the non-conducting case.

First of all it is embarrassing. Secondly, the incorrect statement is (almost certainly) the first one - the field lines will bend in some manner near the boundary in a non-conductive case.

Well, the intuition is sometimes misleading :(

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.