A dielectric sphere of radius R with uniform dielectric constant ε has an azimuthally symmetric density charge σ = σ0 cos θ placed on the surface. Outside the sphere is vacuum.

  • (a) Obtain the electrostatic potential inside the sphere, φin.
  • (b) Obtain the electrostatic potential outside the sphere, φout.
  • (c) What is the electric field inside the sphere?

I'm not sure if my approach was correct. I assume not because part c) says to find the electric field, yet I found it without needing to find the potential. I'm sure I went wrong somewhere.

My attempt:

I draw a Gaussian sphere around the given sphere. Then I find $\vec{D}$, the electric displacement.

$$\int \vec{D} \cdot \vec{dA}=Q_{free,enc.}$$ $$D=\frac{\sigma_0 \cos \theta R^2}{r^2}, \qquad r>R$$

Using this, I can find the electric field by $\vec{D}=\epsilon \vec{E}$, but this doesn't feel right. Normally, with these types of problems there's a reason to solve the steps in order.

Update: Still really confused. This is what I tried:

$$\phi _{in} = \sum A_l r^l P_l \cos \theta\quad\text{ and }\quad\phi _{out} = \sum B_l r^{-(l+1)} P_l \cos \theta.$$

At the surface, the potentials will be equal:

$$\sum A_l r^l P_l \cos \theta - \sum B_l r^{-(l+1)} P_l \cos \theta = \sigma_0 \sin \theta$$

After this, I'm dreadfully lost.

  • 3
    $\begingroup$ The thing is what did you take as $\vec{dA}$. The charge distribution is not radially symmetrical. So you cannot just willy nilly take the usual $\int r^2\sin\theta d\theta d\phi$. This is why you have to go through all the rigmarole of using legendre polynomials technique $\endgroup$ Oct 23, 2016 at 5:22

1 Answer 1


Try doing the problem without the dielectric filling the sphere first as a warmup. Here's why that's useful:

Note that the definition of the displacement field is:$$\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P},$$ where $\mathbf{P}$ is the dipole moment density. Taking the curl of both sides of the equation in the static state gives: $$\nabla \times \mathbf{D} = \nabla \times \mathbf{P}.$$ Taking the divergence gives: $$\nabla \cdot \mathbf{D} = \rho + \nabla \cdot \mathbf{P}.$$ We usually define the bound charge density as $\rho_{\mathrm{bound}} \equiv -\nabla \cdot \mathbf{P}$, and the remaining charge density is then called 'free', $\rho_{\mathrm{free}}\equiv \nabla \cdot \mathbf{D}$.

The divergence and curl equations above can be converted into the boundary conditions for the three vector fields across surfaces. Applying Gauss's law to the divergence equations gives: $$\begin{align}\Delta \mathbf{P} \cdot \hat{n} & = \sigma_{\mathrm{bound}}, \\ \Delta \mathbf{D} \cdot \hat{n} & = \sigma_{\mathrm{free}},\ \mathrm{and} \\ \Delta \mathbf{E} \cdot \hat{n} & = \frac{\sigma_{\mathrm{bound}}}{\epsilon_0}, \end{align}$$ where $\hat{n}$ is a unit vector perpendicular to the surface and the $\Delta$s are understood to be a difference on one side of the surface from the other. These conditions are frequently expressed in terms of the components of the field perpendicular to the surface (eg $\mathbf{E}_\perp$), as implied by the dot product. In words, these mean that the change perpendicular components of the fields are dictated by their associated surface charge density.

Unlike the electric field, though, the displacement and polarization can have solenoidal components, as you can see from the curl equation above. Applying Stoke's theorem at the boundary surface for them gives: $$\begin{align}\Delta \mathbf{D}\times \hat{n} &= \Delta \mathbf{P}\times \hat{n}, \ \mathrm{and} \\ \Delta \mathbf{E} \times \hat{n} &= 0.\end{align}$$ Like the conditions on the perpendicular components, this can be expressed as a condition on the components of the fields parallel to the surface (eg $\mathbf{E}_{||}$). In words, the parallel components of the electric field must be continuous, and the discontinuity in the parallel components of the polarization must be balanced by a discontinuity in the parallel components of the displacement field.

Note that I don't think you actually need to find $\mathbf{D}$ to solve this problem - it should be sufficient to solve the dielectricless problem, use that to infer the polarization in the sphere, and from that calculate how the net surface charge density is modified by the bound charge.

All of that said, you should find that the empty sphere is a pure dipole field on the inside and the outside. Keep in mind that a dipole field appropriate for $r\rightarrow 0$, ie 'inside', has a different structure than one for $r\rightarrow \infty$, 'outside'. Your textbook should have a section on solving such problems using Legendre polynomials, as mentioned in a comment by @PrasadMani. The multipole fields are separated by the fact that they behave under rotations in orthogonal fashions, and it shouldn't surprise you to find that adding a rotationally invariant dielectric sphere won't cause a mixing of multipole field types.

  • $\begingroup$ So I reasoned that $V_{inside}=\sum Ar^l P_l cos \theta$ and $V_{outside}=\sum Br^{-(l+1)} P_l cos \theta$. Then, I said that the 2 definitions for the potential must be equal at the surface, when $r=R$. But after this, I don't see how to determine the $A/B$ coefficients. $\endgroup$ Oct 23, 2016 at 21:00
  • $\begingroup$ Calculate the electric field from those potentials, and use the boundary conditions on the fields I described in the post. $\endgroup$ Oct 23, 2016 at 21:44

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