A problem from Griffiths asks us to find the electric potential inside and outside a spherical shell. The potential on the shell is specified to be $V(R, \theta) = k \cos(3\theta)$. There is no charge anywhere else.

The solution (I think) is $$V_{in}(r,\theta) = \frac{k}{5} \left(-3\left(\frac{r}{R}\right)P_1(\cos\ \theta) + 8\left(\frac{r}{R}\right)^3P_3\cos\ \theta\right)$$

The field is not zero inside this shell.

But if I apply Gauss's law over any surface fully inside this shell, $$\oint \mathbf{E}\ \cdot d\mathbf{a}\ = \frac{Q_{enc}}{\epsilon_0}$$ $Q_{enc}$ is zero because there are no charges anywhere. And since the above equation holds for every closed surface inside the shell, $\mathbf{E} = 0$ everywhere inside the shell. Which is not true according to the potential I computed.

What am I missing?

  • 1
    $\begingroup$ Gauss' law only states that the integral of the electric field over the surface is zero. It does not state that the electric field is zero at every point on the surface. $\endgroup$ – Crimson Dec 30 '20 at 13:35
  • 1
    $\begingroup$ You can verify your solution by calculating the corresponding electric field and integrate this over any desired surface. $\endgroup$ – Crimson Dec 30 '20 at 13:36
  • $\begingroup$ In addition, your solution does not seem to satisfy the boundary condition you indicate for $V(r =R)$. For $r \rightarrow R$, your potential seems to produce $k \cos (3\theta)$. $\endgroup$ – secavara Dec 30 '20 at 13:54
  • $\begingroup$ @Crimson Oh yes! I understand now. And verified. The surface integral is indeed zero even if the field isn't. Thank you. $\endgroup$ – VB0904 Dec 30 '20 at 15:51
  • $\begingroup$ @secavara You are right! Edited the post. Thank you. $\endgroup$ – VB0904 Dec 30 '20 at 15:53

If the charge inside a Gaussian surface is zero, then every field line which enters the surface must leave at some other point. That does not require a zero field inside.


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