There is this problem in section 3.3, of Jackson's classical electrodynamics, which relates to solving Boundary- Value Problems with Azimuthal Symmetry, we have a Conducting sphere with hemispheres at different potentials, I've found the inside and outside potentials

$\phi(r,\theta)$=V \begin{cases}\frac{3}{2}\left(\frac{r}{a}\right) P_{1}(\cos \theta)-\frac{7}{8}\left(\frac{r}{a}\right)^{3} P_{3}(\cos \theta)+\frac{11}{16}\left(\frac{r}{a}\right)^{5} P_{5}(\cos \theta) \cdots, & \text { for } r<a \\ \frac{3}{2}\left(\frac{a}{r}\right)^{2} P_{1}(\cos \theta)-\frac{7}{8}\left(\frac{a}{r}\right)^{4} P_{3}(\cos \theta)+\frac{11}{16}\left(\frac{a}{r}\right)^{6} P_{5}(\cos \theta) \cdots, & \text { for } r>a\end{cases}

And I want to find the surface charge density on the sphere, I don't have any idea on how to approach it.

Your help will be very much appreciated.

  • $\begingroup$ Are you aware of the boundary conditions required for any electric field? $\endgroup$
    – Newbie
    Commented Jan 15, 2022 at 12:48
  • $\begingroup$ @Newbie what boundary conditions may I ask? $\endgroup$ Commented Jan 15, 2022 at 13:09
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Newbie
    Commented Jan 15, 2022 at 13:10
  • 1
    $\begingroup$ @Newbie okey thanks, the difference between gradiants of outside and inside potentials equal the surface charge/epslilon, right? $\endgroup$ Commented Jan 15, 2022 at 13:23
  • $\begingroup$ If you want the two hemispheres to be at different potentials, they will need to be insulated from each other. $\endgroup$
    – R.W. Bird
    Commented Jan 15, 2022 at 17:58

1 Answer 1


If you have found the potential inside and outside the hemispheres, extending Jackson (1.17), $$ \vec E_{above} - \vec E_{below}= \frac{\sigma}{\epsilon_0}\hat n \\ \nabla V_{above} - \nabla V_{below}= -\frac{\sigma}{\epsilon_0}\hat n \\ $$ Therefore the charge density can be found by, $$ \frac{\partial {V_{above}}}{\partial n} - \frac{\partial {V_{below}}}{\partial n} = -\frac{\sigma}{\epsilon_0} $$

  • $\begingroup$ This much has been already communicated to the OP. $\endgroup$
    – Newbie
    Commented Jan 15, 2022 at 19:53

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