0
$\begingroup$

enter image description here

Consider a grounded spherical shell conductor (radius $R$) with a point charge $q$ inside a distance $a$ from the center. In order to calculate the electric field inside we can place an image charge with charge $q'=-q\frac{R}{a}$ and position $d=\frac{R^2}{a}$ but if we apply Gauss law around the sphere, the electric field outside is zero $\rightarrow$ flux is zero $\rightarrow$ net charge inside is zero. That means the total charge induced on the sphere is $-q$. But how does that make sense, shouldn't the charge on the shell be the same as the point image charge we used to calculate the potential inside? So we got $q'=-q$ instead of $q'=-q\frac{R}{a}$. Where is the contradiction?

$\endgroup$
0
0
$\begingroup$

The method of images is only used here to determine the electric field is inside the sphere. In particular, a uniform surface charge density on the sphere produces no field at all inside the sphere, by the shell theorem, so the value of that uniform density is not determined by the image charge argument. Instead, its value takes whatever is necessary to make the total charge zero, for the reason you pointed out.

$\endgroup$
4
  • $\begingroup$ Does that mean it doesn't always happen that the charge on the conductor is the same as the image charge we used, or did I not understand you correctly? Because in the simple example of a plane it does happen. $\endgroup$ – Darkenin May 9 '20 at 9:54
  • $\begingroup$ @Darkenin Yes, that logic doesn’t work in general. $\endgroup$ – knzhou May 9 '20 at 17:02
  • $\begingroup$ Does this happen because of the grounding? $\endgroup$ – Darkenin May 9 '20 at 18:53
  • $\begingroup$ @Darkenin No, the grounding just fixes the total charge of the point charge and sphere to zero. You could also do the same problem but say that the sphere has a total charge $Q$ on it. In that case, the extra uniform charge density would be whatever is necessary to get a total charge of $Q$ on the sphere. $\endgroup$ – knzhou May 9 '20 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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