I have a conducting sphere with radius $R_1$ and charge $Q_1$ inside a conducting thick spherical shell with inner radius $R_2>R_1$ and outer radius $R_3>R_2$. Both the spheres have the same center.


In the first experiment, the outer side of the spherical shell is grounded. In the second experiment (which is performed independently to the first one), the inner side of the spherical shell is grounded. I need to calculate the charge both on the inner and outer side of the shell in each experiment.

I am not sure what is the difference between the two experiments, since in both of them I get that the charge of the inner side is $Q_2=-Q_1$ and the charge of the outer side is $Q_3=0$:

The field inside the shell is $0$, and thus $Q_2=-Q_1$.

In the first experiment:

$$ \frac{KQ_1}{R_3}+\frac{KQ_2}{R_3}+\frac{KQ_3}{R_3}=0 $$ $$ \frac{KQ_1}{R_3}-\frac{KQ_1}{R_3}+\frac{KQ_3}{R_3}=0 $$ $$ \frac{KQ_3}{R_3}=0 $$ $$ Q_3=0 $$

And in the secondexperiment:

$$ \frac{KQ_1}{R_2}+\frac{KQ_2}{R_2}+\frac{KQ_3}{R_3}=0 $$ $$ \frac{KQ_1}{R_2}-\frac{KQ_2}{R_2}+\frac{KQ_3}{R_3}=0 $$ $$ \frac{KQ_3}{R_3}=0 $$ $$ Q_3=0 $$

I am not sure if I am right or not, and if I am, then is there a difference between the experiments?

  • $\begingroup$ Does this answer your question? physics.stackexchange.com/q/292024 $\endgroup$ Sep 19, 2021 at 11:29
  • $\begingroup$ @AmirhoseinRezaee Not exactly, since my question is about a thick shell and its two sides. I added an image to illustrate it. $\endgroup$
    – Daniel
    Sep 19, 2021 at 12:06
  • $\begingroup$ @AmirhoseinRezaee This changes the question since the electric charge can move from the inner side of the shell to its outer side during the grounding process. $\endgroup$
    – Daniel
    Sep 19, 2021 at 21:47
  • $\begingroup$ I'm not getting clarity on your description of your experiment. It may be helpful if you would edit your question to specify. When you say "ground the outside of the spherical shell" I assume you mean put a grounding wire on the surface at $r = R_3$, and when you say "ground the inside of the spherical shell" I assume you mean put a grounding wire on the surface at $r = R_2$. But I'm not sure -- hence, please edit your question for clarity. $\endgroup$
    – TimWescott
    Jul 6, 2023 at 17:20

2 Answers 2


If you only charge the inner sphere , the inside of the outer sphere has the same charge, the outside therefore the same charge as the innermost. If you now ground at R3 the charge there goes to earth so it will be zero. R2 and R1 stay with Q+ and Q-. If you charge R1 and then you ground it, all charges go away, so everything is without charge. So I don't understand the second statement. if you first ground R3, and then R1 R3 will have the charge opposite to the charge R1 had.

  • $\begingroup$ My question is about grounding the outer side (at distance $R3$) in the first experiment, and the inner side (at distance $R2$) in the first experiment. $\endgroup$
    – Daniel
    Sep 19, 2021 at 18:41

If no current is flowing, then all points on your conducting outer sphere will be at the same potential -- regardless of whether you are grounding the surface at $r = R_2$ or $r = R_3$.

So the results will be identical -- the charges in that outer sphere will arrange themselves however they need to be arranged to reach equilibrium, and because the potentials are the same in both cases, the charges will be arranged the same in both cases.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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