A thin, metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell and another charge q' is placed outside it. All the three charges are positive. The net force on the charge at the centre and the force due to shell on this charge is?

In my opinion the force on the central charge will be due to outside charge q' plus the force due to the shell. Now the force due to outside charge is 0 due to electrostatic shielding. Besides, the force due to shell can be seen in a two tier way.

In accordance with Gauss law the inner surface of the shell must have been induced with — q charge and the charge remaining on outer surface would be Q+q. This Q+q charge would be distributed non uniformly due to presence of q'. So force on q due to the shell can be seen as force due to two shells with charge — q distributed uniformly on one, and Q+q distributed non uniformly on the other.

The field due to these shells in the interior is 0 as can be explained by Gauss law. So the final answer I arrive at is 0 in both the cases. But this is only correct for the first part as force on q due to shell is towards right if the centre of the shell is positioned at (0,0,0).

I have explained my approach at length and think that I have got a problem with my concepts with regard to conductors. Any help would greatly be appreciated.


2 Answers 2


In the absence of charge q, the field inside the sphere, due to Q or due to q', would be zero, since the only way to create a field inside a conductive shell is to place a charge inside it.

When we put charge q inside the sphere, its field may rearrange Q or q', but those charges will still remain external to the sphere and, therefore, they would still have no contribution to the field inside the sphere.

Hence, charge q should experience no force.

Adding the answer to the second part of the question regarding the force on q due to the shell alone.

Since the configuration of the charge on the shell is pretty complex (besides the initial charge Q, it will have charge redistributions induced by q' and by q), we can take advantage of the fact that the forces on q due to the shell and due to the external charge q' should have the same magnitudes and opposite signs (to yield zero net force).

Since the force on q due to q' is $k_e\frac {qq'} {r^2}$, where $r$ is distance between q and q', the force due to the shell must be $-k_e\frac {qq'} {r^2}$

  • $\begingroup$ By what about the second part? $\endgroup$ Oct 21, 2018 at 12:07
  • $\begingroup$ @MohdKhan The field inside the sphere due to any charges other than the charge q placed inside the sphere is going to be zero. If you accept that, there is no need to go into details for every specific charge configuration. $\endgroup$
    – V.F.
    Oct 21, 2018 at 12:37
  • $\begingroup$ @MohdKhan You should review Gauss's Law $\endgroup$
    – Bob D
    Oct 21, 2018 at 16:34
  • $\begingroup$ @Bob D It says that the net Flux through a closed gaussian surface is equal to the charge enclosed /epsilon knot times $\endgroup$ Oct 22, 2018 at 1:00
  • $\begingroup$ @MohdKhan It goes a little beyond Gauss's law. We know that there should be no field inside a conductor - otherwise free electrons inside the conductor would move to kill it. We also can say that there are no excessive charges inside a conductor (they all reside on the surface) - if there was an excessive charge inside a conductor, there would be a non-zero flux around it and, therefore non-zero electric field, which we just have just shown should be zero. $\endgroup$
    – V.F.
    Oct 22, 2018 at 2:43

The isolated charge, q, at the center of the sphere will reappear as a uniformly distributed charge on the outside of the sphere. So the external field due to the interior charge is the same whether the sphere is present or not.

The fact that the sphere has its own charge, Q, can be treated the same way, except that that charge gets redistributed by the presence of the exterior charge, q'. However that redistribution can be handled separately by considering an image of the exterior charge as seen in the spherical mirror surface of the sphere. The image appears inside the sphere at a distance R^2/r' from the center and has magnitude q'' = -q'R/r. Here, R is the radius of the sphere and r' is distance of q' from the center of the sphere. Since the overall charge on the sphere is unchanged, it must be represented as a uniform charge of Q-q'' plus the interior image, q''.

So the exterior charge, q', will see forces from charges q and Q-q'' both effectively at the center of the sphere plus the image charge, q'', positioned inside the sphere as described above.

In contrast, the isolated charge, q, at the center of a metallic sphere will feel no forces since it is centrally located inside a spherical Faraday shield.


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.