While solving some exercise questions of electrostatics, I come across a problem given in the image:

enter image description here

Also what will the force on central charge only due to the shell?

I thought that there is already a charge $Q$ which is uniformly distributed on surface and due to charge $q1$ shell surface may aquire some induced charge and the net charge on surface will be sum of that induced charge and charge $Q$ and this may be non-uniformly distributed on the outer surface of shell. But whatever be the charge on outer surface, there will be a charge of equal magnitude but opposite sign on the inner surface which will be uniformly distributed over inner surface. So net force from these charges on the central charge $q$ will be zero.

But it was given in solution that force on charge $q$ due to both shell & charge outside is zero but only due to the shell was towards right. If we consider only the shell neglecting $q1$, then also the charge $-Q$ will be uniformly distributed on inner walls and will exert net force zero. Then why rightward? Is there anything which I am assuming wrong?


No, in the presence of the point charge $q_1$, the shell will no longer be uniformly charged: it will change its distribution so as to keep the electric field inside the shell zero. This is what the charge distribution will look like:

enter image description here

So the sum of the electric field due to both the shell and $q_1=0$. The force on the charge at the center is zero. The force due to $q_1$alone is $\frac{kqq_1}{r^2}$ to the left so the force due to he shell alone must have the same magnitude and be directed to the right. Whe they say "consider only the shell neglecting $q_1$" they mean find the force due to the shell alone in the same scenario, not to forget that $q_1$ is there at all, and the effect its presence has.

  • $\begingroup$ But I was taught that if a charge (here +q1) is brought close to any body (here the metallic hollow sphere) then there will be uneven distribution of charge on the outer surface (here it will be-q1) as in your figure, but will be uniformly distributed on the inner surface (here +q1 is diributed on inner surface). So this inner charges will have equal forces from all sides contributing net zero force. So in second case also force only due to shell will also be zero. $\endgroup$
    – Avi
    Mar 14 '17 at 7:03
  • $\begingroup$ Even if the charge was distributed that way on the shell and the force due to the charge on the inner surface was--although you really don't need to think out "inner" and "outer" surfaces--don't you still have to take into account the force due to the unevenly distributed charge q1 on the outer surface? $\endgroup$
    – GeeJay
    Mar 14 '17 at 7:11

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