In the shown figure the conductor is uncharged and a charge q is placed inside a spherical cavity at a distance a from the centre(C).Point $P$ and a charge $+Q$ are shown.

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I want to find electric field due to induced charges on the inner surface of cavity at point $P$.

I feel the $q$ charge inside the shell will induce $-q$ charge on the inner surface of the shell.So the inner surface of the shell will induce an electric field of magnitude $\dfrac{q}{4 \pi \epsilon_o c^2}$ at point $P$.Is my assumption correct?

Also,I want to find the electric potential due to charges on the inner surface of the cavity and $q$ at $P$.

I think it should be $\dfrac{k(-q)}{c}+\dfrac{k(q)}{x}$ where $x$ is distance between charge $q$ and point $P$.But the answer given in my textbook is $0$.I've no idea why!Its seems very strange.Can someone clarify?


2 Answers 2


For problems of this kind you need Gauss' law, saying that the electric flux through a closed surface is proportional to the charges contained inside that surface. Then you also need that inside of a conductor, there cannot be any electric field.

Taking these things together give you a route towards your first question. Please think about it yourself before continue to read.

Put the surface inside the conductor and around the cavity. The total electric flux must be zero because the whole surface is inside the conductor. This means that the charge on the inner surface is just $-q$. Your feeling is correct. Using a similar approach can give you the electric field inside of the cavity. For the Gaussian surface choose a sphere around the point charge. From spherical symmetry we know that the electric field is the same in every direction and also radially. This simplifies the integral to a simple product of sphere surface area and the electric field. Solving for the electric field then gives you the potential which is also what you have derived.

For the second part it is not clear to me whether the blob of conducting material is uncharged before or after the point charge is put into the cavity. From the solution of your problem I think that the conductor is has been grounded shortly after the point charge was implemented.

You can again use Gauss' law to derive the result. Alternatively one can use that a spherical distribution of charges act to things on the outside like if the charges were concentrated in the middle.

We noticed already that there is a charge $-q$ on the inside surface of the cavity such that the charge $q$ is screened. Concentrating both charges to the middle gives $Q = 0$, therefore there is no electric field on the outside. This assumes that there are no charges on the outside of the conducting blob. If the conductor had been neutral before the point charge was implemented, there would be a net charge $q$ on the outside now because $-q$ charges went to the inside. This would then create a new field which can be detected at point $P$. Because the blob has such an undefined shape, it will be extremely hard to compute the actual field there. This is a hint that the field is zero :-).

  1. Electric field due to induced charge$(-q)$ on the inner cavity surface 'alone' cannot be calculated as the charge distribution is not known(since $q$ is not at the centre).

(If you want to find out the net electric field at $P$, then it is not due to the charge $q$ and the induced charges in the inner surface of the cavity(They cancel each other out; this is a property of conductors). It is actually due to $+q$ induced on the outer surface of the conductor due to the induced charge $-q$ on the spherical cavity. Also, you have to consider the charge induced to $Q$ on the outer surface of the conductor to find the net electric field at $P$)

  1. Potential is zero(due to $q$ and the induced $(-q$) because of the property of the conductor I mentioned(the charges inside a cavity inside the conductor are virtually hidden from the outer world; their field is always zero outside the conductor; their effect is seen only due to induced charges on the outer surface of the conductor)

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