V(center) of charged metal sphere inside a grounded shell

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Potential at the center of conducting charged sphere surrounded by a grounded shell

If I take another sphere of charge –q of radius a with uniform charge density, then the potential on the spherical region from the radius a to b is same to that of the original question.

According to Uniqueness theorem, the V at the center of the sphere due to the above system should be equal to that of the original question.

Since the conducting sphere is an equipotential, the potential due to this sphere at center is same to that of the surface = $\frac { q}{4 \pi \epsilon _0 R}$ .The potential due to the image sphere of charge –q at the center is $\frac { -q}{4 \pi \epsilon _0 a}$ .

So, the total potential at the center is $\frac { q}{4 \pi \epsilon _0 R}$ - $\frac { q}{4 \pi \epsilon _0 a}$ , option (d).

 

[kuruman]

Looks OK, but you don't need to invoke the Uniqueness Theorem and image charges. Note that grounding the sphere effectively brings the reference point from infinity to $r=a$. Then just use Gauss's Law to find the E-field in the region $R < r < a$ and then use $V(R)-V(a)=-\int_a^R{E~dr}$.

 

[Pushoam]

Note that grounding the sphere effectively brings the reference point from infinity to r=a.

I didn't note this.
I was in the habit of taking reference point at infinity, so I thought of solving it that way even when an easier approach was near.

Thanks for pointing it out.

 

[haruspex]

didn't note this.

If you are prepared to trust the list of options then, noting that, you can get to the answer very quickly. It means that the radius b cannot be relevant, so we rule out options a and c.
As radius a tends to R, the induced charge cancels the charge on the inner sphere. This leaves d as the only option.