- #1

cwbullivant

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## Homework Statement

A metal sphere with radius a is supported on an insulating stand at the center of a hollow, metal spherical shell with radius b. There is charge +Q on the inner sphere and charge -Q on the outer shell. Take the potential V to be zero at infinite separation.

Calculate the potential V(r) for:

i) r < a

ii) a < r < b

iii) r > b

## Homework Equations

1) $$ E = -\frac{\partial V}{\partial r} $$

2) $$ V = \int E d\ell $$ (along radial direction)

## The Attempt at a Solution

(iii) I already know; V is obviously 0 outside both shells (since E is also zero outside both, as a Gaussian sphere drawn beyond r(b) will enclose Q-Q = 0 charge).

I derived (ii) as a problem in lecture last week.

This leaves only (i) as the problem. The attempt was to use the second of the two equations up top, first finding the electric field so it can be applied:

Applying Gauss' law, we can find that the Electric field inside the inner sphere is:

$$ \frac{q'}{q} = \frac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi a^{3}} $$

$$ q' = q\frac{r^{3}}{a^{3}} $$

From there, using a Gaussian sphere with a radius r < a:

$$ E(4\pi r^{2}) = \frac{q\frac{r^{3}}{a^{3}}}{\epsilon_{0}} $$

$$E = \frac{1}{4\pi \epsilon_{0}} \frac{qr}{a^{3}} $$

And then applying the equation for potential difference:

$$ \int E d\ell = -\Delta V $$

Where dl = dr, choosing a radial path from infinity to r.

$$ \int \frac{1}{4\pi \epsilon_{0}} \frac{qr}{a^{3}} dr = -\Delta V $$

$$ \frac{1}{4\pi \epsilon_{0}} \frac{qr^{2}}{2a^{3}} = \Delta V = V_{r} - V_{\infty} = V_{r}$$

I'm suspicious of this solution though, as it doesn't reduce to

$$ \frac{1}{4\pi \epsilon_{0}} \frac{q}{a} $$

For the case where r = a.

Did the mistake lie in how I calculated the electric field, the technique I used to determine potential, or something else altogether?