# Electric Field minimum in a spherical capacitor.

1. Oct 20, 2009

### Kizaru

1. The problem statement, all variables and given/known data

The potential difference $$\Delta\phi$$ between the plates of a spherical capacitor is kept constant. Show that then the Electric Field at the surface of the inner sphere will be a minimum if $$a = \frac{1}{2}b$$

2. Relevant equations
$$E = \frac{Q}{4\pi\epsilon_{0}r^{2}}$$
between the plates.

3. The attempt at a solution
Not really sure. I know that
$$\Delta\phi = - \int E dl$$
and that it is constant. But not really sure how to proceed from here. I think I need to find some sort of new equation for E, then find the minima? Answer in the back of the book is E = 4 (delta)(phi) / b

The answer latex didn't work, sorry.

Last edited: Oct 21, 2009
2. Oct 21, 2009

### turin

Your line integral can be taken along a radial line, from a to b.

BTW, I would use $\Delta\phi$ instead of $\nabla\phi$ for the total potential difference. It's just a notational convention, but it is a very common one, and I was a little bit confused at first.

3. Oct 21, 2009

### Kizaru

Err, yes I meant delta phi. Taking the line integral doesn't help with the solution. If I did take the line integral, I get the electric potential difference. But this isn't helpful because
$$\Delta\phi = \frac{Q}{4\pi\epsilon_{0}}\left(\frac{1}{b}-\frac{1}{a}\right)$$

And then how would I find the minimum of the electric field from there?

I already passed in the problem--probably done incorrectly. I used the mean value theorem to integrate over E from b to a, then divide by (b-a). Then I took the derivative with respect to a, and tried to set equal to 0 but there were no values for which that occurred. I did manage to get rid of one of the two terms when a = (1/2)b.

4. Oct 21, 2009

### turin

Consider Q as a function of a (or b, or their ratio).