# Minimum value of the electric field

1. Apr 6, 2015

### Jason Williams

1. The problem statement, all variables and given/known data
The potential difference Δϕ between the plates of a spherical capacitor is kept constant. Show that the electric field at the surface of the inner sphere will be a minimum at a = b/2, where a is the radius of the inner sphere and b is the radius of the outer sphere.

2. Relevant equations

E = Q/(4*pi*ϵ0*r^2) between the inner and outer shells and E = sigma/ϵ0 at the surface of a conductor... sorry guys, I'm in a rush and I'm new to Physics Forums so please bear with me noobiness.

3. The attempt at a solution

I tried using E = -dV/dr, setting it equal to 0 and then solving for r, but that clearly doesn't give us the answer. I'm really just confused about how I should even be approaching this problem, and how to extract the E-field from the potential difference.

Thanks!

2. Apr 6, 2015

### TSny

Hello Jason, Welcome to PF!

Can you express $\Delta \phi$ in terms of $a, b$ and $Q$?

Hint: Think about the integral relationship between $\Delta \phi$ and $E$.

3. Apr 6, 2015

### Jason Williams

Well we know $$\Delta \phi= Q*(1/a - 1/b)/(4*pi*epsilon)$$, which we get from doing the integral of E.ds

But here's where I'm getting stuck. I'm trying to reason it out and I'm thinking that maybe the electric field is a minimum when the potential difference between point "a" and point "b" is at a minimum?

4. Apr 6, 2015

### TSny

OK, good.

Note that $\Delta \phi$ is fixed. It's assumed to be held constant as you vary $b$. Does $Q$ vary as $b$ varies?

Last edited: Apr 6, 2015
5. Apr 6, 2015

### Jason Williams

Right so if $$\Delta \phi$$ is fixed, the only thing that can be changed are $$a$$ and $$b$$, which then alter the electric field. I don't think Q would change as we vary $$b$$ because we set the inner conductor to have charge $$Q$$ in the beginning.

6. Apr 6, 2015

### TSny

The potential difference $\Delta \phi$ and $a$ are held fixed. You then want to vary $b$ so that Eb is a minimum, where Eb is the field at the outer conductor.

[CORRECTION: See post #13.]

Q is not fixed. You can think of the capacitor as connected to a battery that keeps $\Delta \phi$ constant. As you vary the separation between the plates, the battery will change Q in order to keep the potential difference fixed.

Can you think of a way to relate Q to b and Eb?

Last edited: Apr 7, 2015
7. Apr 6, 2015

### Jason Williams

Okay the battery analogy makes sense. But isn't the question asking us to show that the electric field at the surface of the inner sphere will be a minimum if a = b/2?

8. Apr 6, 2015

### TSny

Yes. Can you find Eb as a function of b? You have all the necessary equations.

9. Apr 6, 2015

### Jason Williams

Yes, E_b = Q/(4*pi*ϵ0*b^2). I'm sorry for being so slow, but I still don't understand how that helps. Using that, equation, we have two things that change, both Q and b.

10. Apr 6, 2015

### TSny

You have another equation that relates Q to $\Delta \phi$ and b.

11. Apr 6, 2015

### Jason Williams

Okay, so using $$E_b = \frac{Q}{4\pi\epsilon_0b^2}$$ and $$Q = \Delta \phi \frac{(4\pi\epsilon_0)}{(\frac{1}{a} - \frac{1}{b})}$$, giving us $$E = \frac{\Delta \phi}{(\frac{b^2}{a} - b)}$$.

And then we find the minimum from there. Okay, I think I got it now. Thank you so much for all your help and patience!

12. Apr 6, 2015

### TSny

Good.

13. Apr 7, 2015

### TSny

OOPS! Sorry, Jason. Somehow I misread the problem as minimizing the field at r = b. Instead, you want to minimize the field at r = a. Looks like you need to assume b is fixed and vary a.

Last edited: Apr 7, 2015
14. Apr 7, 2015

### Jason Williams

Ahh okay, yeah that's what I figured and changed it in my homework. Thanks so much again!