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!