Minimum value of the electric field

[Jason Williams]

Homework Statement

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.

Homework 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.

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!

 

[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 $E$.

 

[Jason Williams]

Thanks for the reply and so glad to be here!

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?

 

[TSny]

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

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?

 

[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.

 

[TSny]

The potential difference $\Delta \phi$ and $a$ are held fixed. You then want to vary $b$ so that E_b is a minimum, where E_b 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 E_b?

 

[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?

 

[TSny]

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

 

[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.

 

[TSny]

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

 

[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!

 

[TSny]

Good.

 

[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.

 

[Jason Williams]

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