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Minimum value of the electric field

  1. Apr 6, 2015 #1
    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. jcsd
  3. Apr 6, 2015 #2

    TSny

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    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##.
     
  4. Apr 6, 2015 #3
    Thanks for the reply and so glad to be here!

    Well we know [tex]\Delta \phi= Q*(1/a - 1/b)/(4*pi*epsilon)[/tex], 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?
     
  5. Apr 6, 2015 #4

    TSny

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    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
  6. Apr 6, 2015 #5
    Right so if [tex]\Delta \phi[/tex] 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.
     
  7. Apr 6, 2015 #6

    TSny

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    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
  8. Apr 6, 2015 #7
    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?
     
  9. Apr 6, 2015 #8

    TSny

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    Yes. Can you find Eb as a function of b? You have all the necessary equations.
     
  10. Apr 6, 2015 #9
    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.
     
  11. Apr 6, 2015 #10

    TSny

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    You have another equation that relates Q to ##\Delta \phi## and b.
     
  12. Apr 6, 2015 #11
    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!
     
  13. Apr 6, 2015 #12

    TSny

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    Good.
     
  14. Apr 7, 2015 #13

    TSny

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    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
  15. Apr 7, 2015 #14
    Ahh okay, yeah that's what I figured and changed it in my homework. Thanks so much again!
     
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