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Spherical capacitor (Irodov 3.101.)

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the capacitance of an isolated ball-shaped conductor of radius R1 sorrounded by an adjacent concentric layer of dielectric with permitivity ε and outside radius R2.


    2. Relevant equations



    3. The attempt at a solution

    The official solution says something like:

    Difference of potentials is --- int(R1,R2)(E*dr) + int(R2,∞)(E*dr)

    Sorry, but I don't know how to write equations nicely.
    I dont quite understand what is the second integral here for. Can someone explain?
     
  2. jcsd
  3. Apr 30, 2012 #2
    The electric field only exists between the two conductors, and they carry equal by magnitude, but opposite by sign charges [itex]Q[/itex], and [itex]-Q[/itex], respectively. Using Gauss's Law, you should get:
    [tex]
    E(r) = \frac{Q}{4 \pi \epsilon} \, \frac{1}{r^2}
    [/tex]
    and the direction is along the radius.

    The potential difference is:
    [tex]
    V_1 - V_2 = \int_{R_1}^{R_2}{E(r) \, dr} = \frac{Q}{4 \pi \epsilon} \, \int_{R_1}^{R_2}{\frac{dr}{r^2}}
    [/tex]

    To perform this integral, I suggest you use the following subsitution:
    [tex]
    r = A \, x^\alpha \Rightarrow dr = A \, \alpha \, x^{\alpha - 1} \, dx
    [/tex]
    Then, the integrand becomes:
    [tex]
    \frac{A \, \alpha \, x^{\alpha - 1} \, dx}{A^2 \, x^{2\alpha}} = \frac{\alpha}{A} \, x^{-1 - \alpha} \, dx
    [/tex]
    What would be the most convenient choice for [itex]\alpha[/itex]? Does it depend what you choose for [itex]A[/itex] (look at how the bounds of integration are changing)?
     
  4. May 1, 2012 #3
    Thank you for the reply.

    I know how to solve integrals. What I dont know is... Why are there two integrals in the official solution for potential difference? One from R1 to R2 and the other from R2 to infinity. If you dont have it, the soultions for irodov are available online so if you could check it out and tell me what the second integral is there for, that would be great.
     
  5. May 2, 2012 #4
    I haven't seen the solution online. If you know the link, post it here.

    There is an integral from [itex]R_2[/itex] to [itex]\infty[/itex] because the electric field is not zero, since there is no exterior grounded spherical shell. I misread the formulation of the problem.

    By the way, you should use Gauss's Law for the dispalcement vector [itex]\vec{D}[/itex], which is

    Flux of D = total enclosed free charge

    and then find [itex]E = D/\epsilon[/itex] in the corresponding region. You can see that the electric field is discontinuos at the boundary.

    Effectively, you have two spherical capacitors connected in series.
     
    Last edited: May 2, 2012
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