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Electrostatic energy of spherical shell.

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the electrostatic energy, W, of a spherical shell of radius R with total charge q, uniformly distributed. Compute it with the following methods:

    a) Calculate the potential V in spherical shell and calculate the energy with the equation:

    W = (1/2) * ∫σVda

    b) Calculate the electric field inside and outside, and calculate the energy with the equation:

    W = (1/2) *εo ∫ E^2 dV




    3. The attempt at a solution

    a)
    Electric filed inside = 0


    Electric field outside = (1/4piεo) * Q/r^2


    The Potential is v(r)-v(infinity) = - ∫(infinity to R) (Electric field outside) dr

    = (1/4piεo) * Q/R

    W = (1/2) * ∫σVda


    because dq = σda

    W = (1/2) * ∫(1/4piεo) * q/R * dq = (1/8piεo) * Q^2/R


    b) The electric field is already calculated...


    W = (1/2) *εo ∫ E^2 dV

    = (1/2) *εo ∫ (1/16pi^2εo^2) * q^2/r^4 dV

    (1/2) *εo ∫(0 to 2pi) ∫(0 to pi) ∫ (0 to R ) [(1/16pi^2εo^2) * q^2/r^4 ] * r^2 sinθ dr dθ dβ


    = -(1/8piεo) * Q^2/R




    The problem is, the difference in the signal...isn't it supposed to be equal??
     
  2. jcsd
  3. Oct 31, 2012 #2
    You undid the simplification that was made for you :)
    W = 1/2 ∫ V dq = 1/2 ∫ σV da

    The q inside your integral is the constant Q, not little q that varies, but I think you made a typo since the constants are still right.


    Didn't you say in part a that the E field inside the sphere was everywhere 0?
     
  4. Nov 1, 2012 #3

    So the energy is zero?
     
  5. Nov 1, 2012 #4
    the integration must be from R to infinity,not from 0 to R.
     
  6. Nov 1, 2012 #5
    yeh, it makes sense. And the result is positive.

    Thanks for the help.
     
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