Electrostatic energy of spherical shell.

  • Thread starter Fabio010
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  • #1
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Homework Statement



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




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??
 

Answers and Replies

  • #2
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W = (1/2) * ∫σVda

because dq = σda

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


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
Didn't you say in part a that the E field inside the sphere was everywhere 0?
 
  • #3
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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?

So the energy is zero?
 
  • #4
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the integration must be from R to infinity,not from 0 to R.
 
  • #5
83
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the integration must be from R to infinity,not from 0 to R.
yeh, it makes sense. And the result is positive.

Thanks for the help.
 

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