Energy of a configuration of two concentric spherical charged shells

  • #1
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Homework Statement:

What is the energy configuration of a system consisting of two concentric shell?
The internal shell has radius a and charge q
The external shell radius b and charge -q

Relevant Equations:

All below
I found the total work done is:
##\frac{q^2}{8\pi \varepsilon a} + \frac{q^2}{8\pi \varepsilon b} + \epsilon \int E_{1}.E_{2} dv##
The third is a little troublesome i think, but i separated into threeregions, inside the "inside" shell, between both shell and outside both.
Inside => ##E_{1}.E_{2} = 0##
Between => ##E_{1}.0 = 0##
Outside => ##\epsilon \frac{(4\pi (kq)^2)}{b}## = ##\frac{(q)^2}{4\pi \epsilon b}##
so

##\frac{3q^2}{8\pi \epsilon b} + \frac{q^2}{8\pi \varepsilon a}##

is this right?
 

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  • #2
TSny
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I found the total work done is:
##\frac{q^2}{8\pi \varepsilon a} + \frac{q^2}{8\pi \varepsilon b} + \epsilon \int E_{1}.E_{2} dv##
Can you state the starting point from which you derived the above expressions?
The third is a little troublesome i think, but i separated into threeregions, inside the "inside" shell, between both shell and outside both.
Inside => ##E_{1}.E_{2} = 0##
Between => ##E_{1}.0 = 0##
Outside => ##\epsilon \frac{(4\pi (kq)^2)}{b}## = ##\frac{(q)^2}{4\pi \epsilon b}##
For the outside region did you take into account the directions of the two fields? Do they point in the same direction or in opposite directions in this region?
 
  • #3
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Can you state the starting point from which you derived the above expressions?
For the outside region did you take into account the directions of the two fields? Do they point in the same direction or in opposite directions in this region?
1600214824536.png
THis is where i started.

About the outside, yes, i take. Their product would be negative, but one charge is positive and the other negative, so we have negative again. Integrate r^(-2) would lead us to another negative sign, but we need to integrate from b to infinity, so this will let we with other negative, so we have
---- = +
 
  • #4
TSny
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I think it's easier to evaluate ##\frac{\epsilon_0}{2}\int \mathbf E \cdot \mathbf E d\tau## using the total field ##\mathbf E## rather than break it up as ##\mathbf E = \mathbf {E_1} + \mathbf{E_2}##.

However, it's certainly ok to split the field as you did.
About the outside, yes, i take. Their product would be negative, but one charge is positive and the other negative, so we have negative again. Integrate r^(-2) would lead us to another negative sign, but we need to integrate from b to infinity, so this will let we with other negative, so we have
---- = +
Since the scalar product ##\mathbf{E_1} \cdot \mathbf{E_2}## is negative in the outer region, the integral ##\frac{\epsilon_0}{2}\int \mathbf{E_1} \cdot \mathbf{E_2} \, d\tau## just adds up a bunch of negative quantities. So, the result has to be negative. In a sense, you have "double-counted" the effect of the opposite signs of the charges. The opposite directions of ##\mathbf{E_1}## and ##\mathbf{E_2}## is due to the opposite signs of the charges. So, the negative scalar product of the fields already accounts for the opposite signs of the charges.
 
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