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Consider a point charge $q$ placed at the center of a neutral spherical conducting shell extending from radius $a$ to $b$. What is the work needed to move the charge out to infinity?

The simple solution is obvious: The charge $-q$ is induced on the inside of the shell. The work to move this charge in place is $-q^2/(4\pi\epsilon_0a).$ The charge $+q$ is induced on the outer surface, but because the electric field outside of the inner surface now is zero, it takes zero work to bring it in place. So the energy of the initial configuration is $-q^2/(4\pi\epsilon_0a)$. The energy of the final configuration is zero, so the required work is $q^2/(4\pi\epsilon_0a).$

But what if we instead look at the electric field and the energy density $\epsilon_0/2|\vec{E}|^2$. I know that if we compute this energy for the field due to a point particle, the integral diverges. But I think we should be able to cancel it because it appears in both the initial configuration and in the final configuration.

Initial: Let $\hat{r}$ point from the origin (where the charge is placed). Then $$\vec{E} = q/(4\pi\epsilon_0)\hat{r}/r^2, \ r<a $$ $$\vec{E} = \vec{0}, \ a<r<b $$ $$\vec{E} = q/(4\pi\epsilon_0)\hat{r}/r^2, \ b<r $$ Hence the energy of this configuration is $$\frac{q^2}{8\pi\epsilon_0}\left( \int_0^a\frac{1}{r^2}dr+\int_b^\infty \frac{1}{r^2}dr\right). $$

Final: Let $\hat{r}$ point from the where the charge is placed. Then $$\vec{E} = q/(4\pi\epsilon_0)\hat{r}/r^2.$$ Hence the energy of this configuration is $$\frac{q^2}{8\pi\epsilon_0}\left( \int_0^a\frac{1}{r^2}dr+\int_a^\infty \frac{1}{r^2}dr\right). $$

The difference in energy is $$\frac{q^2}{8\pi\epsilon_0}\left( \int_a^b\frac{1}{r^2}dr\right) = \frac{q^2}{8\pi\epsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right),$$ which is obviously not the same as the previous result.

Any input on this?

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    $\begingroup$ "because the electric field outside of the inner surface now is zero, it takes zero work to bring it in place." The field vanishes inside the conductive material, but the effective field acting on the charges on the surface of the sphere does not; any surface charge element experiences force of repulsion from the other surface charge elements. So the first method is not valid. $\endgroup$ – Ján Lalinský Nov 15 '16 at 22:08
  • $\begingroup$ @JánLalinský Ah, yes you're right. So if I understand you correctly, I would need to account for the energy to assemble the outer shell of charge $+q$ assuming no external field. This I can find by integrating $|\vec{E}|^2$, giving $q^2/(8\pi\epsilon_0b)$. Correct? However, the energy $-q^2/(4\pi\epsilon_0a) $ must now also be incorrect because it failed to account for the repulsion of charges on the shell. Maybe it's best to do the calculation by integrating the electric field as I did. BTW, this is a problem from Griffiths EM which (incorrectly) gives the answer $q^2/(4\pi\epsilon_0a)$. $\endgroup$ – Étienne Bézout Nov 15 '16 at 22:46
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The energy can also be calculated in this way: replace the point charge by a small charged sphere of radius $R$; then energy is equal to electrostatic energy

$$ \int \frac{1}{2}\rho(\mathbf x) \varphi(\mathbf x)\,d^3\mathbf x $$

This evaluates to

$$ \frac{q^2}{8\pi\epsilon_0}(1/R + 1/b - 1/a). $$

When energy of the separated configuration is subtracted, we get

$$ \frac{q^2}{8\pi\epsilon_0}(1/b - 1/a) $$

which is negative (the conductive shell attracts the charge).

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  • $\begingroup$ Thanks! I recently developed some doubt however about the energy cost of a assembling a spherical shell due to repulsion. Consider the inner surface of the conductor to be a conductor with charge $-q$. We can imagine this surface to initially be very large. We want to shrink it down to radius $a$. You're saying it takes work to shrink the sphere due to repulsion of charges on the sphere. Correct? However, a conductor is an equipotential surface, so moving around charges on the conductor costs no work. Correct? $\endgroup$ – Étienne Bézout Nov 17 '16 at 17:11
  • $\begingroup$ @ÉtienneBézout, if the charges are brought closer together, the work does not depend on whether they are on a surface of a conductor or isolated particles in vacuum. The charge elements repel each other. In case of conductor, it takes zero work to move charge fixed on a sphere, but it does take non-zero work to shrink the sphere itself. $\endgroup$ – Ján Lalinský Nov 18 '16 at 2:34
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Well you are doing a very basic blunder. In first method, what you are finding out is the potential energy of the charge at centre which must be: Kq^2(1/b-1/a) It's is equivalent to the work done in moving a charge from the center of 2 concentric shells having charges plus and minus q. And identify of the shells is retained after removing the charge.

However question demands something else. Here identity of shell is destroyed as charge is removed to infinity. And system becomes void after charge removal. So technically you need to find out the energy of electric field . Which can be found by second method. After charge removal , no electric field exists and therefore go for second method.

For an example , check this question out. Your concepts will be much clear.

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  • $\begingroup$ Please do not post pictures of text. Pictures are not accessible to all users. Please replace your pictures with the relevant text you want posted. For equations and variables, use MathJax $\endgroup$ – BioPhysicist Jan 20 '20 at 13:41

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