Work: Moving point charge from center of conducting shell to infinity

Question

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?

Answer 1

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).

Answer 2

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.