I am trying to solve the following problem:

A charge $Q$ is brought in from infinity to the neighbourhood of an isolated uncharged conductor. The fields due to the induced charge distribution on the conductor do an amount of work $W$ on $Q$ as it is brought in. We now "freeze" the surface charge in place on the conductor and remove $Q$ back to infinity. How much energy is stored in the remaining electric field distribution?

My reasoning so far is that when the charge $Q$ is moved from infinity near the conductor, it's going to polarize the conductor. Specifically, I know that for a grounded conductor I can place an image charge inside the conductor to represent the induced charge, and then I can remove the ground, and place an excess charge on the surface of the conductor until I zero out the total charge to recreate the uncharged conductor as in the problem statement. Using that logic, it seems like I should be able to calculate the dipole moment between the image charge and the charge that is then uniformly distributed over the surface of the conductor to make the total charge zero, using the equation:

$\vec{p} = \iiint \vec{r} \rho(\vec{r})dV.$

It seems though that I am making this more complicated than it should be, because I am running into some issues with calculating $\vec{p}$; also it seems like there should be no polarization field inside of the conductor since the electric field inside of the conductor is $\vec{0}$... Can someone offer insight as to how to more intuitively calculate how much energy will be stored in the field?

My gut feeling is that I should be trying to approximate the electric field outside of the conductor as a dipole, and then integrate using

$U = \frac{\epsilon_0}{2}\iiint_{all\,space} E^2 dV$

But again, I'm not sure if this is the correct approach, and it seems like I'm really over complicating things.



1 Answer 1


Notice that $W > 0$ since the displacement of $Q$ has the same direction with electric field, which is attractive. The energy of the conductor and $Q$ is $-W$ because we need to do mechanical work of $-W$, which would be stored in the system, to overcome the attractive force between the conductor and $Q$. If we denote the field at charge $Q$ as $\phi_{0}$, then we have \begin{align} U = -W = \frac{1}{2}\int \rho \phi dV = \frac{1}{2}(q_{\text{ind}}\phi_{\text{cond}} + Q\phi_{0}) = \frac{1}{2}Q\phi_{0} \end{align} since conductor is equipotential and sum of induced charge vanishes.

After removing charge $Q$ while fixing induced charges, the potential of conductor is no longer constant but \begin{align} \phi(\vec{r}) = \phi_{\text{cond}} - \frac{Q}{|\vec{r} - \vec{a}|} \end{align} where $\vec{a}$ is the initial location of charge $Q$. The energy of this distribution can be obtained as \begin{align} U = \frac{1}{2}\int \rho \phi dV = \frac{1}{2} \int \sigma \left(\phi_{\text{cond}} - \frac{Q}{|\vec{r} - \vec{a}|}\right) dA \end{align} Notice that the first term vanishes since $\phi_{\text{cond}}q_{\text{ind}} = 0$ and the second term is $Q$ times potential of induced charge at $\vec{a}$, or \begin{align} U = -\frac{1}{2}Q\phi_{0} = W \end{align} Therefore, the energy stored in the remaining distribution is $W$.


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