Consider a point charge $q$ a distance $z$ from an infinite grounded conducting surface in the $xy$-plane. Using the method of images, we know that the potential $V$ can be found in the region $z>0$ by using an image charge at $-z$. However in this situation, how do we work out the force the charge experiences (since the image charge is not stationary)?
Constraining the problem to just the $z$-axis, if we find $F = – \frac{\partial W}{\partial z}$ from first principles we get:
$$
F = – \lim_{\delta z \to 0} \frac{W(z+\delta z) – W(z)}{\delta z} = \frac{q^2}{4 \pi \epsilon_0} \lim_{\delta z \to 0}\frac{1}{\delta z} \Big( \frac{1}{z+2\delta z} – \frac{1}{z}\Big) = \frac{q^2}{4 \pi \epsilon_0} \Big( \frac{-2}{z^2})
$$
where the first term has it's denominator increased by $2\delta z$ since, if the real charge moves, so does the point charge. But this isn't correct and seems to have an extra factor of 2, why? Can we not use the same definition for force from work?
Edit:
And what happens in the situation when we have two free point charges (i.e. the image charge is now a real charge) and they both fall towards each other?
Best Answer
The resolution is that the image charge doesn't exist, so it doesn't take any "work" to "move" it.
More specifically, the image charge represents the effect of all the surface charges distributed on the grounded plane. "Moving" the image charge actually physically corresponds to moving charge on the plane, which occurs in response to the real charge's motion. But since the plane is always grounded, moving the charge on the plane always takes no work, because you're just moving charge from zero potential to zero potential.
Remember that the formula $F = - dU/dx$ applies when $F$ is the force on a particle, $U(x)$ is the potential energy of that particle, and $x$ is its position. In situations where you have two particles, the potential energy depends on both particles' positions, so we actually have $$F_1 = - \frac{\partial U(x_1, x_2)}{\partial x_1}, \quad F_2 = - \frac{\partial U(x_1, x_2)}{\partial x_2}.$$ In this particular case we have $$U(x_1, x_2) = \frac{q^2}{4 \pi \epsilon_0} \frac{1}{|x_1 - x_2|}$$ which you can confirm gives the right force on each particle, with no extra factors of $2$.