Consider a point charge located above an infinite, grounded, conducting plane. The potential everywhere above the plane is the same as that of the image charge (located below it). Why is it wrong to say the total energy of the original charge is just its charge times the potential at that location? I'm aware there's a factor of one half, I just don't understand what's the problem with this specific argument, since the there isn't such a factor for the potential itself
Electrostatics – Understanding Energy Due to Image Charges
electrostaticsenergymethod-of-images
Related Solutions
The problem is addressed in Electrodynamics Fulvio Melia Example 2.1 page 39, using Green function techniques. However, I cannot fully understand his final solution. Let us define the plane held at a constant potential $V_0$ to be the $z$-$y$ plane. The charge $q$ is at the distance $a$, namely at $(a,0,0)$. The potential at large distances is zero. The solution (Eq. 2.54) is
$$V(x,y,z) = V_{image} (grounded) + V_0 f(x,y,z).$$
The first term is the usual image method solution, for the case where the conducting plane is grounded, namely
$$V_{image}(grounded) = q\left[\left[(x-a)^2 + y^2 + z^2\right]^{-1/2} - \left[(x+a)^2 + y^2 + z^2\right]^{-1/2}\right],$$
which gives zero when $x=0$. In other words this is the solution of the Poisson equation, but the boundary conditions are not those required.
The second term, $V_0 f(x,y,z)$ is the solution of Laplace equation, and $f(x,y,z)$ is $V_0$ independent.
According to the text book Eq. (2.54) there
$$f(x,y,z) = (x/2 \pi) \int dz' \int dy' \left[x^2 + (y-y')^2 + (z-z')^2\right]^{-3/2}.$$
The integrals over the primed variables stretch from minus infinity to plus infinity and this function is V_0, q and the length scale a independent.
The solution is mysterious for me. Namely, we can get rid of $y$ and $z$ in the integral (for any finite $y$ and $z$) by change of variables of integration. Then the integral can be solved (for $x>0$) and the solution is $f(x,y,z)= x/ x = 1$. Well, that is wrong (or subtle) since that would mean that we only need to add to the usual image solution a constant $V_0$. This does not satisfy the boundary condition at large distances from the plane where the potential is zero.
Maybe the integral for the dimensionless function f(x,y,z) should be regularised somehow.
Since the image part of the solution, $V_{image}(grounded)$, takes care of the charge in the Poisson equation, we can find a solution that is a linear combination of $V_{image}(grounded)$ and some other function, the latter solves the Laplace equation. Then let the boundary at $x=L$ be grounded and we want to take $L\to \infty$ and on x=0 we have V_0 (this seems like a legitimate trick). The Laplace equation for this problem gives
$$V(x,y,z) = -V_0 x/L + V_0,$$
and if we take $L\to \infty$ we get the result just mentioned (for any finite $x$) namely a constant (which does not satisfy the boundary condition at infinity). So I think the limit of a large system ($L\to \infty$) is creating a certain non-trivial problem, which ultimately is related to the Green function technique used by Fulvio Melia. Any ideas?
Of-course if f(x,y,z) is unity for any finite distance x, this would mean that switching from grounded plane to non-grounded one (compared to zero potential at infinity) has practically no effect. This hypothesis was postulated in some of the previous discussions.
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.
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?
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 $2$.
Best Answer
There is only one charge. You include the second charge to make the problem similar by including the image charge in such a way which satisfies the boundary condition for the problem of a point charge and an infinite plane. So, when we consider the energy of the plane and the charged particle, we are actually considering only half of the energy of the system with one real and one image charge. That's why the actual energy is going to be half of it.
Please let me know if you have any doubts.