# Work done in moving a charge, from near a conducting plane to infinity

A point charge $q$ is located at a distance $l$ from the infinite conducting plane. What amount of work has to be performed in order to slowly remove this charge very far from the plane? (Irodov 3.55)

I understand there is a method of images for solving such problems. I did use it, but I have three reasonable ways to use it, yet only one is giving correct answer.

Method 1: $W_{\text{conservative force}}=-\Delta U_{\text{of that force field}}$

So, since our charge will be at rest finally at infinite distance ("slowly remove"), so by Work-Energy theorem, $W_{\text{electrostatic force}}+W_{\text{external agent}}=0$, hence, $W_{\text{external agent}}=\Delta U_{\text{electrostatic}}=0-\frac{-q^2}{4\pi\epsilon_0\cdot(2l)}=\frac{q^2}{8\pi\epsilon_0l}$

But this is the incorrect answer.

Method 2: $W_{\text{external agent}}= \int\text{Force}\cdot{\text{Displacement}}=\int^\infty_{2l}\frac{q^2}{4\pi\epsilon_0\cdot(x^2)}dx=\frac{q^2}{4\pi\epsilon_0}(\frac 1{2l})=\frac{q^2}{8\pi\epsilon_0l}$

Same as the answer in method 1, and still wrong.

Method 3: $W_{\text{external agent}}= \int\text{Force}\cdot{\text{Displacement}}=\int^\infty_{l}\frac{q^2}{4\pi\epsilon_0\cdot((2x)^2)}dx=\frac{q^2}{16\pi\epsilon_0}(\frac 1{l})=\frac{q^2}{16\pi\epsilon_0l}$

In this attempt, I only changed the variable of integration, and it gave the correct answer.

My question:

Method 1 and 2 are completely logical according to me and they should give a correct answer. Yet, only 3 gives the correct answer. So, what is the logical mistake in method 1 and 2?

• Thanks for your answer! A clarification: the other side of the conducting plane (the one not facing the point charge $q$) will also have certain induced charge. Correct? But, it will not have any electrostatic potential energy because there is no point charge on the other side (since for potential energy, we need at least two charged bodies, but on the other side of the plane, we only have one) Am I correct? Thanks! – Gaurang Tandon Jan 27 '18 at 3:15