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?
 A: For the first two methods you have missed out an important idea.  
The electric potential energy is "stored" in the electric field.  
With your first two methods you have found the energy stored by the system of two charges in the electric field which occupies the "whole of space".
The infinite conducting plates cut the "whole of space" in half and so with the infinite conducting plate the energy stored in the electric field is half that found using your first two methods.
A: Method of images is being used only to solve the problem, there is not any real point charge behind the plane. So in order to find the work done in dismantling the system if we follow the usual process by which we dismantle the two point charges placed at a distance, if we move one charge with respect to origin while keeping  mirror image charge fixed then we have altered the original problem as the infinite plane  does not remain  conducting plane then. So  we have to dismantle in a way (imaginary)  such that by moving charge $q$ by $dx$ amount the distance between  two charges increase by $2\ dx$. Work done /PE is half of that in case of two real point charges.
