Energy in a point-plane symmetry problem I was faced with a following problem:
A point charge $q$ is at a distance $l$ from an infinite conducting plane. Find the energy $W$ of interaction between this charge and the charges induced on the plane.
My thought process mostly revolved around localization of energy idea i.e. integrating the upper half of the space, but plane being infinite, as well as  having in mind that that would calculate the energy as a whole rather than an interaction.
The author's idea was to take $q$ to infinity and reason that the energy needed for that is equal to what we want (at this point the energy of the system is zero-$q$ is to far away to induce anything on the plate). However wouldn't energy of  plate itself and of the point-charge also take the hit?In which case  we only conserve $W_1+W_2+W_{12}$ rendering the idea useless?
Could someone point out if the idea is wrong or better yet I am?
 A: The author is correct.
The interaction energy $W_0$ is the electrostatic potential energy of the system of charges. This is the work required to assemble a charge distribution, or the negative of that required to remove the charges until they are infinitely far apart where they no longer interact.
As the point charge $+q$ is withdrawn to an infinite distance from the plate, the -ve charges induced on the plate automatically disperse at the same time until they also are infinitely far apart. The movement of $+q$ causes the movement of the induced charges. So the work done $W_0$ to move $+q$ to infinity includes the potential energy $W_2$ which is released by the induced charges as they disperse.
This makes sense because, as you pointed out, the final configuration has no electrostatic potential energy.
If somehow the induced charge distribution had been frozen in place so that it did not move as $+q$ was withdrawn, then the work done $W_1$ to remove $+q$ would be higher than $W_0$, and there would be some electrostatic energy $W_2$ remaining in the charge distribution on the plate. That is : $$W_1=W_0+W_2$$

To calculate $W_0$ note that the induced charge distribution is equivalent to having a mirror-image of the charge $+q$ (see Method of Images). 
