# Work required to bring a charge near a grounded, conductive infinite plane

Whenever there is a grounded plane and we need to calculate the work done in order to bring a charge $$q$$ a distance $$R$$ from the infinite plane, we can use the method of images and place an imaginary charge on the other side of the plane that agrees with the boundary conditions of the electric potential and Poisson equation. Now, what I have a hard time understanding is why calculating the work as $$q\phi$$ where $$\phi$$ is the electric potential at a distance $$R$$ from the plane is not correct, because the answer is always with a factor of $$\frac{1}{2}$$, meaning $$\frac{1}{2}q\phi$$. I know this factor comes when we calculate each force twice when calculating the work done to construct a configuration of charges, but what is the reason here? Is it somehow connected to counting forces twice? I know it has something to do with us not putting any work to bring the imaginary charge but I don't think I completely get it.

The work done in bringing the charge $$q$$ from infinity to a height $$R$$ above the plane is the same as the work done in bringing the charge $$q$$ to a distance $$2R$$ from the image charge $$-q$$. This is because the image charge is also at a height $$R$$ below the plane, which makes the distance between the charges $$2R$$. Hence, the factor of $$1/2$$ arises.
Alternatively, if you choose to think of work as being "stored" in the electric field, the actual scenario with the charge and plane only has electric field in the space above the plane. The space below the plane has no electric field. However, in the image scenario, there is also a "mirror" electric field in the space below where the plane should be, with the plane as the plane of symmetry. Thus it has twice the amount of electric field. Hence, the actual situation only has half the work stored, thus the factor of $$1/2$$.