# Work to induce charge on grounded conductors

I'm currently studying Method of Images in Griffiths book and in section 3.2 he introduces the method of images for a point charge at a distance $$d$$ from a grounded conducting plane at potential $$V = 0$$. In subsection 3.2.3, Griffiths compute the energy of the real system and the image charge system and obtain results differing by a factor of 2, The explanation of griffiths is that in the image problem, we do work on both charges bringing them from infinity to a distance $$2d$$ apart from each other. However in the real problem, griffiths says that we do work only on the point charge $$q$$, since the induced charge on the conductor moves along an equipotential. My problem is: if the potential inside the grounded conductor still $$V=0$$ even with the external electric field due to the point charge, how charges can be induced, since no force acts on them? I've tried to re-read conductors section in griffiths chapter 2, but as long as I remember, he makes no mention to this problem.

I've already read some questions related to this topic but none of them answered my question\

Also sorry for possible typos, I'm still learning english

Interesting question. This stumped me for a second, but I think the following argument explains it:

Imagine moving the point charge $$q$$ a tiny bit closer to the grounded plate (a displacement $$dx$$, say). Before this displacement, the charges in the plate are in equilibrium and the plate is at $$V=0$$. After the displacement, then instantaneously, the charges in the grounded plate will feel a force (and $$V$$ will not be uniformly zero), and very quickly rearrange themselves into a new equilibrium position.

However, since the displacement of the charge $$q$$ was infinitesimal, the charges in the plate will only move an infinitesimal distance $$dr$$ due an infinitesimal force $$dF$$ in order to reach their new equilibrium, so that the total work done on the charges in moving from one equilibrium to the other is of the order $$(dr)(dF)$$, which is a product of differentials so can be ignored, i.e., the work done is zero!

The reason this isn't usually talked about is because conductors are usually treated in electrostatics, where moving charges aren't part of the theory. Since there are ways to argue that the field in a conductor is zero, discussions of how the charges got to their equilibrium positions in the first place are usually avoided. But for this example, I think the above explanation works.

Hope this helps.

• Tottaly clarifying answer. Thanks for your time
– Geni
Dec 8, 2020 at 22:31