# Confusion with Electrostatic Potential Difference

Suppose that you have a positively charged plate at a plane (xy-plane, for example). Therefore this plate exerts an electric field with a magnitude of σ/2ε₀. Due to this, there will be the same potential at equidistant points from the plane: V = Ed (choosing as a reference point infinity). Therefore the potential difference going from a plane P parallel to the xy-plane will just be V = Ed (where E is σ/2ε₀) if we choose the potential at the surface of the plate to be 0, which would be just the work per positive unit charge that must be done from the plane P to the charged plate.

Now, if I place another plate at the plane P (not charged) such that it is parallel to the positively charged plate, due to induction, electrons on the second plate will displace to the layer of the plate that is closer to the first plate, leaving protons behind. Supposing electrostatic conditions, what is the potential difference now from the first plate (+σ) to the second plate (-σ) which is located at the plane P? Is it still V = Ed? And what if the plate was already negatively charged with (-σ)? Would the situation change?

If I said something that is wrong, please correct me. Thanks very much in advance.

• Does this answer your question? What happens to potential when a charged body is placed next to an uncharged conductor? Commented Jul 28, 2021 at 21:38
• FYI, the potential difference will be $V=ED$ regardless of where zero potential is assigned. The potential will be $V=ED$ if we choose the potential at the surface to be zero. Commented Jul 28, 2021 at 21:39
• @BobD it did clear things up. I didn't think of the conducting body being induced by the charges on the induced uncharged body. Thanks! Commented Jul 29, 2021 at 1:28