Potential difference between charged parallel plates

[Robin64]

So we know that the E-field between two parallel plates is constant and that the potential difference between the plates is just the E-field times the distance between the plates. Let's say we're moving a positive charge from a negatively charged plate to a positively charge plate ( or near), and instead of using the E-field and distance between the plates to calculate the potential difference, we instead just want to use the formula for electric potential, V=kq/r. How do we deal with the r=0 that inevitably crops up?

 

[andrewkirk]

The V=kq/r formula is for forces between two point particles, not two plates. To use it you need to focus on a positively charged particle, call it p, on the positive plate, and the forces between that and the moving positive particle, call it q. But since the plate is conductive, as q nears the plate, p, as well as all other positive particles, will move away from the point on the plate that q is approaching, so we will never get to a situation where r is zero. Even if we could somehow attach q immovably to a point on the plate, a new equilibrium would be achieved in which all the other positive particles arrange themselves at a distance from p.

 

[Robin64]

That makes sense. I completely forgot to consider that the charge redistribution that would occur with a charge near a plate with like charge.

Thanks.