"A positively-charged particle is held at point A between two parallel metal plates. The plate on the left has a net positive charge q and the plate on the right has a net negative charge -q. The particle is then moved to point B. How does the electric potential energy at point A compare with that at point B?

Explanation

A positively charged particle produces a positive electric potential while a negatively

charged particle produces a negative electric potential.

Given two parallel plates one is positive and the other is negative and a positively charge particle is place at point A.

The electric potential V at point P is the algebraic sum of the electric potentials contributed by the two parallel plates charges

V=Kq1/r+ Kq2/r

Since q1 is positive and q2 is negative and they have the same magnitude q.

Then, at position A,

Let assume the total distance between the parallel plate is x

Then, the positive plate should be at distance 'a' from, the other is at x-a.

V= Kq/a-kq/(x-a)

V=kq(1/a-1/(x-a))

V=kq(x-a-a/a(x-a))

V=kq(x-2a)/(ax-a^2)

Then, at position B

Let assume the total distance between the parallel plate is x

Then, the positive plate should be at distance 'b' from, the other is at x-b.

V= Kq/b-kq/(x-b)

V=kq(1/b-1/(x-b))

V=kq(x-b-b/b(x-b))

V=kq(x-2b)/(bx-b^2)

Close to the positive plate, the

potential has very large positive values. Close to the negative charge, the potential has very large negative values.

They have different potential.