"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.

the answer would be a