Question

A charge of +Q is fixed in space. A second charge of +q was first placed at a distance r1 away from +Q. Then it was moved along a straight line to a new position at a distance R away from its starting position. The final location of +q is at a distance r2 from +Q.
What is the change in the potential energy of charge +q during this process?
(A) kQq/R
(B) kQqR/r12
(C) kQqR/r22
(D) kQq((1/r2)-(1/r1))
(E) kQq((1/r1)-(1/r2))

Answer 1

Answer:

Option (D)

Explanation:

The formula for the potential energy between the two charges is given by

$U=\frac{KQq}{r}$

where, r is the distance between the two charges.

In first case the distance between the two charges is r1.

The potential energy is

$U_{1}=\frac{KQq}{r_{1}}$

In first case the distance between the two charges is r2.

The potential energy is

$U_{2}=\frac{KQq}{r_{2}}$

The change in potential energy is

$\Delta U = U_{2}-U_{1}$

$\Delta U=\frac{KQq}{r_{2}}-\frac{KQq}{r_{1}$

$\Delta U=KQq \times \left ( \frac{1}{r_{2}} -\frac{1}{r_{2}} \right )$