Question

What is the change in potential energy of a particle of charge +q that is brought from a distance of 3r to a distance of 2r by a particle of charge q?

• A. $k{q}^2/r$
• B. $-k{q}^2/6r$
• C. $k{q}^2/r^2$
• D. $-k{q}^2//4r^2$
• E. $8k{q}^2/r^2$

Answer 1

The correct option is B $-k{q}^2/6r$

As the $+q$ is brought near by charge $q$ so there is an attraction force it means q is a negative charge, equal to $-q$.

The electric potential energy of a system of two point charges ( $+q$ , $-q$) sepatrated by a distance $3r$ is given by ,

$U=k\frac{q\times -q}{3r}$ ,

when the separation is reduced to $2r$ then potential energy will become ,

$U^{\prime }=k\frac{q\times -q}{2r}$

therefore change in potential energy is

$\Delta U=U^{\prime }-U$

or $\Delta U=-k\frac{q\times q}{2r}+k\frac{q\times q}{3r}$

or $\Delta U=k{q}^2(1/3r-1/2r)=-k{q}^2/6r$