Question

Two positive point charges of $18\mu C$ and $15\mu C$ are $1m$ apart. What is the work done in bringing them $0.4m$ apart?

• A. $15\times 10^{-10}J$
• B. $3.6\times 10^{-6}J$
• C. $4\times 10^{-2}J$
• D. $2\times 10^{-14}J$

Answer 1

Answer: (B)

$\Delta U=U_2-U_1$
$\Delta U=\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}_2}-\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}_1}$
$\therefore \Delta U=\frac{q_1{q}_2}{4\pi {\epsilon }_0}\left[\frac{1}{0.4}-\frac{1}{1}\right]$
$\Delta U=18\times 15\times 10^{-12}\times 9\times 10^9(1.5)$
$\Delta U=3645\times 10^{-3}J$
$W=\Delta U=3.6\times 10^{-6}J$