Question

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

$\Delta U = U _{2}- U _{1}$
$\Delta U =\frac{ q _{1} q _{2}}{4 \pi \varepsilon_{ 0 } r _{2}}-\frac{ q _{1} q _{2}}{4 \pi \varepsilon_{ 0 } r _{1}}$
$\therefore \Delta U=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}}\left[\frac{1}{0.4}-\frac{1}{1}\right]$
$\Delta U =18 \times 15 \times 10^{-12} \times 9 \times 10^{9}(1.5)$
$\Delta U =3645 \times 10^{-3} J$
$W =\Delta U =3.6 \times 10^{-6} J$