Two positive point charges of 0.2 μC and 0.01 μC are placed 10 cm apart. The work done in reducing the distance to 5 cm will be

$1.8 × 10^{-3} J$

The correct answer is Option (1) → $1.8 × 10^{-4} J$

Given charges: $q_1=0.2\ \mu C=0.2\times10^{-6}\ C$, $q_2=0.01\ \mu C=0.01\times10^{-6}\ C$

Initial separation: $r_i=0.10\ \text{m}$

Final separation: $r_f=0.05\ \text{m}$

Work done = change in potential energy:

$W=\frac{1}{4\pi\epsilon_0}\ q_1q_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right)$

Using $\frac{1}{4\pi\epsilon_0}=9\times10^9\ \text{N·m}^2/\text{C}^2$:

$W=9\times10^9\cdot(0.2\times10^{-6})(0.01\times10^{-6})\left(\frac{1}{0.05}-\frac{1}{0.10}\right)$

$W=9\times10^9\cdot2\times10^{-14}\cdot(20-10)$

$W=9\times10^9\cdot2\times10^{-14}\cdot10$

$W=1.8\times10^{-3}\ \text{J}$

Answer: $1.8\times10^{-3}\ \text{J}$