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
Answer & explanation
Correct answer: option 1
The correct answer is Option (4) → $1.8 × 10^{-4} J$
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 $\frac{1}{4\pi\epsilon_0}=9\times10^9\ \text{N·m}^2/\text{C}^2$:
$W=9\times10^9\cdot(0.2\times10^{-6})(0.01\times10^{-6})\left(\frac{1}{0.05}-\frac{1}{0.10}\right)$
$W=9\times10^9\cdot2\times10^{-14}\cdot(20-10)$
$W=9\times10^9\cdot2\times10^{-14}\cdot10$
$W=1.8\times10^{-4}\ \text{J}$
Answer: $1.8\times10^{-4}\ \text{J}$