Two point charges 2 μC and 3 μC are placed at a distance of 5 cm from each other. The magnitude of the electrostatic force between them is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 21.6 N
Coulomb's law: $F = k \frac{|q_1 q_2|}{r^2}$
Given: $q_1 = 2\ \mu C = 2 \times 10^{-6}\ \text{C}$, $q_2 = 3\ \mu C = 3 \times 10^{-6}\ \text{C}$, $r = 5\ \text{cm} = 0.05\ \text{m}$, $k = 9 \times 10^9\ \text{N·m²/C²}$
Substitute values:
$F = 9 \times 10^9 \cdot \frac{(2 \times 10^{-6})(3 \times 10^{-6})}{(0.05)^2}$
$F = 9 \times 10^9 \cdot \frac{6 \times 10^{-12}}{0.0025}$
$F = 9 \times 10^9 \cdot 2.4 \times 10^{-9} = 21.6\ \text{N}$
Final Answer: $F = 21.6\ \text{N}$