NCERT Solutions

The Question is asking to calculate the electrostatic force between two charged spheres with given charges separated by 30 cm.

Concept: Coulomb's Law

$$F_{BA} = \text{force on }q_B \text{due to } q_A = \frac{k q_A q_B}{{r_{BA}}^2}$$ $$k = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 N m^2 C^{-2}$$

Given:

$$q_A = 2 \times 10^{-7} C$$ $$q_B = 3 \times 10^{-7} C$$

Equation Setup:

$$F_{AB} = F_{BA} = k \frac{q_A q_B}{{r_{AB}}^2}$$ $$F_{AB} = F_{BA} = 9 \times 10^9 \frac{2 \times 10^{-7} \times 3 \times 10^{-7}}{0.3^2} N$$

Calculation:

$$F_{AB} = F_{BA} = 6 \times 10^{-3} N$$

Results:

Charges are of same sign, so the force of magnitude $6 \times 10^{-3} N$ would be repulsive in nature