Solution
The electrostatic force between any pair of point charges is governed by Coulomb’s Law:
$$F = \frac{k q_1 q_2}{r^2}$$From this, we see that for a fixed set of charges, the force scales inversely with the square of the distance:
$$F \propto \frac{1}{r^2}$$In the first case, the charges are on a cube of edge length $a$. In the second case, the same charges are placed on a cube of edge length $b$. This means the geometry is identical, but all distances have been scaled by a factor of $\frac{b}{a}$.
If the original force was $F$ corresponding to distance $a$, the new force $F’$ corresponding to distance $b$ is:
$$\frac{F’}{F} = \frac{(1/b^2)}{(1/a^2)} = \frac{a^2}{b^2}$$ $$F’ = F \left( \frac{a^2}{b^2} \right)$$Therefore, the magnitude of the net force is:
Answer: (c) $\frac{a^2 F}{b^2}$