Question

A total charge Q is broken in two parts $Q_{1}$ and $Q_{2}$and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

• A. $\mathbf{Q}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{R}}\mathbf{,}\mathbf{Q}_{\mathbf{1}}\mathbf{= Q -}\frac{\mathbf{Q}}{\mathbf{R}}$
• B. $\mathbf{Q}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}}\mathbf{,}\mathbf{Q}_{\mathbf{1}}\mathbf{= Q}\mathbf{-}\frac{\mathbf{2Q}}{\mathbf{3}}$
• C. $\mathbf{Q}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}}\mathbf{,}\mathbf{Q}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{3Q}}{\mathbf{4}}$
• D. $\mathbf{Q}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{2}}\mathbf{,}\mathbf{Q}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{2}}$

Answer 1

Answer: (D)

$\mathbf{Q}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{2}}\mathbf{,}\mathbf{Q}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{2}}$

Answer 2

Force between charges $Q_{1}$ and$Q_{2}$

$F = k\frac{Q_{1}Q_{2}}{R^{2}} = k\frac{Q_{1}\left( Q - Q_{1} \right)}{R^{2}}$

For F to be maximum, $\frac{dF}{dQ_{1}} = 0$ i.e.,

$\frac{d}{dQ_{1}}\left\{ k\frac{\left( Q_{1}Q - Q_{1}^{2} \right)}{R^{2}} \right\} = 0$ or $Q - 2Q_{1} = 0,Q_{1} = \frac{Q}{2}$

Hence $Q_{1} = Q_{2} = \frac{Q}{2}$