First slide

Electrostatic force

Question

A total charge Q is broken in two parts Q₁ and Q₂ and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

Easy

A

$Q_{2} = \frac{Q}{R}, Q_{1} = Q - \frac{Q}{R}$
B

$Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{2 Q}{3}$
C

$Q_{2} = \frac{Q}{4}, Q_{1} = \frac{3 Q}{4}$
D

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

Solution

$Q_{1} + Q_{2} = Q$ ….(i) and $F = k \frac{Q_{1} Q_{2}}{r^{2}}$ ….(ii)
From (i) and (ii) $F = \frac{{kQ}_{1} (Q - Q_{1})}{r^{2}}$
For F to be maximum $\frac{dF}{dQ} = 0 \Rightarrow Q_{1} = Q_{2} = \frac{Q}{2}$

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