SATHEE: Electrostatics Question 168

Electrostatics Question 168

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 [MP PET 1990]

Options:

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

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Answer:

Correct Answer: C

Solution:

$Q_{1} + Q_{2} = Q$ ….. (i) and $F = k \frac{Q_{1} Q_{2}}{r^{2}}$ …..(ii)

From (i) and (ii) $F = \frac{k Q_{1} (Q - Q_{1})}{r^{2}}$

For F to be maximum $\frac{d F}{d Q_{1}} = 0$

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

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Electrostatics Question 168

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 [MP PET 1990]

Options:

Answer:

Solution:

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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 [MP PET 1990]