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Q.

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

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a

Q2=QR, Q1=Q−QR

b

Q2=Q4, Q1=Q−2Q3

c

Q2=Q4, Q1=3Q4

d

Q1=Q2, Q2=Q2

answer is D.

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Detailed Solution

Q1+Q2=Q          ….(i)  and F=kQ1Q2r2       ….(ii)

From (i) and (ii) F=kQ1(Q−Q1)r2

For F to be maximum dFdQ=0⇒Q1=Q2=Q2

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