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
A
Q2=QR,Q1=Q−QR
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B
Q2=Q4,Q1=Q−2Q3
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C
Q2=Q4,Q1=3Q4
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D
Q1=Q2,Q2=Q2
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Solution
The correct option is DQ1=Q2,Q2=Q2 Q1+Q2=Q .....(i) and F=kQ1Q2r2 ....(ii) From (i) and (ii) F=kQ1(Q−Q1)r2 For F to be maximum dFdQ1=0⇒Q1=Q2=Q2