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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

  • Option 1)

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

  • Option 2)

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

  • Option 3)

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

  • Option 4)

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

 

Answers (1)

best_answer

As we learned

Coulombic force -

F\propto Q_{1}Q_{2}=F\propto \frac{Q_{1}Q_{2}}{r^{2}}=F=\frac{KQ_{1}Q_{2}}{r^{2}}

- wherein

K - proportionality Constant 

Q1 and Q2 are two Point charge

 

 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_{1}}=0\Rightarrow Q_{1}=Q_{2}=\frac{Q}{2}

 


Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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