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Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases 4.5 times in comparison with the initial value. The ratio of the initial charges of the balls is/??

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let initial charges on them are Q_{1} and  Q_{2}

force is repulsive means Q_{1} and Q_{2} are of same sign. 

let initial separation be r.

F_{initial}=\frac{kQ_{1}Q_{2}}{r^{2}}

after touching each other charge on both divides equally means both have the charge    \frac{Q_{1}+Q_{2}}{2}

final separation is half of the initial means r/2.

F_{final}=\frac{k(\frac{Q_{1}+Q_{2}}{2})^{2}}{(r/2)^{2}}

\frac{F_{final}}{F_{initial}}=4.5

\frac{(Q_{1}+Q_{2})^{2}}{Q_{1}Q_{2}}=4.5

\Rightarrow Q_{1}^{2}+Q_{2}^{2}+2Q_{1}Q_{2}=4.5Q_{1}Q_{2}

\Rightarrow 2\left (\frac{Q_{1}}{Q_{2}} \right )^{2}-5\frac{Q_{1}}{Q_{2}}+2=0

\Rightarrow \frac{Q_{1}}{Q_{2}}=2 \ or \ \frac{1}{2}

 

Posted by

Vishwash Tetarwal

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