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  • Tell me? 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 repulsi

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

  • Option 1)

    2

  • Option 2)

    3

  • Option 3)

    4

  • Option 4)

    6

 

Answers (1)

best_answer

As we learned

By conduction -

When two conducters brought in contact.

- wherein

i.e The charge will spread over both the conducters.

 

 Suppose the balls having charges Q1 and Q2 respectively.

F=\frac{k\left ( \frac{Q_{1}+Q_{2}}{2} \right )^{2}}{\left ( \frac{r}{2} \right )^{2}}=\frac{k(Q_{1}+Q_{2})^{2}}{r^{2}}

It is given that 

\\*F=4.5F\; so\; \frac{k(Q_{1}+Q_{2})^{2}}{r^{2}}=4.5k.\frac{Q_{1}Q_{2}}{r^{2}}\\*\Rightarrow (Q_{1}+Q_{2})^{2}=4.5Q_{1}Q_{2}.\\*On \; solving\; it\; gives\; \: \frac{Q_{1}}{Q_{2}}=\frac{2}{1}

 


Option 1)

2

Option 2)

3

Option 3)

4

Option 4)

6

Posted by

Aadil

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