two point charges repel each other with the force of 100 N.one of the charge is increased by 10% and other decreased by 10% .the new force of repulsion at the same distance would be___N

Question

Vikas TU · Answer

Dear Student,

For a pair of charges the force of interaction between them is directly proportional to the product of the charges if the distance is constant.( From coulombs' law)

that is hence if initially the charges have a magnitude q

and the force of interaction between them is F1

now when the charges are altered the charges become 0.9 q and 1.1 q so the new force of interaction between the new pair of charges becomes F2

writing the above statement in proportion:

we get,

hence F1/ F2 = ( q x q )/ (1.1q x 0.9q)

or, F1/ F2 = 1/(1.1 x 0.9)

F1/ F2 = 1/0.99 = 1.01

so F2 = F1 / 1.01

given F1 = 100 N

so F2 = (100/ 1.01) N

or F2 = 99 N

Cheers!!

Regards,

Vikas (B. Tech. 4^th year
Thapar University)

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two point charges repel each other with the force of 100 N.one of the charge is increased by 10% and other decreased by 10% .the new force of repulsion at the same distance would be___N

two point charges repel each other with the force of 100 N.one of the charge is increased by 10% and other decreased by 10% .the new force of repulsion at the same distance would be___N

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