Question

Two charges repel each other with a force of 200N. One of the charges is increased by 20% and the other is decreased by 10% . The new repulsive force at the same distance would be

Answer 1

  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
 TDear Students is the new repulsive force between the charges after alteration.

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