Question

Two point charges repel each other with a force 100 N. If one charge is increased by 10% and the other is reduced by 10% then the force of repulsion at the same distance is

A:- 100 N
B:- 99 N
C:- 121 N
D:- 141 N

Answer 1

GIVEN:
FORCE=F=100N
Let q be the magnitude of two charges.
According to Coulomb's law :
For constant DISTANCE,
Force is directly proportional to product of charges
F1= qxq
If one charge is increased by 10% and the other is reduced by 10%
Then New charges will be 0.9q and 1.1 q
So New force will be F2
F2=0.9q x 1.1q
Take the ratio of F1/F2
F1/F2=qxq/0.9qx1.1q
=1/0.99
=1.01
F2=F1/1.01
=100/1.01
=99N
Therefore, the force of repulsion at the same distance is 99N
So option B - 99N is correct answer.

Answer 2

given f= 100 n

let the charges is 100 then

according to the ques

one charge is increased by 10% and the other is reduced by 10%

q1= 90 q2 = 110

f2 = q1 ×q2 = 9900c

where f1 = 100 n given that

f3 = f2 /f1 = 9900/100

f3 = 99 newton ans

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