An electricity and magnetism problem by Razing Thunder

Two point charges separated by a distance attract each other with a force of 10 N 10\text{ N} . If one of the charges is increased by 10 % 10\% and the other is decreased by 10 % 10\% , then find the new value of the force of interaction at the same distance.

9.9 N 11N 10 N 9 N

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1 solution

Vilakshan Gupta
Jun 22, 2020

Force between 2 2 charges is given by F = q 1 q 2 4 π ϵ 0 r 2 F=\dfrac{q_{1}q_{2}}{4\pi\epsilon_{0} r^2} , where q 1 q_{1} and q 2 q_{2} are charges and r r is the distance between the charges.

Now, if q 1 q_{1} is made 1.1 q 1 1.1q_{1} and q 2 q_{2} is made 0.9 q 2 0.9q_{2} , their product becomes 0.99 q 1 q 2 0.99q_{1}q_{2} .

Hence, the new force between them is 0.99 0.99 times the original value which is 9.9 N \boxed{9.9~N}

I was thinking: 10 % 10 % = 1 % , 10 1 % = 9.99 10\% * 10\% = 1\%, 10 - 1\% = \fbox{9.99} .

A Former Brilliant Member - 11 months, 3 weeks ago

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