Two point charges +2 C and +6 C separated b | vedclass

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(A) According to Coulomb's Law,the force between two charges $q_1$ and $q_2$ separated by distance $r$ is given by $F = k \frac{|q_1 q_2|}{r^2}$.Initi

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$6 \ N$ (attractive)

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(A) According to Coulomb's Law,the force between two charges $q_1$ and $q_2$ separated by distance $r$ is given by $F = k \frac{|q_1 q_2|}{r^2}$.Initially,$q_1 = +2 \ C$ and $q_2 = +6 \ C$. The force is $F_1 = k \frac{(2)(6)}{r^2} = 18 \ N$.This implies $k \frac{12}{r^2} = 18$,so $k/r^2 = 18/12 = 1.5$.Now,$-4 \ C$ is added to each charge:New charge $q_1' = +2 \ C - 4 \ C = -2 \ C$.New charge $q_2' = +6 \ C - 4 \ C = +2 \ C$.The new force is $F_2 = k \frac{|q_1' q_2'|}{r^2} = k \frac{|(-2)(2)|}{r^2} = k \frac{4}{r^2}$.Substituting $k/r^2 = 1.5$,we get $F_2 = 1.5 	imes 4 = 6 \ N$.Since the charges are of opposite signs ($-2 \ C$ and $+2 \ C$),the force is attractive.

Two point charges $+2 \ C$ and $+6 \ C$ separated by a distance '$r$' repel each other with a force of $18 \ N$. If a negative charge of $-4 \ C$ is added to each of these charges,then the force experienced by them is:

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