A charge of 1 mu C is divided into two parts | vedclass

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(B) Given total charge $Q = 1 \mu C = 10^{-6} \ C$.The charge is divided in the ratio $2:3$.Let the two charges be $q_1 = 2x$ and $q_2 = 3x$.Since $q_

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$0.00216$

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(B) Given total charge $Q = 1 \mu C = 10^{-6} \ C$.The charge is divided in the ratio $2:3$.Let the two charges be $q_1 = 2x$ and $q_2 = 3x$.Since $q_1 + q_2 = 1 \mu C$,we have $5x = 1 \mu C$,so $x = 0.2 \mu C$.Thus,$q_1 = 2 	imes 0.2 \mu C = 0.4 	imes 10^{-6} \ C$ and $q_2 = 3 	imes 0.2 \mu C = 0.6 	imes 10^{-6} \ C$.The distance between them is $r = 1 \ m$.The electrostatic force $F$ is given by Coulomb's Law:$F = \frac{k q_1 q_2}{r^2} = \frac{9 	imes 10^9 	imes (0.4 	imes 10^{-6}) 	imes (0.6 	imes 10^{-6})}{1^2}$$F = 9 	imes 10^9 	imes 0.24 	imes 10^{-12} = 2.16 	imes 10^{-3} \ N = 0.00216 \ N$.

$A$ charge of $1 \mu C$ is divided into two parts such that their charges are in the ratio of $2: 3$. These two charges are kept at a distance of $1 \ m$ apart in vacuum. Then,the electric force between them (in $N$) is

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