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 charges are divided in the ratio $2:3$. Therefore,$q_1 = \frac{2}{2+3} 	imes 10^{-6} \ C = 0.

Vedclass · Accepted Answer

$0.00216$

Vedclass · Answer

(B) Given total charge $Q = 1 \mu C = 10^{-6} \ C$. The charges are divided in the ratio $2:3$. Therefore,$q_1 = \frac{2}{2+3} 	imes 10^{-6} \ C = 0.4 	imes 10^{-6} \ C$ and $q_2 = \frac{3}{2+3} 	imes 10^{-6} \ C = 0.6 	imes 10^{-6} \ C$. The distance between them is $r = 1 \ m$. The electrostatic force $F$ is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$. Substituting the values: $F = (9 	imes 10^9) 	imes \frac{(0.4 	imes 10^{-6}) 	imes (0.6 	imes 10^{-6})}{1^2}$. $F = 9 	imes 10^9 	imes 0.24 	imes 10^{-12} = 2.16 	imes 10^{-3} \ N = 0.00216 \ N$.

$A$ 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 $1 \ m$ apart in vacuum. Then,the electric force between them (in $N$) is

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