Two charges each equal to 2,mu C are 0.5,m ap | vedclass

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(A) According to Coulomb's Law,the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a vacuum is given

Vedclass · Accepted Answer

$0.144$

Vedclass · Answer

(A) According to Coulomb's Law,the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a vacuum is given by: $F = k \cdot \frac{q_1 q_2}{r^2}$Where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.Given: $q_1 = q_2 = 2 	imes 10^{-6} \, C$ and $r = 0.5 \, m$.Substituting the values:$F = 9 	imes 10^9 	imes \frac{(2 	imes 10^{-6}) 	imes (2 	imes 10^{-6})}{(0.5)^2}$$F = 9 	imes 10^9 	imes \frac{4 	imes 10^{-12}}{0.25}$$F = 9 	imes 10^9 	imes 16 	imes 10^{-12}$$F = 144 	imes 10^{-3} = 0.144 \, N$.

Two charges each equal to $2\,\mu C$ are $0.5\,m$ apart. If both of them exist inside vacuum,then the force between them is.......$N$

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