The repulsive force between two particles of | vedclass

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(D) Given that the electrostatic repulsive force is equal to the weight of the particle:$F_e = F_g$$\frac{k q^2}{r^2} = mg$Rearranging for $r^2$:$r^2

Vedclass · Accepted Answer

$1.18$

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(D) Given that the electrostatic repulsive force is equal to the weight of the particle:$F_e = F_g$$\frac{k q^2}{r^2} = mg$Rearranging for $r^2$:$r^2 = \frac{k q^2}{mg}$Substituting the given values:$r^2 = \frac{9 	imes 10^9 	imes (1.6 	imes 10^{-19})^2}{1.66 	imes 10^{-27} 	imes 10}$$r^2 = \frac{9 	imes 2.56 	imes 10^{9} 	imes 10^{-38}}{1.66 	imes 10^{-26}}$$r^2 = \frac{23.04 	imes 10^{-29}}{1.66 	imes 10^{-26}}$$r^2 = 13.8795 	imes 10^{-3} = 1.38795 	imes 10^{-2}$Taking the square root:$r = \sqrt{1.38795} 	imes 10^{-1} \ m$$r \approx 1.178 	imes 10^{-1} \ m$Rounding to two decimal places,we get $r \approx 1.18 	imes 10^{-1} \ m$.

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