The Coulombian repulsive force between two al | vedclass

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(C) The electric charge of an $\alpha$-particle is $q = 2e = 2 	imes 1.6 	imes 10^{-19} \ C = 3.2 	imes 10^{-19} \ C$.Given distance $r = 3 \ cm =

Vedclass · Accepted Answer

$1.024 	imes 10^{-24} \ N$

Vedclass · Answer

(C) The electric charge of an $\alpha$-particle is $q = 2e = 2 	imes 1.6 	imes 10^{-19} \ C = 3.2 	imes 10^{-19} \ C$.Given distance $r = 3 \ cm = 3 	imes 10^{-2} \ m$.The Coulombian repulsive force is given by $F = \frac{k q_1 q_2}{r^2}$.Since $q_1 = q_2 = q$,we have $F = \frac{k q^2}{r^2}$.Substituting the values:$F = \frac{9 	imes 10^9 	imes (3.2 	imes 10^{-19})^2}{(3 	imes 10^{-2})^2}$$F = \frac{9 	imes 10^9 	imes 10.24 	imes 10^{-38}}{9 	imes 10^{-4}}$$F = 10.24 	imes 10^{9 - 38 + 4}$$F = 10.24 	imes 10^{-25} \ N = 1.024 	imes 10^{-24} \ N$.

The Coulombian repulsive force between two alpha particles kept at a distance of $3 \ cm$ in air is . . . . . . .

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