The repulsive force between two point charges | vedclass

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(A) In the first case,the charges are point charges,so the distance between them is $d = 1 \, m$. The force is $F = k \frac{q_1 q_2}{d^2}$.In the seco

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$F \propto \frac{1}{d^2}$

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(A) In the first case,the charges are point charges,so the distance between them is $d = 1 \, m$. The force is $F = k \frac{q_1 q_2}{d^2}$.In the second case,the charges are distributed on spheres of radius $R = 25 \, cm = 0.25 \, m$. Due to mutual repulsion,the charges on the spheres shift away from each other.As shown in the figure,the effective distance between the centers of charge becomes $d' > d$.Since the force $F$ is inversely proportional to the square of the distance $(F \propto \frac{1}{r^2})$,as the effective distance $d'$ increases,the repulsive force decreases according to the relation $F \propto \frac{1}{d^2}$.

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