Force of attraction between two point charges | vedclass

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(A) For point charges,the force is given by Coulomb's law: $F_e = \frac{k Q^2}{d^2}$.When the charges are placed on spheres of radius $R = 0.3 \, d$,t

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Greater than $F_e$

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(A) For point charges,the force is given by Coulomb's law: $F_e = \frac{k Q^2}{d^2}$.When the charges are placed on spheres of radius $R = 0.3 \, d$,the distance between the centers is $d$.Since the spheres are close to each other ($R$ is a significant fraction of $d$),the electrostatic induction effect causes the charges to redistribute on the surfaces of the spheres.Specifically,the positive charge on one sphere is attracted towards the negative charge on the other sphere,causing the charges to shift closer to each other on the inner faces of the spheres.Because the effective distance between the centers of charge becomes less than $d$,the force of attraction increases.Therefore,the new force is greater than $F_e$.

Force of attraction between two point charges $Q$ and $-Q$ separated by $d \, \text{m}$ is $F_e$. When these charges are placed on two identical spheres of radius $R = 0.3 \, d$ whose centres are $d \, \text{m}$ apart,the force of attraction between them is

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