When two spheres having 4 Q and -2 Q charge a | vedclass

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Question

(A) Initial force between the spheres is given by Coulomb's law: $F = \frac{k(4Q)(2Q)}{r^2} = \frac{8kQ^2}{r^2} \quad \dots(1)$ When the spheres are c

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$\frac{F}{2}$

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(A) Initial force between the spheres is given by Coulomb's law: $F = \frac{k(4Q)(2Q)}{r^2} = \frac{8kQ^2}{r^2} \quad \dots(1)$ When the spheres are connected by a conducting wire,the total charge is redistributed equally between them: $Q_{new} = \frac{4Q + (-2Q)}{2} = \frac{2Q}{2} = Q$ Now,the new distance is $r' = \frac{r}{2}$. The new force $F'$ is: $F' = \frac{k(Q)(Q)}{(r/2)^2} = \frac{kQ^2}{r^2/4} = \frac{4kQ^2}{r^2}$ From equation $(1)$,we know $\frac{kQ^2}{r^2} = \frac{F}{8}$. Substituting this into the expression for $F'$: $F' = 4 	imes \left(\frac{F}{8}ight) = \frac{F}{2}$

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