Two insulated charged copper spheres A and B | vedclass

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(B) Distance between the spheres $A$ and $B$,$r = 0.5 \; m$.Initially,the charge on each sphere,$q = 6.5 	imes 10^{-7} \; C$.When sphere $A$ is touch

Vedclass · Accepted Answer

$5.7 	imes 10^{-3} \; N$.

Vedclass · Answer

(B) Distance between the spheres $A$ and $B$,$r = 0.5 \; m$.Initially,the charge on each sphere,$q = 6.5 	imes 10^{-7} \; C$.When sphere $A$ is touched with an uncharged sphere $C$,the charge $q$ is shared equally. Hence,the charge on each of the spheres $A$ and $C$ becomes $q_A = q/2$.When sphere $C$ (now with charge $q/2$) is brought in contact with sphere $B$ (with charge $q$),the total charge is shared equally between them. The new charge on each is $(q + q/2) / 2 = 3q/4$. So,$q_B = 3q/4$.The new force of repulsion between $A$ and $B$ is given by Coulomb's law:$F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_A q_B}{r^2} = (9 	imes 10^9) \cdot \frac{(q/2) \cdot (3q/4)}{r^2} = (9 	imes 10^9) \cdot \frac{3q^2}{8r^2}$.Substituting the values: $F = (9 	imes 10^9) \cdot \frac{3 	imes (6.5 	imes 10^{-7})^2}{8 	imes (0.5)^2} = 5.703 	imes 10^{-3} \; N$.

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