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Question

(B) According to Coulomb's Law,the force between two point charges is given by $F = k \frac{Q_1 Q_2}{R^2}$.Initially,the charges are $Q_1 = +10\,\mu C

Vedclass · Accepted Answer

$-8:1$

Vedclass · Answer

(B) According to Coulomb's Law,the force between two point charges is given by $F = k \frac{Q_1 Q_2}{R^2}$.Initially,the charges are $Q_1 = +10\,\mu C$ and $Q_2 = -20\,\mu C$. The force is $F_1 = k \frac{(10)(-20)}{R^2} = -200 \frac{k}{R^2}$.When the spheres are brought into contact,the total charge is redistributed equally because the spheres have equal radii. The new charge on each sphere is $Q' = \frac{Q_1 + Q_2}{2} = \frac{10 - 20}{2} = -5\,\mu C$.After separation to the same distance $R$,the new force is $F_2 = k \frac{(-5)(-5)}{R^2} = 25 \frac{k}{R^2}$.The ratio of the forces is $\frac{F_1}{F_2} = \frac{-200 \frac{k}{R^2}}{25 \frac{k}{R^2}} = \frac{-200}{25} = -8$.Thus,the ratio $F_1 : F_2$ is $-8:1$.

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