The force between two conducting spheres of s | vedclass

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(C) Initially,the force between the two spheres is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2} = k \frac{|(+8 \mu C)(-4 \mu C)|}{r^2} = k \fr

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$F/8$

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(C) Initially,the force between the two spheres is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2} = k \frac{|(+8 \mu C)(-4 \mu C)|}{r^2} = k \frac{32 \mu C^2}{r^2}$.When the two identical conducting spheres are connected by a wire,the charges redistribute equally because they have the same radius.The new charge on each sphere will be $q' = \frac{q_1 + q_2}{2} = \frac{+8 \mu C - 4 \mu C}{2} = +2 \mu C$.The new force $F'$ between the spheres is $F' = k \frac{|q' q'|}{r^2} = k \frac{|(+2 \mu C)(+2 \mu C)|}{r^2} = k \frac{4 \mu C^2}{r^2}$.Comparing the two forces: $\frac{F'}{F} = \frac{4}{32} = \frac{1}{8}$.Therefore,$F' = \frac{F}{8}$.

The force between two conducting spheres of same radius having charges $+8 \mu C$ and $-4 \mu C$ separated by some distance in air is $F$. If the spheres are connected by a conducting wire and after some time the wire is removed,then the magnitude of the force between the two conducting spheres is

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