The force between two identical spheres charg | vedclass

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(A) Initially,the force between two identical spheres each with charge $q$ separated by distance $r$ is given by Coulomb's law:$F = \frac{K q^2}{r^2}$

Vedclass · Accepted Answer

$\frac{3F}{4}$

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(A) Initially,the force between two identical spheres each with charge $q$ separated by distance $r$ is given by Coulomb's law:$F = \frac{K q^2}{r^2}$When $50\%$ of the charge from the first sphere is transferred to the second sphere,the new charges on the spheres become:$q_1 = q - 0.5q = 0.5q = \frac{q}{2}$$q_2 = q + 0.5q = 1.5q = \frac{3q}{2}$The new force $F_{	ext{new}}$ between the spheres at the same distance $r$ is:$F_{	ext{new}} = \frac{K q_1 q_2}{r^2} = \frac{K (q/2) (3q/2)}{r^2}$$F_{	ext{new}} = \frac{3}{4} \left( \frac{K q^2}{r^2} ight)$Since $F = \frac{K q^2}{r^2}$,we have:$F_{	ext{new}} = \frac{3}{4} F$

The force between two identical spheres charged with the same charge $q$ is $F$. If $50\%$ of the charge from one sphere is transferred to the second sphere,then the new force will be:

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