Two charges of equal magnitudes and at a dist | vedclass

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(D) According to Coulomb's Law,the force between two charges $Q_1$ and $Q_2$ separated by a distance $r$ is given by $F = k \frac{Q_1 Q_2}{r^2}$.Given

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$F / 16$

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(D) According to Coulomb's Law,the force between two charges $Q_1$ and $Q_2$ separated by a distance $r$ is given by $F = k \frac{Q_1 Q_2}{r^2}$.Given that the charges are equal in magnitude,let $Q_1 = Q_2 = Q$. Thus,the initial force is $F = k \frac{Q^2}{r^2}$.When the charges are halved,the new charges become $Q' = Q/2$. When the distance is doubled,the new distance becomes $r' = 2r$.The new force $F'$ is given by $F' = k \frac{(Q/2)(Q/2)}{(2r)^2}$.$F' = k \frac{Q^2 / 4}{4r^2} = \frac{1}{16} \left( k \frac{Q^2}{r^2} \right)$.Since $F = k \frac{Q^2}{r^2}$,we have $F' = F / 16$.

Two charges of equal magnitudes and at a distance $r$ exert a force $F$ on each other. If the charges are halved and the distance between them is doubled,then the new force acting on each charge is

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