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(C) According to Coulomb's Law,the electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = \frac{1}

Vedclass · Accepted Answer

$\frac{F}{4}$

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(C) According to Coulomb's Law,the electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$.Given that the charges are equal,let $q_1 = q_2 = q$. Thus,$F = \frac{1}{4\pi \epsilon_0} \frac{q^2}{r^2}$.When the distance is doubled,the new distance $r' = 2r$.The new force $F'$ is given by $F' = \frac{1}{4\pi \epsilon_0} \frac{q^2}{(2r)^2}$.$F' = \frac{1}{4\pi \epsilon_0} \frac{q^2}{4r^2} = \frac{1}{4} \left( \frac{1}{4\pi \epsilon_0} \frac{q^2}{r^2} \right)$.Therefore,$F' = \frac{F}{4}$.

If the distance between two equal point charges is doubled,what would happen to the force between them?

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