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

Vedclass · Accepted Answer

$d/4$

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(B) According to Coulomb's Law,the force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a medium is given by $F = \frac{k q_1 q_2}{r^2}$,where $k$ is the electrostatic constant for that medium.Given the initial state: $F = \frac{k q_1 q_2}{d^2}$.Let the new distance be $d'$ such that the new force $F' = 16F$.So,$16F = \frac{k q_1 q_2}{(d')^2}$.Substituting the value of $F$ from the first equation: $16 \left( \frac{k q_1 q_2}{d^2} \right) = \frac{k q_1 q_2}{(d')^2}$.Canceling $k q_1 q_2$ from both sides,we get: $\frac{16}{d^2} = \frac{1}{ (d')^2 }$.Taking the square root of both sides: $\frac{4}{d} = \frac{1}{d'}$.Therefore,$d' = \frac{d}{4}$.

In a medium,the force of attraction between two point charges,distance $d$ apart,is $F$. What distance apart should these point charges be kept in the same medium,so that the force between them becomes $16F$?

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