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(B) According to Coulomb's Law,the force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = k \frac{q_1 q_2}{r^2}$.G

Vedclass · Accepted Answer

$4F$

Vedclass · Answer

(B) According to Coulomb's Law,the force $F$ 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 particles are identical,let $q_1 = q_2 = q$. Initially,the distance is $r_1 = Y$,so $F_1 = k \frac{q^2}{Y^2} = F$.When the distance is reduced to half,the new distance is $r_2 = \frac{Y}{2}$.The new force $F_2$ is given by $F_2 = k \frac{q^2}{(Y/2)^2} = k \frac{q^2}{Y^2 / 4} = 4 \left( k \frac{q^2}{Y^2} \right)$.Substituting $F_1 = F$,we get $F_2 = 4F$.

$F$ is the force between two identical charged particles placed at a distance $Y$ from each other. If the distance between the charges is reduced to half the previous distance,then the force between them becomes:

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