Two particles with charges $q_{1}$ and $q_{2}$ are kept at a certain distance to exert force F on the other. If the distance is reduced to one-fifth, then the force between them is :

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Solution

Let the distance between them be $d$
Force $F = \frac{k q_{1} q_{2}}{d^{2}}$
$⟹ F \propto \frac{1}{d^{2}}$
Si, if the distance between tem is reduced by a factor of $5$ , force between them gets increased by a factor of $25$ .
So, new force $F^{'} = 25 F$

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Two particles with charges q1 and q2 are kept at a certain distance to exert force F on the other. If the distance is reduced to one-fifth, then the force between them is :

Let the distance between them be dForce F=kq1q2d2⟹ F∝1d2Si, if the distance between tem is reduced by a factor of 5, force between them gets increased by a factor of 25.So, new force F′=25F

Two particles with charges $q_{1}$ and $q_{2}$ are kept at a certain distance to exert force F on the other. If the distance is reduced to one-fifth, then the force between them is :

Let the distance between them be $d$
Force $F = \frac{k q_{1} q_{2}}{d^{2}}$
$⟹ F \propto \frac{1}{d^{2}}$
Si, if the distance between tem is reduced by a factor of $5$ , force between them gets increased by a factor of $25$ .
So, new force $F^{'} = 25 F$