Question

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

Answer 1

As per the question two charges are given as   $q_{1}\ and\ q_{2}$

Let the separation distance between them is r.

Hence the Coulombic force of repulsion or attraction between them is -

                              $F =\frac{1}{4\pi \epsilon} \frac{q_{1}q_{2}} {r^2}$

$Here\ \epsilon\ is\ the\ \ permittivity\ of\ the\ medium\ in\ which\ charges\ are\ present$

The distance is reduced to one fifth.Hence, the separation distance between them will be now r/5.

Now the coulombic force between them will be-

                                  $F'=\frac{1}{4\pi \epsilon} \frac{q_{1} q_{2}} {[\frac{r}{5}]^2}$

                                  $F'=\frac{1}{4\pi \epsilon} \frac{25q_{1} q_{2}} {r^2}$

Hence, from above we see that the force is increased by 25 times.

Answer 2

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distance is reduced to 25times the previous one...