Question

If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled and the distance is cut in half, what is the new value of the force of repulsion, F2? Express your answer in terms of F1.

Answer 1

Answer:$F_2=16\times F_1$

Explanation:

Given

Force of repulsion between two charge particle is given by force F

Electrostatic force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ is the charges  of particle

r=distance between charge particle

$=\frac{kq^2}{r^2}$when charges are doubled and distance is reduced to half

i.e. q become 2 q and r becomes 0.5 r

$F_2=\frac{k(2q)^2}{(0.5r)^2}$

$F_2=\frac{kq^2}{r^2}\times 4\times 4$

$F_2=16\times F_1$