Question

What will happen to the force if wedecrease r by a factor of 2 ?​

Answer 1

Answer:

The new force will become 1/4 times of the initial force.

Explanation:

The Coulomb force between two charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them.

It can be given by :

$F=k\dfrac{q_1q_2}{r^2}$ ....(1)

r is distance

If we decrease the r by a factor of 2, r'=2r.

New force will become,

$F'=k\dfrac{q_1q_2}{r'^2}\\\\=k\dfrac{q_1q_2}{(2r)^2}\\\\=k\dfrac{q_1q_2}{4r^2}\\\\=\dfrac{1}{4}\times \dfrac{kq_1q_2}{r^2}\\\\F'=\dfrac{F}{4}$

So, the new force will become 1/4 times of the initial force.