Question

by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one third​

Answer 1

Answer:

The force is now 9 times the original force

Explanation:

Coulomb's Law

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

$\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}$

$\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}$

$\displaystyle F'=9k\frac{q_1q_2}{d^2}$

$\displaystyle F'=9F$

The force is now 9 times the original force