Question

The electrostatic force between two small charged particles is 5.2 x 10^-7 N at a distance of 2 meters. What will be the new force if the distance is
increased to 4 meters apart?

Answer 1

Answer:

$F_2=1.3\times 10^{-7}\ N$

Explanation:

Given that,

Initial force, $F=5.2\times 10^{-7}\ N$

distance, d = 2 m

New distance, d' = 4 m

We need to find the new force. The electrostatic force between two charges is given by :

$F=k\dfrac{q_1q_2}{d^2}$

So,

$\dfrac{F_1}{F_2}=(\dfrac{d_2}{d_1})^2\\\\\dfrac{F_1}{F_2}=(\dfrac{4}{2})^2\\\\\dfrac{F_1}{F_2}=4\\\\F_2=\dfrac{F_1}{4}\\\\F_2=\dfrac{5.2\times 10^{-7}}{4}\\\\F_2=1.3\times 10^{-7}\ N$

So, the new force is equal to $1.3\times 10^{-7}\ N$.