Question

Two point charges are separated by 6 cm. The attractive force between them is 20 N. Find the force between them when they are separated by 12 cm. (Why can you solve this problem without knowing the magnitudes of the charges

Answer 1

Answer:

The force between them when they are separated by 12 cm is 5 N.

Explanation:

Distance between two point charges, d = 6 cm

The attractive force between them is 20 N, F = 20 N

Let F' is the force between them when they are separated by 12 cm, d' = 12 cm

The force between point charges is given by the formula as :

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

$F\propto \dfrac{1}{d^2}$........(1)

New force,

$F'\propto \dfrac{1}{d'^2}$............(2)

From (1) and (2) :

$\dfrac{F}{F'}=\dfrac{d'^2}{d^2}\\\\\dfrac{20}{F'}=\dfrac{12^2}{6^2}\\\\F'=5\ N$

So, the force between them when they are separated by 12 cm is 5 N. Hence, this is the required solution.