Question

36. 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 two charges when they are separated by distance 12 cm is  $5N$  

Explanation:

Here lets us consider two charges $q_{1}$ and $q_{2}$ .

Let $F_{1}$  be the force between them when they are separated by distance 6 cm

     $F_{2}$ be the force between them when distance between them = 12 cm

            $F_{1} = \frac{KQ_{1}Q_{2}}{6^{2}}$  =20 N     ..................1

            $F_{2} = \frac{KQ_{1}Q_{2}}{12^{2}}$                 ...................2

 Dividing both equations , we get

        $\frac{20}{F_{2}} =\frac{12^{2}}{6^{2}}$

    $F_{2} = 5N$.

Here we need not to know the values of charges , since they get cancelled while dividing.