Question

two charges exert a force of 10 Newton on each other when separated by a distance 0.2 metre in year when they are placed in another medium of dielectric constant K is equal to 4 and separated by distance R then they are same force the distance equals to​

Answer 1

Let q and Q are two charges are separated 0.2m apart.

from Coulomb's law,

Force between them, F = kqQ/(0.2)²

given, F = 10N

so, 10 = kqQ/(0.04)

or, kqQ = 10 × 0.04 = 0.4 ......(1)

now both charges are placed in another medium of dielectric constant K equal to 4 and separated by distance R.

then, force between them, F' = kqQ/4R²

[ as you force is inversely proportional to dielectric constant so, F' = kqQ/R² × 1/4 ]

F' = F = kqQ/4R² [ according to question same force acts on both conditions ]

from equation (1),

10 = 0.4/4R²

or, R² = 0.01 = (0.1)² => R = 0.1 m

hence, distance between them is 0.1m

Answer 2

Answer:

In diff med k= four

F =

Explanation: