Question

Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force 'F'. The electrostatic force between them in vacuum at the same distance r will be-

Options

• 5F

• F

• `"F"//2`

• `"F"//5`

Answer 1

K = 5 F = `1/(4πε_0"k") ("Q"_1"Q"_2)/"r"^2`

The force in the air charges is

F' =  `1/(4πε_0) ("Q"_1"Q"_2)/"r"^2`

K F = 5F