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

A. 5F
B. F
C. `(F)/(2)`
D. `(F)/(5)`

Answer

A

Explanation

`F_(0)=(1)/(4piepsi_(0))(q_(1)q_(2))/(r^(2))` [ Force in vacuum]
`F=(1)/(4pi epsi_(0)K)(q_(1)q_(2))/(r^(2))` [Force in medium]