Question

A Proton and electron are placed 1.6 CM apart in free space find the magnitude of electrostatic force between them and the nature of the force

Answer 1

This question can be solved using coulombs law.

Charge of proton =1.6*10^-19 coulomb
Charge of electron= -1.6*10^-19 coulomb
The distance between the proton and electron = 1.6cm
= 1.6*10^-2 m
= 0.016 m
According to coulombs law,
electrostatic force = ( k*q1*q2)/d^2
Where ,
k = proportionality constant, which is approximately 9*10^9 N
q1 = charge of proton
q2 = charge of elrctron
d = distance between electron and proton in meters.
substituting the values in the above equation,
F = ( K*q1*q2) /d^2
= (9*10^9)*(1.6*10^-19)*(1.6*10^-19)/(0.016^2)
= 9*10 ^-25 joules
The nature of electrostatic force is attractive.