Question

At what separation distance can a proton hover above another proton due to electrostatic repulsion? Hint: Compare gravity to electrostatic repulsion. Please show step by step and explain how you got the numbers. Thanks

Answer 1

Solution:

Mass and charge of a proton are as folows:

Mass = m = 1.6726 x 10^-27 kg

Charge on a proton = q = 1.6 x 10^-19 C

Gravitational constant = G = 6.67 x 10^-11 Nm^2 /kg^2

Gravitational force between any two masses is Fg = G m1m2/r^2

Fg= (6.67 x 10^-11) ( 1.6726 x 10^-27) (( 1.6726 x 10^-27) / (r^2)

where r is the distance between the two protons

The electroststic repulsive force between the two protons is given by the Coulomb's law.

Fe = kq1q2/r^2

k = 8.99 x10^9 Nm^2/C^2

Fe = (8.99 x 10^9)(1.6 x 10^-19) (1.6 x 10^-19)/ r^2

When gravity is compared to electroststic force, i.e. when there is an equilibrium between these two forces,

Fg = Fe

(6.67 x 10^-11) ( 1.6726 x 10^-27) (( 1.6726 x 10^-27) / (r^2) = (8.99 x 10^9)(1.6x 10^-19)(1.6x 10^-19) / r^2

=> 1.64 x 10^-18 = 2.301 x 10^-28 ) / r^2

=> r = sqrt [( 2.301 x 10^-28) / (1.64 x 10^-18)]

        = 0.0000194 m