Question

At time =0, a proton is at rest at location <4.7, 1.8, 0> x 10^-6m. An electron is at <2.9,8.7,0> x 10^-6 m. and is moving with a momentum of <3.6,-7.8,0> x 10^-27 kg*m/s. the proton and electron are in outer space, far from any other matter. The proton has a charge of 1.6 x 10^-19 C, and the elctrons charge is exactly the negative of this value.

A.) what is the magnitude of the gravitational force on the electron by the proton?

B.) what is the magnitude of the electric force on the electron by the proton. By what factor is it larger than the gravitational force?

C.) what is the electric force vector on the electron by the proton?

D.) after time 2.0 x 10^-10 s, what is the momentum of the electron?

Answer 1

Given that,

location of proton = (4.7 , 1.8, 0 ) x 10^-6 m

location of electron = (2.9, 8.7, 0) x 10^-6 m

their distnace along x will be : 4.7 - 2.9 = 1.8 x 10^-6 m

along y will be: 8.7 - 1.8 = 6.9 x 10^-6 m

So, the actual distance between electron and proton is = d = sqrt [(1.8x 10^-6 m)² + (6.9x 10^-6 m)²] = 7.13 x 10^-6 m

(A) We know that the gravitational force is given by:

F(gravity) = G m1 m2 / d²

Where, G is the Universal gravitation constant, m1 and m2 are the masses and d is the distance between them.

m1 = me = 9.1 x 10^-31 kg and m2 = mp = 1.67 x 10^-27 kg ; G = 6.67 x 10^-11

F(gravity) =  6.67 x 10^-11 x 9.1 x 10^-31 x 1.67 x 10^-27 / ( 7.13 x 10^-6)² = 1.99 x 10^-57 N

Hence, F(gravity) = 1.99 x 10^-57 N

(B) We know that, F(electric) = k q1 q2 /  d²

F(electric) = 9 x 10⁹ x 1.6 x 10^-19 x 1.6 x 10^-19 / ( 7.13 x 10^-6)² = 0.45 x 10^-17 N

Hence, F(electric) = 0.45 x 10^-17 N

F(electric) / F(gravity) =  0.45 x 10^-17 N / 1.99 x 10^-57 N = 0.226 x 10⁴⁰

Hence, F(electric) is larger by a factor of  0.226 x 10⁴⁰

(C) The force will be at an angle of = tan -1 (1.8/6.9) = 14.64 Deg

F(electric)x = 0.45 x 10^-17 N x cos 14.64 = 0.44 x 10^-17 N

F(electric)y = 0.45 x 10^-17 N x sin 14.64 = 0.11 x 10^-17 N

F = sqrt [(0.44)² + (0.11)² ] x 10^-17 = 0.44 x 10^-17 N