Question

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron would the proton have to be? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C, mproton = 1.67 × 10-27 kg, melectron = 9.11 × 10-31 kg)

Answer 1

r = 5.08 m

Step-by-step explanation:

The electric force of attraction or repulsion is given by :

`F=(kq_1q_2)/(r^2)`

We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.

So, the force from the proton is balanced by the mass of the electron.

`(kq_pq_e)/(r^2)=mg`

r is distance

`r=\sqrt{(kq_pq_e)/(mg)} \\\\r=\sqrt{(9* 10^9* (1.6* 10^(-19))^2)/(9.11* 10^(-31)* 9.8)} \\\\r=5.08\ m`

So, proton have to be at a distance of 5.08 meters above the electron.