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

The value is
`r = 5.077 \ m`

Step-by-step explanation:

From the question we are told that

The Coulomb constant is
`k = 9.0 *10^(9) \ N\cdot m^2 /C^2`

The charge on the electron/proton is
`e = 1.6*10^(-19) \ C`

The mass of proton
`m_(proton) = 1.67*10^(-27) \ kg`

The mass of electron is
`m_(electron ) = 9.11 *10^(-31) \ kg`

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

`m_(electron) * g = ( k * e^2 )/(r^2 )`

`(9*10^9 * (1.60 *10^(-19))^2 )/(r^2 ) = 9.11 *10^(-31 ) * 9.81`

`r = √(25.78)`

`r = 5.077 \ m`