Calculating the Voltage for a Proton Floating Between Two Boards

[Mushroom79]

Homework Statement

A proton just floating in a field in a between two boards (g=9,82 m/s^2)
What is the voltage U?

distance d = 2 m
charge for protone Q = 1,602*10^-19

Homework Equations

E = F/Q

E = U/d

The Attempt at a Solution

I've tried two different ways of which I don't know is correct
In both cases I assume that F = 9,82 as something must be lifting it upp to make up for the gravitation

Attempt 1.

F = g = 9,82

E = F/Q = 9,82/(1,602*10^-19) = 6,13

E = U/d --> U = E*d = 6,13*2 = 2,26 V

Attempt 2.

E = U/d --> U = E*d = 19,64 V

 

[gneill]

Careful, you're using g (acceleration due to gravity) as a force. The force due to gravity on the proton would be its weight, m*g.

 

[Mushroom79]

Careful, you're using g (acceleration due to gravity) as a force. The force due to gravity on the proton would be its weight, m*g.

Oh, I forgot about that
Probably not the right solution considering the answer:

F = m*g = 1,672*10^-27 * 9,82 = 1,642 * 10^-26

E = F/Q = (1,672*10^-26)/(1,602*10^-19) = 1,044*10^-7

E = U/d --> U = E*d = (1,044*10^-7)*2 = 2,088*10^-7

 

[gneill]

Looks okay to me.

 

[Mushroom79]

Looks okay to me.

Okay, thank you for the help