Eletric field, acceleration

1. Apr 30, 2015

Zbud

1. The problem statement, all variables and given/known data
I dont have any statement, i have to find the problem statement from the answers
a) E = Q / [ Eo* (0.020 m)^2 ]
b) a = E (1.60*10-19 C)/(1.67*10^-27 kg) = 2.0 *10^12 m/s^2
R=0,020 m
q=1.60*10-19 C
m=1.67*10^-27 kg
2. Relevant equations
F=ma => a=qE/m
Gauss Law Integral of (E,ds)=Q/Eo
3. The attempt at a solution
So given E constant everywhere i can assume that E*(integral of ds)=Q/Eo, so if this was a sphere it would be
4piR^2*E=Q/Eo so its out, my equation must be something like E*R^2=Q/Eo , so my area should be a a square ?,
also E and the trajectory of my q must be parallel right ?
Any helpful hints are really appreciated ;d

2. May 1, 2015

ehild

3. The attempt at a solution
So given E constant everywhere i can assume that E*(integral of ds)=Q/Eo, so if this was a sphere it would be
4piR^2*E=Q/Eo so its out, my equation must be something like E*R^2=Q/Eo , so my area should be a a square ?,
also E and the trajectory of my q must be parallel right ?
Any helpful hints are really appreciated ;d[/QUOTE]
You guess well, it might be the constant electric field of a a charged square-shaped plate, or rather the electric field between the plates of a planar capacitor with charge Q, where the plates of the capacitor are squares of sides 0.02 m.
The trajectory of the particle with charge q need not be parallel to E, but its acceleration has to be. What do you think the particle is?

3. May 1, 2015

Zbud

Given its mass and charge it must be certainly a proton, i have been thinking about the capacitor but truly as we have not talked about it once during the lectures so I wasnt keen on that idea,I'm grateful for your insight into this unusual problem

4. May 1, 2015

ehild

That is correct, it must be a proton. Well done!
In case of single charged plate, the electric field is Q/(2Aε0), but it is Q/(Aε0) in case of a capacitor.