Electric Field and Acceleration for a Single Charged Plate and Capacitor

AI Thread Summary
The discussion revolves around calculating the electric field and acceleration for a charged plate and capacitor. The electric field (E) is derived from the formula E = Q / [Eo * (0.020 m)^2], while the acceleration (a) is calculated using a = E (1.60*10^-19 C)/(1.67*10^-27 kg), resulting in a value of 2.0 * 10^12 m/s^2. The problem hints at a constant electric field, suggesting it may be related to a charged square plate or a planar capacitor. It is clarified that the trajectory of the charged particle does not need to be parallel to E, but its acceleration must be. The particle in question is identified as a proton based on its mass and charge.
Zbud
Messages
2
Reaction score
0

Homework Statement


I don't 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

Homework Equations


F=ma => a=qE/m
Gauss Law Integral of (E,ds)=Q/Eo

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
 
Physics news on Phys.org
Zbud said:

Homework Statement


I don't 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

Homework Equations


F=ma => a=qE/m
Gauss Law Integral of (E,ds)=Q/Eo

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

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?
 
  • Like
Likes 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
 
Zbud said:
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
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top