Electric Field and Acceleration for a Single Charged Plate and Capacitor

In summary: So your equation for E is correct. Your idea about the area being a square is also correct. And yes, the trajectory of the particle does not have to be parallel to the electric field, but its acceleration must be. A good start to solving this problem.
  • #1
Zbud
2
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
 
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  • #2
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?
 
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  • #3
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
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.
 

Related to Electric Field and Acceleration for a Single Charged Plate and Capacitor

What is an electric field?

An electric field is a physical field that is created by electrically charged particles. It is a force field that exerts a force on other charged particles that enter its region.

How is an electric field measured?

Electric fields are measured using a device called an electric field meter. This meter measures the strength and direction of the electric field at a specific point in space. The unit of measurement for electric field strength is volts per meter (V/m).

What factors affect the strength of an electric field?

The strength of an electric field is affected by two main factors: the magnitude of the charge creating the field and the distance from the charge. The greater the magnitude of the charge, the stronger the electric field will be. The farther away from the charge, the weaker the electric field will be.

How does an electric field cause acceleration?

An electric field can cause acceleration by exerting a force on a charged particle. This force can either attract or repel the charged particle, causing it to move in the direction of the electric field. The strength of the electric field will determine the magnitude of the acceleration.

What are some real-life applications of electric fields?

Electric fields have many practical applications in our daily lives. Some examples include the functioning of electronic devices, such as computers and cell phones, the attraction and repulsion of magnets, and the transportation of electricity through power lines. Electric fields are also used in medical procedures, such as electrocardiograms and electroencephalograms.

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