Electric field and electric forces

In summary, an electric field is a physical field created by the presence of electrically charged particles and exerts a force on other charged objects. It differs from electric force, which is the force that acts on charged particles within the field. The strength of an electric field is measured in volts per meter and decreases with distance according to the inverse square law.
  • #1
Olgapoollamas
1
0

Homework Statement


An electron is projected with an initial speed v0 = 2.00×10^6 m/s into the uniform field between the parallel plates, as shown in figure. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates.

<img src=http://session.masteringphysics.com/problemAsset/1053977/4/YF-21-033.jpg>


Homework Equations



a). If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

b)Suppose that in the figure the electron is replaced by a proton with the same initial speed v0. Would the proton hit one of the plates?

c)If the proton would not hit one of the plates, what would be the magnitude of its vertical displacement as it exits the region between the plates?



The Attempt at a Solution



Part a)
ok so i found the acceleration by using basic mechanics

x=v0 *t
then, y=(1/2)*a*t^2
where t=x/v0, so
y= (1/2)*a*(x/v0)^2
then solved for a:

a=(2*y*v0^2)/x^2

where y=.5 cm
x=2 cm, and v0=2.0*10^6 m/s



since F=ma and F=Eq solved for E
E=ma/q
Then plugged in a
E=(m/q)((2*y*v0^2)/x^2)
The mass and the charge of the electron is known:
Mass of e=9.019*10^-31
Charge of e=1.602*10^-19

I keep getting the same answer when i plug numbers back into the equation and the computer is telling me that is wrong. I don't know if there's anything wrong with my calculations or the equation. Please let me know if I'm on the right track.

Part b)
no it will not (correct answer). Same force with more mass, the proton will go past.

Part c)

F=qE

a=F/m(mass of a proton) where acceleration we already know from part a.
So,

(2*y*v0^2)/(2 cm)^2 = F/m

then solved for y, and I still can't get the answer. Anyone sees a mistake? Please let me know.

Thank you so much.
 
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  • #2
Your expression for the electric field is correct. If the computer is telling you that the answer is wrong, the reason could be one of the following
1. You keep making the same calculational mistake when you are substituting the numbers or you are making a unit conversion error. Without knowing what answer you got, we cannot tell if this is the case.
2. The magnitude of the electric field is what you have, but the computer is expecting a negative sign in front of it. Without a figure, we cannot tell if this is the case.

For future reference, we cannot tell where you wnet wrong if you don't tell us what you got in more detail.
 
  • #3
How come there are several recent replies to +10 year old questions?
 
  • #6
drmalawi said:
How come there are several recent replies to +10 year old questions?
Some post-season spring cleaning.
 

1. What is an electric field?

An electric field is a physical field that surrounds an electrically charged object and exerts a force on other charged objects within its vicinity.

2. How is an electric field created?

An electric field is created by the presence of electrically charged particles. These particles can be either positive or negative and interact with each other through the exchange of virtual particles called photons.

3. What is the difference between electric field and electric force?

The electric field is a physical field, while electric force is a force that acts on charged particles within that field. The electric field determines the strength and direction of the electric force on a charged particle.

4. How is the strength of an electric field measured?

The strength of an electric field is measured in units of volts per meter (V/m). This represents the amount of force that a unit charge would experience when placed in the electric field.

5. How does distance affect the strength of an electric field?

The strength of an electric field decreases as the distance from the source of the field increases. This is known as the inverse square law, which states that the strength of an electric field is inversely proportional to the square of the distance from the source.

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