Velocity of an electron in an electric field

In summary, the electrons will achieve a speed of 8.36*10^7 m/s when traveling a distance of 0.2 m from rest in an electric field of 100 kV / m.
  • #1
zeralda21
119
1

Homework Statement



In a picture tube (eg. a TV set) electrons are accelerated of an electric field. If the electric field strength is 100 kV / m and the electrons travel a distance 0.2 m from rest, what speed will they achieve?

Homework Equations


E=U/d where E is electric field intensity, U is the voltage and d is the distance.

W=qU

F=qE

(mv^2)/2 = kinetic energy


The Attempt at a Solution



I think that the potential electrical energy of the electron will be converted into kinetic energy. Hence:

W=qU=(mv^2)/2 or v=sqrt((2qU)/m). The charge and mass of an electron is known and so is the electrical field intensity. If I insert these values in the equation above I will get:

sqrt((2*1.602*10^(-19)*100*10^3)/9.11*10^(-31))=1.87..*10^8 m/s. The correct answer is 8*10^7 m/s.
 
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  • #2
hi zeralda21! :smile:
zeralda21 said:
In a picture tube (eg. a TV set) electrons are accelerated of an electric field. If the electric field strength is 100 kV / m and the electrons travel a distance 0.2 m from rest …

sqrt((2*1.602*10^(-19)*100*10^3)/9.11*10^(-31))=1.87..*10^8 m/s. The correct answer is 8*10^7 m/s.

what about the distance? :wink:
 
  • #3
tiny-tim said:
hi zeralda21! :smile:


what about the distance? :wink:

You are right, I didn´t take that into consideration. Here is another try:

If W=qU and E=U/d, then of course W=qEd. If I set this equal to the kinetic energy and solve for v i end up with:

W=qEd=(mv^2)/2 ----> v=sqrt((2qEd)/m). When I insert values for q,E,d and m(which are all known) the answer is 6*10^7 m/s, much more accurate but I think the error is too large for an "approximation error". No? What do you guys say?
 
  • #4
??

you got 18.7 (times 107) before, and you're multiplying by √(0.2) …

how do you get 6 ? :confused:
 
  • #5
tiny-tim said:
??

you got 18.7 (times 107) before, and you're multiplying by √(0.2) …

how do you get 6 ? :confused:

I am ashamed :redface:. The answer is 8.36*10^7 m/s. Thanks tiny-tim :smile:
 
  • #6
zeralda21 said:
I am ashamed :redface:

you are humble yet contrite

o:) go in peace! o:)
 

1. What is the velocity of an electron in an electric field?

The velocity of an electron in an electric field is dependent on several factors, such as the strength of the electric field, the charge of the electron, and any other forces acting on the electron. Therefore, the velocity can vary greatly and cannot be accurately determined without specific information about the conditions.

2. How does the strength of the electric field affect the velocity of an electron?

The strength of the electric field directly affects the velocity of an electron. As the strength of the electric field increases, the force acting on the electron also increases, causing the electron to accelerate and increase in velocity.

3. Can the velocity of an electron in an electric field be negative?

Yes, the velocity of an electron in an electric field can be negative. This indicates that the electron is moving in the opposite direction of the electric field, which is possible if there are other forces acting on the electron.

4. How do other forces, such as magnetic fields, affect the velocity of an electron in an electric field?

Other forces, such as magnetic fields, can influence the velocity of an electron in an electric field. These forces can either add to or counteract the force of the electric field, resulting in a change in the electron's velocity.

5. Is there a maximum velocity that an electron can reach in an electric field?

According to classical physics, there is no maximum velocity that an electron can reach in an electric field. However, in the quantum realm, there is a limit to the velocity an electron can attain, known as the speed of light. This limit is determined by the properties of the electron and the electric field it is in.

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