Velocity of an electron in an electric field

[zeralda21]

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.

 

[tiny-tim]

hi zeralda21!

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?

 

[zeralda21]

hi zeralda21!


what about the distance?

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?

 

[tiny-tim]

??

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

how do you get 6 ? ​

 

[zeralda21]

??

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

how do you get 6 ? ​

I am ashamed . The answer is 8.36*10^7 m/s. Thanks tiny-tim

 

[tiny-tim]

I am ashamed

you are humble yet contrite

go in peace! ​