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Fields in Physics assignment question about electron beams in a cathoe ray tube

  1. Feb 12, 2012 #1


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    1. The problem statement, all variables and given/known data

    Ok so we have been given a diagram of a cathoe ray tube, with relevant data on it. The second question asks to "calculate the speed of the electron, using data from the first question". I need to make sure the answer is correct as it leads into many different questions.

    I am unable to attach the picture, but here are all details:
    Distance between metal plates: 5cm
    Length of metal plates: 10cm
    Charge on upper plate: 500V
    No charge on lower plate
    Charge on anode: 1000V

    2. Relevant equations

    The first question was to calculate the kinetic energy gained by an electron travelling from cathode to anode. What i calculated was:

    E_k=-1.6×〖10〗^(-16) N

    3. The attempt at a solution

    I used two different methods to work this out, but got two different answers:


    E_k=1/2 mv^2


    v=√((2×-1.6×〖10〗^(-16))/(9.11×〖10〗^(-31) ))

    v=-1.87 〖ms〗^(-1)



    s=((2×-1.6×〖10〗^(-16)×500))/(9.11×〖10〗^(-31) )

    s= -1.76 〖ms〗^(-48)
  2. jcsd
  3. Feb 12, 2012 #2


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    Homework Helper

    Welcome to PF!
    The first speed calc is correct except it works out to 1.87 x 10^7 on my calculator.
    I don't understand the second - what is C? I don't recognize the formula.
  4. Feb 12, 2012 #3


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    Thank you!
    I have put it into my calculator again a few times but still getting the same answer: 1.87x10^-24 .... I put it all in separately, so work out what 2x 1.6x10^-16 then divide it...then work out the square root. would doing this alter my answer?

    In the second equation the C represents the charge on the electron. I found it on a website so i'm not totally sure it is suitable for this problem.

    Also what do i do about the fact that the electron has a negative charge on it? because i'm unable to square root a negative number.
  5. Feb 12, 2012 #4


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    Homework Helper

    There is no negative number - the energy is positive. Technically, the polarity is oriented so the V is negative, too, so the qV is positive.

    v=√((2×-1.6×10^(-16))/(9.11×10^(-31) ))
    Say you separate the numbers and powers: 2x1.6/9.11 = 0.35
    The power of 10 is -16-(-31) = 15. So under the square root, 0.35 x 10^15
    Square root of that is 1.87 x 10^7.
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