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Electric Field Strength - Assignment Question

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data
    The Nucleus of a helium ion contains two protons approximately 5x10-15m apart. They can be considered to be surrounded by one orbiting electron approximately 5x10-11m away. Calculate the following:

    a) The electrostatic force between the two protons
    b) The electrostatic force on the electron caused by the nucleus
    c) The electrical potential energy of the electron

    2. Relevant equations
    Equations & Values
    F=kQq/r2
    E=kQ/r2
    V=kQ/r


    e=-1.6x10-19C
    k=8.99x109Nm2C-2
    r (between protons)=5x10-15m
    r (electron)=5x10-11m


    3. The attempt at a solution

    I am confident at my attempt at the answer for part a

    F=(8.99x109)(-1.6x10-19)(-1.6x10-19)/(5x10-15)2
    F=9.2N (repulsive)

    For part b, I am slightly unsure as to what I am doing as I don't particularly understand what is it asking me. As the nucleus has the two protons within it, does this mean that it has double the charge?

    e.g. e(nucleus) = 2(-1.6x10-19)C = -3.2x10-19C

    so

    F=(8.99x109)(-3.2x10-19)(-1.6x10-19)/(5x10-11)2
    F=1.8x10-7N (repulsive)

    Or am i missing something I should be doing to calculate the charge of the nucleus?

    For part C, my immediate thought is that:

    V=(8.99x109)(-1.6x10-19)/5x10-11
    V=-2.9x10-7V (attractive)

    But, given that this answer is worth 3 marks, I am thinking this answer is too simple and I must be missing something here.

    I am sorry for this simplistic nature of this question, but in all honesty we have not been told anything on this subject and I am trying to fill in the blanks left by our tutors.

    Thank you in advance.
     
    Last edited: Mar 6, 2016
  2. jcsd
  3. Mar 6, 2016 #2
    In part b you have two charges placed at the centre -so you are right as they are like charges they can not make a dipole.
    but your conclusion about nature of force is questionable-you have overlooked a -ve sign.

    this part needs correction your equation was correct - but your placement of charge is in error.
    think about potential due to charge.
     
  4. Mar 6, 2016 #3
    Hi drvrm,

    Thanks for the reply.

    As far as part b is concerned, I am struggling to see where I have overlooked the -ve.

    If I am told that both protons and the electron have the charge -1.6x10-19C and I am correct in my assumption that the charge of the nucleus is simply the combination of the charge of the two protons, therefore the charge of the nucleus is -3.2x10-19C that means that both of the charges in question are -ve.

    If both charges are -ve then it must be repulsive as only opposites attract?

    Or is this regarding the wording of the question? As the question is asking for the force on the electron caused by the nucleus does this mean that I have to assume that the charge on the proton is +ve and as such the force acting upon the proton by the nucleus is attractive?

    Forgive my naivety but no matter how hard I look, I can't see where I have overlooked the sign. Obviously I don't expect you to tell me where as the aim is for me to figure it out for myself, but can you point me in an area where I should be looking?

    As for part c, I will look further at this once I have got my head around part b :)

    Thank you again for your help here.
     
    Last edited: Mar 6, 2016
  5. Mar 6, 2016 #4
    we know that proton carries a positive charge equal in magnitude to electronic charge and electron's charge is negative otherwise the atomic stability will be in question.
    you can see any data book on charge of an electron and proton-physics text book.

    regarding part c one has to calculate the potential due to nucleus on the electron which should be K Qq/r as KQ/r will be only potential/work done due to bringing an unit positive charge from infinity to that point
     
  6. Mar 7, 2016 #5
    Thanks for that,

    You are absolutely right re: nucleus. I was confusing myself as we were specifically given a negative value for the charge and completely overlooked that the charge must be positive in a nucleus, so as for part b I get:

    Q=3.2x10-19C
    q=-1.6x10-19C
    r=5x10-11m

    so...

    F=kQq/r2

    F=(8.99x109)(3.2x10-19)(-1.6x10-19)/(5x10-11)2

    F=-1.8x10-7N (attractive)

    For part c (calculate the electrical potential energy of the electron) it is then a case of:

    V=kQ/r

    V=(8.99x109)(3.2x10-19)/(5x10-11)

    V=57.536V

    PE=Vq

    PE=(57.536)(-1.6x10-19)

    PE=-9.2x10-18V

    Have I finally got my head around this now?

    Once again, thank you so much for your assistance.
     
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