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Homework Help: Volume charge density across a potential difference

  1. Aug 30, 2014 #1
    1. The problem statement, all variables and given/known data

    A 1.0 μA proton beam is accelerated across a potential difference of 1.0 kV. Assume the beam has uniform current density over a diameter of 2.0 mm, and zero outside.

    Find: volume charge density in the beam, (HINT use λ=I/v where λ= charge/ unit length)

    The radial electric field intensity inside and outside of the beam.

    2. Relevant equations

    λ=I/v, λ=Q/unit length, ρ_v=Q_total/V, KE=1/2mv^2

    3. The attempt at a solution

    I approached the first part by finding kinetic energy using know values for mass and charge then rearranging to find velocity. I then inserted velocity into the hint equation to find total charge then used volume density= total charge over potential difference. However I don't think this last approach is correct.

    For the second question I'm so confused. I know E=-∇V but V is a scalar so my other approach was to show that it had zero electric field due to no charge.

    Any help would be so appreciated! Thanks!
  2. jcsd
  3. Aug 30, 2014 #2


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    The volume charge density is simply the charged contained in a certain volume divided by that volume. If you want your calculation double checked, give more details about what you did.

    For the second part, you need the E field produced by a uniform line of charge. One way to obtain it is by using Gauss' law. If you have not covered this, maybe you have learned the formula for the E field from an infinite line of charge?
  4. Aug 30, 2014 #3
    Hi thanks for replying!

    So given the values for mass and charge of a proton (m=1.67×〖10〗^(-26) kg,Q=1.6×〖10〗^(-19) C), velocity can be determined:
    QV=(1.6×〖10〗^(-19) )(10×〖10〗^3 )
    =1.6×〖10〗^(-15) J

    Then using QV=E=KE I was able to find a velocity,
    E=1/2 mv^2
    =1.38×〖10〗^6 m/s^2

    I then used the hint equation, taking Q to mean total Q,

    Q_tot=(unit length)*I/v
    Q_tot=1.45×〖10〗^(-15) C

    Then using equation ρ_v=Q_tot/Potential
    ρ_v=(1.45×〖10〗^(-15))/(1×〖10〗^3 )
    = 1.45×〖10〗^(-18) C/m^3

    However I don't think this last bit is right. The question does not give a a shape so I don't know how to determine its volume.
  5. Aug 30, 2014 #4


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    Watch out, 1kV = ## 10^3 V ##, not ##10 \times 10^3 V##
    Ok. What does "unit length" represents here? What value did you use in your calculation?

    No, the density of charge is the charge divided by the volume. Since they give a diameter, they want you to take the beam as being a cylinder.
  6. Aug 30, 2014 #5
    Whoa good spotting!

    I used the given diameter :/ ahh so wrong. I honestly have no idea what to use here as I'm so confused about its shape. Whether I'm looking at the shape of the beam or the shape of the p.d.

    That makes sense considering it is a beam. I have length but I do not have a radius so should I leave the answer in terms of r? If this is the case I get
    and using proton values for Q and the given diameter for h,
    ρ=2.5x10^-17 C/r^2.m^3

    However this approach still is not complete and doesn't use the hint equation. Is the fact that it is a beam lead you to believe it is a cylinder? Is it not possible to be looking at a p.d across a sphere?
  7. Aug 30, 2014 #6


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    Here is the trick: the length you use here is completely up to you. What happens is that in the final calculation that length will cancel out. To see this, just call this length L and leave it in your expression for the charge Q. It makes sense that Q will depend on the length of the beam. L will cancel out at the end.
    Watch out: you know the diameter so you can find the radius. What you do now know is the length (which you call h in your equation, it is what I called L above). Just plug the Q you found previously (which contains L) into the equation you wrote for ## \rho##. You will see that the length L will cancel out and you will get the final answer. By the way, you *are* using the hint equation, in the equation ## \rho = Q/ V$, that V means the volume, not the potential.

    Yes, we can talk about the p.d. across a sphere but that is not needed here.
  8. Aug 30, 2014 #7
    Wow that is a beautiful solution, cancelling out at its finest!

    Just for further understanding, why am I able to use the charge for a proton to find the velocity but then when I am finding the volume charge density I have to use Q=LI/v? Is it right to take this second Q to be the total charge of the beam or the charge of the p.d?

    Thank you so much for your assistance! I have such a better understanding now :rofl:
  9. Aug 30, 2014 #8


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    Good job! I am glad to have helped.

    In the equation Q = L I /v, the charge Q represents the total charge in a section of length L so one cannot use the charge of a single proton.

    To find the speed, you used the formula ## KE = Q \Delta V ##. You could use any charge there and E would be the energy of that total charge. But later you use in the kinetic energy ## KE = 1/2 m v^2 ## the mass of a single proton, right? So that means you need the KE of a single proton. This is why in ## KE = Q \Delta V ## you had to use the charge of a single proton.

    Does that make sense?
  10. Aug 31, 2014 #9
    Yes! Perfect sense!

    Cannot thank you enough for sharing your wisdom!
  11. Aug 31, 2014 #10


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    You are very welcome! You understood very quickly all the tricky aspects. Good job.
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