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Homework Help: Charge Density of wire with potential difference

  1. Jan 25, 2009 #1

    Consider a long charged straight wire that lies fixed and a particle of charge +2e and mass 6.70E-27 kg. When the particle is at a distance 1.91 cm from the wire it has a speed 2.80E+5 m/s, going away from the wire. When it is at a new distance of 4.01 cm, its speed is 3.20E+6 m/s. What is the charge density of the wire?

    Alright so this is how I approached it but didn't get it right:

    The difference in kinetic energy at two given distances should be equal to the opposite difference in potential energy ([tex]\Delta[/tex]K = -[tex]\Delta[/tex]U), which when divided by the given charge of the particle should give the potential difference. Right? ([tex]\Delta[/tex]U = q*[tex]\Delta[/tex]V)

    The potential difference is equal to -E*d where d is the distance between the two given points and E is the electric field of the wire given by: lambda/2*pi*epsilon*r.

    With these you should be able to solve for lambda but I keep getting the answer wrong and I think its because I'm using the wrong r in the equation for the electric field. I thought r should be the distance from the wire to the farthest given point but I wasn't sure...

    Can anyone help?
    Last edited: Jan 25, 2009
  2. jcsd
  3. Jan 25, 2009 #2
    Without showing your work, I would guess you may be finding the potential incorrectly. It looks like you are treating E as a constant with the expression V=-E*d. Recall,

    [tex]V=-\int \mbox{Edr}[/tex]

    where r is the perpendicular distance from the wire. Your approach to the problem appears correct.
  4. Jan 25, 2009 #3
    Yes I think I found my problem, I can't use the simplified V = -Ed equation because E is not a uniform electric field in this case.
  5. Jan 28, 2009 #4
    Can you please explain it in more detail ? How do you calculate the potential difference then if you cant consider it a uniforn electric field ? and what is d ?
  6. Jan 28, 2009 #5


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    Homework Helper

    I'm guessing you don't know calculus. chrisk explained it. You must integrate the electric field as a function of distance. The "d" means "differential". If you don't know calculus, then I don't expect you will have any idea what this means.
  7. Jan 28, 2009 #6
    ok.. nevermind.. i got it.. thanks !!
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