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Infinite line of charge

  1. Aug 25, 2005 #1
    The question I've been attempting:

    From my notes and working etc. I've got:

    E = lambda/2*pi*epsilon*r (where lambda = charge per unit length and epsilon = permittivity of free space)

    F = E * q (where q = the charge of the proton, -1.60*10^-19, as apparently defined by the question)

    I've used the F calculated to get acceleration by

    a = F/m (where m = mass of proton)

    I get a value of -7.88*10^7

    I then use the linear acceleration formula v^2 = u^2 + 2*a*s to try and calculate s.

    I get 0.04627... (4.63*10^-2)

    Apparently this is incorrect however. Can anyone see where I'm messing up?
  2. jcsd
  3. Aug 25, 2005 #2


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    The linear acceleration formula holds only when the acceleration is constant.

    Try using energy conservation instead! :)
  4. Aug 25, 2005 #3
    Hmm, still having trouble,

    1/2*m*v^2 + E*q*y = 0 + E*q*(unknown y value)

    The formula for E that I'm using doesn't make sense to me but appears right in my notes and text book: E = lambda (charge per unit length)/2*pi*r*epsilon(permittivity of free space)

    I wouldn't think E relies on r (for a straight line of charge). Aren't field lines parallel for a line of charge? Meaning E is constant at any point in the field?

    I'm not given the r value for the final state of the particle, so I can't work out either E or the unknown y on the right hand side of that formula.
  5. Aug 25, 2005 #4


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    The "Electric Field Lines" are spread out more (1-dim), farther from the wire.
    That means the E-field strength decreases as 1/r .
    You need substripts to distinguish "E_final" from "E_initial" ... not equal!

    You want to integrate E(r) from y_initial to y_final ...
    or if this isn't for calc-based physics, use Potential.
  6. Aug 26, 2005 #5
    No, still confused out of my brain. Was doing it on Mastering Physics.com, but exceeded attempts and failed the question. I tried a billion random different formulas. Worst thing is having no idea if they were even valid to use in an equation

    The answer was apparently 0.134 m, but I still can't see how. I'll probably have to see tutorial teacher or something :confused::confused:
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