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Electric Potential

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data
    1. A positive charge of magnitude 2 microC is at the origin.
    (b)How much work must be done by an outside agent to bring a 3 microC charge from infinity to r=4, assuming that the 2 microC charge is held fixed at the origin ?
    (c) How much work must be done by an outside agent to bring the 2 microC charge from infinity origin if the 3 microC charge is first placed at r=4m and is then held fixed ?

    2. Protons from a Van de Graaff accelerator are released from rest at a potential of 5 MV and travel through a vacuum to a region at zero potential.
    (a). Finad the final speed of the 5-MeV protons.
    (b). Find the accelerating electric field if the same potential change occurred uniformly over a distance of 2m


    2. Relevant equations

    V=kq/r
    E= -dV/dx
    U=qV
    3. The attempt at a solution

    1. So I found that the electric potential at 4m is 4495 V and at infinity V=0.

    To find the work done, I have to integrate -Udx but I don't know what to put for the lower limit. Do I put infinity ?

    Or How do I do part b and c ?

    2. So I found the speed in part a using energy equation and got 2.64 x 10^7 m/s

    How do I then do part b ? I know that E= -dV/dx but then I cannot solve this because I don't know dV.
     
  2. jcsd
  3. Oct 17, 2009 #2
    (It's been a long time since I last looked at this kind of question, so this post may be from moderately to outrageously wrong)


    First, the electric potential at point r is: [tex]V=\frac{k|Q|}{r^2} V[/tex]

    Assuming we're working on vaccum, [tex]k\approx9\cdot10^9 N\cdot m^2/C^2[/tex] and [tex]V\approx1125 V[/tex].

    Potential energy is [tex]Ep=q\cdot V J[/tex], so [tex]Ep\approx3.375\cdot10^{-3} J[/tex]. That's 1.b.

    1.c follows the same calculations, only exchanging both charges. The potential energy is the same.

    2.a and 2.b: Have no friggin' clue ^^'.
     
    Last edited: Oct 17, 2009
  4. Oct 17, 2009 #3

    ehild

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    Homework Helper
    Gold Member

    The potential is work already, you do not need to integrate. Assuming zero potential at infinity, the potential at a point is the work done by the electric field on a unit positive charge when the charge moves from infinity to that point. So to get work from potential difference, you simply multiply it with the charge.
    If you calculate work by integrating the electric field intensity, the lower limit is the place where you know the potential. It is infinity here.
    For question b, remember that r in the formula for the potential is the distance from the fixed charge. The two charges are 4 m apart again, so r=4 m.

    If the electric field is uniform over a distance D, dV/dx = change of potential / D.

    ehild
     
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