1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Electric Potential-Work Problem

  1. May 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Referring to the figure attached, how much work must be done to bring a particle, of charge Q = +16e and initially at rest, along the dashed line from infinity to the indicated point near two fixed particles of charges q1 = +4e and q2 = -2e?

    Distance d = 1.40 cm, theta 1 = 43 degrees, and theta 2 = 60 degrees.

    2. Relevant equations

    [tex]E = \frac{kQ}{r^2} [/tex]

    [tex]v = Ed = \frac{Fd}{q} = \frac{w}{q}[/tex]

    3. The attempt at a solution

    I've solved for the net E field produced by the 2 still charges. My thought is that I would plug in the magnitude of that value into [tex]Ed = \frac{w}{q}[/tex] along with the charge from the moving particle and that would produce a value for w... But, I do not know what to use for distance... how do I handle the infinity... with a limit?

    Help appreciated.
     

    Attached Files:

  2. jcsd
  3. May 7, 2010 #2

    physicsworks

    User Avatar
    Gold Member

    The work must be done to bring the charge Q from infinity (where potential is supposed to be zero) to some point in space is Q times the potential of the electric field in this point (say point A). All you need to do is to find the potential due to [tex]q_1[/tex] and [tex]q_2[/tex] in A. From attached picture you know the distances from [tex]q_1[/tex] and [tex]q_2[/tex] to A, so it's easy to find the potential due to each of charge and then add them to find the total potential. Note also that the work done by the electric field when YOU are moving the charge in it is MINUS the work perfomed by YOU.
     
  4. May 7, 2010 #3
    I'm not sure I understand,

    Are you saying that

    [tex]q_1 E_{net} = v_1[/tex]

    and that the same is true for q2 and v2... so,

    [tex]v_1+v_2 =\Delta V[/tex]

    Therefore

    [tex]w = Q \Delta V[/tex]
     
  5. May 7, 2010 #4

    physicsworks

    User Avatar
    Gold Member

    I'm saying that the work done is
    [tex]W = - Q \int_{\infty}^{A} \mathbf{E} \mathbf{dl} = Q \left[ \varphi(A) - \varphi(\infty) \right][/tex]
    If you set [tex]\varphi ({\infty) = 0[/tex] (remember that potential is only defined up to an additive constant) then
    [tex]W = Q \varphi (A)[/tex]
    where
    [tex]\varphi (A) = \varphi_{q1} + \varphi_{q2}[/tex]
    that is the sum of two potentials due to each of the charge.
     
    Last edited: May 7, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook