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Field and Potential of 3 charges conceptual problem

  1. Sep 23, 2007 #1
    One positive (+2q) and two negative (-q) charges are arranged as displayed in figure 1. Calculate electric field E and electric potential P at points along the poitive y-axis as functions of their coordinate y. What is the direction of E at those points? In your results, does y-component Ey of the field satisfy Ey = -dP/dy? Is it supposed to satisfy?


    Coulomb's Law
    E = ke*q/r[tex]^{2}[/tex]

    First of all, the x-component of the field at any point along the y-axis is zero, since -q(-) and -q(+) cancel out one another, and 2q provides no field in the x-direction.

    I know the field at any point will be equal to the field from 2q (positive y) minus the two y-components from the other two particles (negative y). The contribution from the positive particle is simple.

    Ey(+) = ke(2q/y[tex]^{2}[/tex])

    The other contributions require a little trigonometry, which I'm hoping I've done correctly..

    Assuming r is equal to the distance from -q to y (hypotenuse), r[tex]^{2}[/tex] = y[tex]^{2}[/tex] + a[tex]^{2}[/tex]. Therefor E(-) = Ke(-2q/y[tex]^{2}[/tex]+a[tex]^{2}[/tex]).

    Now I have to break this down to find the y-component for E(-) which (I think) is just :

    Ey(-) = sin[tex]\alpha[/tex]*E(-)

    Since the ultimate goal here is to define two functions, should I define sin[tex]\alpha[/tex] in terms of y for integration purposes? Could I say sin[tex]\alpha[/tex]= y/r = y/(y[tex]^{2}[/tex]+a[tex]^{2}[/tex])[tex]^{1/2}[/tex]? Oof, things are getting rusty...

    My gut says I will have to integrate the equations for Ey(+) and Ey(-), and the difference between them will be my function for the field. I haven't really started at the potential equation yet... figured I would check to see if I'm headed the right direction first. Any advice?
    Last edited: Sep 23, 2007
  2. jcsd
  3. Sep 23, 2007 #2
    It's much easier the other way around. Obtain an expression for the potential, which is simpler to work with being a scalar.

    You can then obtain the electric field with a gradient operator (in other words Ex=-dV/dx Ey=-dV/dy, etc. (partial derivatives)).
  4. Sep 23, 2007 #3
    Potential = Ke (q/r), but r = [tex]\sqrt{[/tex]y[tex]^{2}[/tex]+a[tex]^{2}[/tex] for the two negative charges.

    So will the total P be Ke(2q/y) + Ke(-2q/[tex]\sqrt{[/tex]y[tex]^{2}[/tex]+a[tex]^{2}[/tex])?

    When I derive this I get something like Ke(dq/y[tex]^{2}[/tex]) -Ke(dq/2(y[tex]^{2}[/tex]+a[tex]^{2}[/tex])[tex]^{3/2}[/tex])

    I see! This is the same as Ey with substituted sin like I was asking. The only issue I see is when I derived I got a 2 in the denominator...
  5. Sep 23, 2007 #4
    Don't forget the chain rule... :wink:
  6. Sep 23, 2007 #5
    hahah... I can't help it, my mind refuses to retain calculus methods.
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