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Electric field vector

  1. Jan 23, 2005 #1
    2 positive charges q are a distance s apart on the y axis
    a) find an expression for the electric field vector at distance x on the axis that bisects the 2 charges

    b)thy these sanity checks on your answer: what is the direction of the electric field for positive and for negative values of x? What is the limit of your expression for E_vector as x --> infinity

    OK, these questions are somewhat simple, yet confusing; I don't get what they're really asking. Please help.

    for a) what I get for a is E_vector net = (q/4piEo){[1/(s-x)^2] - [1/x^s]}j_unit N/C

    for b) I don't get what do they mean by +ve/-ve values of x. Isn't x a distance? Then how could it be negative? What I do know that if the position is at infinity then r^2 would get really large to the point that E is zero.

    What I did in a is assuming that x is somewhere between the 2 charges. Now if they say that x can be either positive/negative, what is that supposed to mean?

    These questions just don't make sense. Please forgive me/don't flame me but it's 4 A.M. and I'm desperate trying to solve these problems when some questions have really poor wordings and are damn confusing, and just don't make much sense to the point I wish whoever make these questions would just stop and let the others who can compose better write up/make the questions instead. :grumpy:
     
  2. jcsd
  3. Jan 23, 2005 #2

    dextercioby

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    Science Advisor
    Homework Helper

    Okay,let's deal correctly with point "a"...I think you need to take another look at the geometry of problem,since it involves an isosceles triangle and the theorem of Pythagora.
    Redo your calculations for point "a"...

    Daniel.
     
  4. Jan 23, 2005 #3

    Doc Al

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    Staff: Mentor

    I believe you have an incorrect mental picture of the geometry of the problem.
    Let the axis that bisects the two charges be the x-axis. Thus one charge will be at point (0, s/2), the other at point (0, -s/2).

    Once you get the correct mental picture you'll see that "x" is the position along the x-axis: one side is positive, the other negative.
     
  5. Jan 23, 2005 #4
    Thanks Daniel and Doctor Al,

    Here are what I got, if you don't mind can you check them?

    for a)

    E_vector net = {q/[4 pi epsilon_0 (x^2 + (0.5s)^2)}cos(theta)*2 i_unit N/C, the j_unit components (vertical y axis) will cancel out, while the i_unit components will complement each other. (So there's only an i/x-axis components, no vertical components.) (Theta is the angle that the line drawn from x to s/s to x makes with the x-axis.)

    BTW, does bisect mean that the axis will divide the distance between the upper and lower charges into 2 "equal" parts? I thought that bisect mean dividing a thing into 2 parts, but not necessary equal.

    for b) The direction of the electric field will be to the right when x is on the positive side, left when x is on the negative side. And as x--> infinity, E_vector field will be zero.

    Are these correct answers, thanks in advance. The prof hasn't covered this in class yet, the assignment is due on Thursday. However, I've got 3 exams to write this coming weekends/next week so I rather get these problems done now.
     
  6. Jan 23, 2005 #5

    Doc Al

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    Staff: Mentor

    Good, but express cos(theta) in terms of x and s.

    Yes.
    Right.
    What they want is how it approaches zero. (For example, 1/x and 1/x^2 both approach zero, but very differently.) When you've rewritten the field as I suggest above, look at the limit as x--> infinity. You should recognize it.
     
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