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Alculate the x-component of the electric field

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Positive charge Q is distributed uniformly along the positive y-axis between y=0 and y=a . A negative point charge −q lies on the positive x-axis, a distance x from the origin

    Calculate the x-component of the electric field produced by the charge distribution Q at points on the positive x-axis.

    Calculate the y-component of the electric field produced by the charge distribution Q at points on the positive x-axis.

    2. Relevant equations
    E = q/4πε0r2
    r2 = x2+y2

    3. The attempt at a solution
    So I was able to get the first part of the question relatively easily by using some techniques from the textbook, which resulted in the correct answer:

    dEx = Q/4πε0x√(x2+a2)

    But when I try to find the y-component, I get the answer:

    dEy = (Q/(8πε0a))*(1-(x/(√x2+a2)))

    This answer results in the following message from the website:
    "Your answer either contains an incorrect numerical multiplier or is missing one."

    I need help finding the problem since I don't know where I may have gone wrong.
     
  2. jcsd
  3. Feb 18, 2015 #2

    Nathanael

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    Just making sure; do you mean, "on the positive y-axis"?

    This gave you the correct answer? The units are incorrect. It might be simply a typo (because it's almost correct) but I want to make sure you understand how to do this before finding the y-component.

    Edit:
    The units are correct sorry... I misread the equation
     
    Last edited: Feb 19, 2015
  4. Feb 19, 2015 #3

    haruspex

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    Assuming it is intended as Ex = Q/(4πε0x√(x2+a2)), the dimensions are correct (but yes, the expression is wrong... it should not tend to infinity as x tends to zero).
     
  5. Feb 19, 2015 #4

    haruspex

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    You don't say how you got that answer. We can't tell you where you went wrong if you don't post your working/logic.
     
  6. Feb 19, 2015 #5

    rude man

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    As pointed out in post #2, this answer is incorrect. Did you do a typo somewhere? And why "dEx"? It's just Ex.
    You should be figuring the problem out for yourself rather than looking for formulas in a textbook.
    If you did that you would have come up with one definite integral over y=0 to y=a for the x component, and a different definite integral, also over y=0 to y=a, for the y component.
    BTW why is q mentioned in the problem?
    Also, both the x and y components of the E field are obviously negative for all x. The answers have to be sign-reversed in x for x < 0.
     
  7. Feb 19, 2015 #6

    haruspex

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    Most likely for a later part of the question.
     
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