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Homework Help: Y component of electric field due to line of charge

  1. Sep 6, 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 y-component of the electric field produced by the charge distribution Q at points on the positive x-axis.

    2. Relevant equations

    3. The attempt at a solution
    I formed the equation dE_y=K*[(lambda)dy/(x^2+y^2)]*(y/(x^2+y^2)^(1/2) and trying to integrate
    E_y=K(lambda)*[integral 0 to a][y/(x^2+y^2)^(3/2)]

    using trig substitution y=xtan(theta) and dy=xsec^2(theta)d(theta)

    to make this a little easier to read I will say K(lambda)=C
    =C[integral]{sin(theta)/x} = C[-cos(theta)/x] cos(theta) = x/[(x^2 +y^2)^(1/2)]
    =C[-1/[(x^2 +y^2)^(1/2)] from 0 to a
    =(-K(lambda))/[(x^2 + a^2)^(1/2)] (lambda) = Q/a

    =-(KQ)/[a(x^2+ a^2)^(1/2)]
    This is the answer I came up with, but apparently its wrong. can anyone explain to me why this is wrong and what is the right way to do it?
    Thank you

    P.S. this is my first post, so idk if theres a "cleaner" way to write these equations, if there is please tell me. Sorry if they are hard to follow.
  2. jcsd
  3. Sep 6, 2015 #2

    rude man

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    OK up to here.

    1. If you click on the Σ in the toolbar you can select all sorts of greek letters and other symbols.
    2. You did not integrate correctly. If you check your units you will see that the dimensions in your answer are incorrect. E has dimensions of kQ/L2 and yours is kQ/L3/2. ALWAYS check dimensions (units)!!! It's the most powerful error-checking tool you have.
    3. Learning the technique of integration is in IMO a big waste of time (can't wait to hear the ripostes on this one!). Instead, use tables or our recent gift from Heaven called Wolfram Alpha!
  4. Sep 6, 2015 #3


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    2017 Award

    Hi Md, welcome to PF :smile: !


    Re equation typesetting:

    We use ##\#\### to start and to end to typeset equations in line with the text.
    For bigger stuff we use ##$$## to start and end to get separate lines centered on the page

    Here's a tutorial. First example:

    you type ##\#\### y= a x^2 + bx + c ##\#\###

    the result will be

    ##y = a x^2 + bx + c ##

    and with dollar signs instead of # you get $$y = a x^2 + bx + c $$

    One other neat thing: to learn a lot more, right-click on equations in threads and pick "show math as TeX Commands"


    Re electric field:

    I agree with Rudy on $$dE_y = k\lambda\; {y \, dy \over \left (x^2+y^2 \right )^{3\over 2} }$$
    I happen not to agree with Rudy about the dimensions. Perhaps he has overlooked that a also has the dimension of length.

    You substitute ##y = tan\theta## which may or may not be fine.

    I notice that ##ydy = {1\over 2} d(y^2)## and that suggests another substitution. See if that gets the same answer ...

    (un)fortunately hyperphysics calculates the other component, otherwise we would have an opportunity to check the answer :rolleyes:

  5. Sep 6, 2015 #4

    rude man

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    Rudy agrees! I goofed there.
    But the OP needs to clean up his/her math; e.g. there are missing differentials in
    =C[integral]{sin(theta)/x} = C[-cos(theta)/x] cos(theta) = x/[(x^2 +y^2)^(1/2)]

    and anyway the integral evaluated between 0 and a is not what he/she states.
    Again, I recommend tables or wolfram alpha, going straight with his/her
    which rewritten symbolically would be
    dEy = kλ y dy/(x2 + y2)3/2
    integrated from 0 to a.
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