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Homework Help: Electric Field and curvature

  1. Mar 12, 2006 #1
    Hey guys was wondering if anyone could help me out :)

    Question))A Positive Charge Q is uniformly distributed around a semicircle of radius a. Find the electric field(magnitude and direction) at the center of curvature P.

    Basically it looks like a unit circle except the radius is A and we need to calculate the electric field at point P which is at the origin.

    This is the approach I took but it is wrong.


    And when I integrated it, I came to 2(pi)(k)(Q)/a^2 for the electric field

    I am not sure...

    The right answer is 2(k)(Q)/((pi)(a^2))

    Can anyone help me ??

    Thanks alot everyone
  2. jcsd
  3. Mar 12, 2006 #2


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    each little piece of charge contributes to the VECTOR E-field.
    by symmetry, only the negative y-component is uncancelled.
    (I'm thinking of this as the top half-ring from +x, thru +y, to -x )

    The y contribution is dE_y = k dQ cos(theta) / a^3 .
    all charge is distance "a" from the origin, correct?

    How much dQ is on length dL? well, Q / a pi = dQ / dL .

    get back if this is not detailed enough
  4. Mar 12, 2006 #3
    This is correct

    Confused from here,

    I thought that dQ would be dQ= 2(pi)(a)da?

    From what you are saying, substituting for dQ I would get dE_y=(k)(Q/((a)(pi))/(a^3)
    which would simplify down to dE_y=kQ/(a^4*pi)??

    I'm not too sure aobut dL and how you got the relationship between dQ followed by how I can apply it to my problem. I do realize that the dE_x will be zero and it will be a downward.

    Thanks any extra help is very appreciated
    Last edited: Mar 12, 2006
  5. Mar 12, 2006 #4


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    the radius of the ring is constant, there IS no variation of "a".
    the source charge Q is spread along a line of length = circumference/2 = pi a.
    If you integrate the dQ along this line, you have to get the entire Q back.

    You eventually will integrate E_y along that line (or replace dL = a d(theta))
    But since E is a vector, you canNOT ignore the a cos(theta) ...
    oh, I'm measuring theta from the y-axis, from theta = pi/2 to theta = - pi/2
    use sin(theta) if measuring from the x-axis from zero to pi.

    OOPS! I see that I had a typo in my first post.
    dE_y = k dQ a cos(theta)/a^3 ... sorry !

    Roughly, we expect E = E_y = kQ/a^2 , except that some of it cancels.
    This integration business is to find out if half is cancelled, or 29% ...
    Last edited: Mar 12, 2006
  6. Mar 12, 2006 #5
    Since theta changes throughout the whole process it eventually reaches zero again?

    kQ/a^2 *Integrate(sin(theta)dO from 0 to pi) which comes to -1+1 = 0 ?

    E_y=kQ/a^2 is what I get if you ignore cos(theta) which from what you said should not happen. The circumference which you mentioned should be the length so E=kQ/(pi*a)^2?

    Sorry been at this one for a couple of hours and getting lost as to how the result is 2kQ/(pi(a^2))

    The way what you and I calculated differs is by (2/pi)
  7. Mar 12, 2006 #6


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    No, sin(theta) is positive the entire way from 0 to pi , so it canNOT cancel itself.

    integral [ sin(theta) d(theta) ] = cos(theta) evaluated =
    = [cos(pi) - cos(0) ] = [ -1 - (+1) ] = - 2 .

    half-circumference is a length of the line that the charge is spread along ...
    The E-field denominator is the DISTANCE from the source charge to the
    place that you're finding the field at (called the "field point"). Not pi a .
  8. Mar 12, 2006 #7
    With that you would still come to -2kQ/a^2...

    the negative denotes that it is downard but how do you utilize pi 2 here?

    Thanks again :)
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