1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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 :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electric Field and curvature
  1. Electric Fields (Replies: 4)