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Electric field calculation

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

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

    http://img132.imageshack.us/img132/8278/electricfield1.png [Broken]

    2. Relevant equations

    Electric field of point charge:

    3. The attempt at a solution

    Since I suspect the linear charge distribution [tex]\lambda=\frac{Q}{\pi a}[/tex] will be troublesome to wirk with here, I will define the angular charge distribution [tex]\alpha=\frac{Q}{\pi}[/tex], divide the semicircle into infinitisimal pieces, calculate the electric field for each one and integrate.

    Since each piece is a distance [tex]a[/tex] away from point [tex]P[/tex], the electric field caused by each piece is [tex]dE=k\frac{dQ}{a^2}[/tex]. [tex]dQ=\alpha d\theta=\frac{Q d\theta}{\pi}[/tex], so [tex]dE=k\frac{Qd\theta}{a^2\pi}[/tex].

    By symmetry, the total electric field in the x-direction is zero, so I only have to find the field in the y-direction, which is given by [tex]dE_y=dE\cdot \sin \theta=\frac{kQ}{a^2\pi}\sin\theta d\theta[/tex].

    Integrating from [tex]\theta=0[/tex] to [tex]\theta=\pi[/tex], the result is
    [tex]E_y=\frac{kQ}{a^2\pi}\int_0^\pi \sin\theta d\theta=\frac{kQ}{a^2\pi}\left[-\cos\theta\right]_0^\pi=\frac{2kQ}{a^2\pi}[/tex].

    Since Q is positive, the field is in the negative y-direction.

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 16, 2009 #2


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    Homework Helper

    What you've done looks perfect to me... any particular reason for asking about it?
  4. Jun 16, 2009 #3
    I've done this sort of problem a number of times with straight lines, and the book did not have an answer key for this paticular problem. I was just wondering if I had the correct answer.
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