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espen180
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Homework Statement
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
Homework Equations
Electric field of point charge:
[tex]\vec{E}=k\frac{Q}{r^2}[/tex]
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.
[tex]\vec{E_P}=-\frac{2kQ}{a^2\pi}\hat{j}[/tex]
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