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find electric field in origin with below density function:

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|dw:1329341783470:dw|\[\lambda=a \cos \theta\] figure is semicircular with radius R & "a" is constant. theta is angle with x-axis
if the charge changes with the sign of cosx the field is 0 at the origin by symmetry if the sign of the charge is the same on each half of the semicircle, then you can get a non-zero answer
if you want proof tell me which it is: is the charge the same sign on both sides or not?

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Other answers:

yes turing cahrge is the same, & independent of the sign of cosine
in that case...
charge is charge density times distance\[dq=\lambda ds=aR\cos\theta d\theta\]|dw:1330208182087:dw|by symmetry the x-components if the field will cancel, so we need only consider the y-component.\[dE_y=\frac1{4\pi\epsilon_0}\frac {dq}{R^2}\sin\theta=\frac1{4\pi\epsilon_0}\frac {a R\cos\theta }{R^2}\sin\theta d\theta\]\[E_y=\frac a{4\pi\epsilon_0R}\cdot 2\int_{0}^{\frac\pi2} \cos\theta\sin\theta d\theta=\frac a{2\pi\epsilon_0R}\int_{0}^{\frac\pi2} \cos\theta\sin\theta d\theta\]notice that we double the integral up to pi/2 to utilize more symmetry
i got that answer is \[a/4R \epsilon _{0}\] right?
I think you dropped the pi...
i wanted to know am i right or not, i'm agree with you but test choice's was mistaken
I got\[E=\frac{a}{4\pi\epsilon_0R}\]
i got that too .thanx for your help

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