## hosein 3 years ago find electric field in origin with below density function:

1. hosein

|dw:1329341783470:dw|$\lambda=a \cos \theta$ figure is semicircular with radius R & "a" is constant. theta is angle with x-axis

2. TuringTest

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

3. TuringTest

if you want proof tell me which it is: is the charge the same sign on both sides or not?

4. hosein

yes turing cahrge is the same, & independent of the sign of cosine

5. TuringTest

in that case...

6. TuringTest

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

7. hosein

i got that answer is $a/4R \epsilon _{0}$ right?

8. TuringTest

I think you dropped the pi...

9. hosein

i wanted to know am i right or not, i'm agree with you but test choice's was mistaken

10. TuringTest

I got$E=\frac{a}{4\pi\epsilon_0R}$

11. hosein

i got that too .thanx for your help

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