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hosein
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1329341783470:dw\[\lambda=a \cos \theta\] figure is semicircular with radius R & "a" is constant. theta is angle with xaxis

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3if 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 nonzero answer

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3if you want proof tell me which it is: is the charge the same sign on both sides or not?

hosein
 3 years ago
Best ResponseYou've already chosen the best response.1yes turing cahrge is the same, & independent of the sign of cosine

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3charge is charge density times distance\[dq=\lambda ds=aR\cos\theta d\theta\]dw:1330208182087:dwby symmetry the xcomponents if the field will cancel, so we need only consider the ycomponent.\[dE_y=\frac1{4\pi\epsilon_0}\frac {dq}{R^2}\sin\theta=\frac1{4\pi\epsilon_0}\frac {a R\cos\theta }{R^2}\sin\theta d\theta\]\[E_y=\frac a{4\pi\epsilon_0R}\cdot 2\int_{0}^{\frac\pi2} \cos\theta\sin\theta d\theta=\frac a{2\pi\epsilon_0R}\int_{0}^{\frac\pi2} \cos\theta\sin\theta d\theta\]notice that we double the integral up to pi/2 to utilize more symmetry

hosein
 3 years ago
Best ResponseYou've already chosen the best response.1i got that answer is \[a/4R \epsilon _{0}\] right?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3I think you dropped the pi...

hosein
 3 years ago
Best ResponseYou've already chosen the best response.1i wanted to know am i right or not, i'm agree with you but test choice's was mistaken

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.3I got\[E=\frac{a}{4\pi\epsilon_0R}\]

hosein
 3 years ago
Best ResponseYou've already chosen the best response.1i got that too .thanx for your help
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