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hosein Group Title

find electric field in origin with below density function:

  • 2 years ago
  • 2 years ago

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  1. hosein Group Title
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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

    • 2 years ago
  2. TuringTest Group Title
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    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

    • 2 years ago
  3. TuringTest Group Title
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    if you want proof tell me which it is: is the charge the same sign on both sides or not?

    • 2 years ago
  4. hosein Group Title
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    yes turing cahrge is the same, & independent of the sign of cosine

    • 2 years ago
  5. TuringTest Group Title
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    in that case...

    • 2 years ago
  6. TuringTest Group Title
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    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

    • 2 years ago
  7. hosein Group Title
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    i got that answer is \[a/4R \epsilon _{0}\] right?

    • 2 years ago
  8. TuringTest Group Title
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    I think you dropped the pi...

    • 2 years ago
  9. hosein Group Title
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    i wanted to know am i right or not, i'm agree with you but test choice's was mistaken

    • 2 years ago
  10. TuringTest Group Title
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    I got\[E=\frac{a}{4\pi\epsilon_0R}\]

    • 2 years ago
  11. hosein Group Title
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    i got that too .thanx for your help

    • 2 years ago
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