Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

hosein

  • 4 years ago

find electric field in origin with below density function:

  • This Question is Closed
  1. hosein
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  2. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  4. hosein
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  5. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    in that case...

  6. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  8. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    I think you dropped the pi...

  9. hosein
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  10. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  11. hosein
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i got that too .thanx for your help

  12. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy