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ivancsc1996

  • one year ago

Anybody can solve this?

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  1. ivancsc1996
    • one year ago
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    \[E _{p}=k _{e}\lambda R (x _{p}\int\limits\limits_{0}^{2\pi}\frac{ d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}-R \int\limits\limits_{0}^{2\pi}\frac{ \cos \theta d \theta }{ (x _{p}^{2}-2Rx _{p}\cos \theta +R ^{2})^{3/2} })\]

  2. fabsam14
    • one year ago
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    |dw:1375125061000:dw|

  3. ivancsc1996
    • one year ago
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    I think it may have something to do with partial fractions. I need somebody to guide me through it

  4. FibonacciChick666
    • one year ago
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    is this physics?

  5. ivancsc1996
    • one year ago
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    Yes

  6. abb0t
    • one year ago
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    OH YEAH RIGHT! LOL

  7. FibonacciChick666
    • one year ago
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    ok so, can I assume that the only variable is theta?

  8. ivancsc1996
    • one year ago
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    Yes, everything is constant except theta and cos theta.

  9. FibonacciChick666
    • one year ago
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    ok give me a second to think

  10. ivancsc1996
    • one year ago
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    Take your time, but please don't just give me the answer. Guide me through it. It is not a homework assignment or anything. Just part of my afternoon physics thinking

  11. FibonacciChick666
    • one year ago
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    \[E _{p}=k _{e}\lambda R (x _{p}\int\limits\limits_{0}^{2\pi}\frac{ d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}-R \int\limits\limits_{0}^{2\pi}\frac{ \cos \theta d \theta }{ (x _{p}^{2}-2Rx _{p}\cos \theta +R ^{2})^{3/2} })\] We can combine the two integrals:\[ E _{p}=k _{e}\lambda R \int\limits\limits_{0}^{2\pi}\frac{ x_p-Rcos( \theta) d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}\]

  12. FibonacciChick666
    • one year ago
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    Then long division

  13. FibonacciChick666
    • one year ago
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    or complex integration

  14. FibonacciChick666
    • one year ago
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    complex would be easier

  15. ivancsc1996
    • one year ago
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    I don't know complex integration. I'm only about to start my highschool senior year. I will give it a shot at long division.

  16. FibonacciChick666
    • one year ago
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    integration by parts would work too you would need two degrees of integration then notice they look the same

  17. FibonacciChick666
    • one year ago
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    good luck I'll take a look later gtg to class

  18. oldrin.bataku
    • one year ago
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    are you sure you didn't screw up deriving this integral?

  19. ivancsc1996
    • one year ago
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    Jajaja. I don't know. Take a look at it: I have a ring with uniform charge Q. Calculate the electric field at any point around the x-axis. |dw:1375329903540:dw| Well, I did it this way: \[\lceil dE _{p} \rceil=k _{e }\frac{ dQ }{ r ^{3} }\lceil r \rceil=k _{e }\frac{ dQ }{ r ^{3} }(\lceil y \rceil+\lceil x _{p}-R \cos \theta \rceil) \]\[dQ=\lambda ds= \lambda R d \theta\]One we integrate, the side with the y vector, will cancel because of symmetry, so in magnitude:\[E _{p}=\int\limits\limits_{0}^{2 \pi} k _{e}\frac{ \lambda R }{ r ^{3} }(x _{p}-R \cos \theta)d \theta=\int\limits\limits_{0}^{2 \pi} k _{e}\frac{ \lambda R }{ ((x _{p}-R \cos \theta)^{2} +R ^{2}\sin ^{2}\theta)^{3/2} }(x _{p}-R \cos \theta)d \theta \]The ceils just represent vetors.

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