ivancsc1996
  • ivancsc1996
Anybody can solve this?
Differential Equations
jamiebookeater
  • jamiebookeater
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ivancsc1996
  • ivancsc1996
\[E _{p}=k _{e}\lambda R (x _{p}\int\limits\limits_{0}^{2\pi}\frac{ d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}-R \int\limits\limits_{0}^{2\pi}\frac{ \cos \theta d \theta }{ (x _{p}^{2}-2Rx _{p}\cos \theta +R ^{2})^{3/2} })\]
anonymous
  • anonymous
|dw:1375125061000:dw|
ivancsc1996
  • ivancsc1996
I think it may have something to do with partial fractions. I need somebody to guide me through it

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FibonacciChick666
  • FibonacciChick666
is this physics?
ivancsc1996
  • ivancsc1996
Yes
abb0t
  • abb0t
OH YEAH RIGHT! LOL
FibonacciChick666
  • FibonacciChick666
ok so, can I assume that the only variable is theta?
ivancsc1996
  • ivancsc1996
Yes, everything is constant except theta and cos theta.
FibonacciChick666
  • FibonacciChick666
ok give me a second to think
ivancsc1996
  • ivancsc1996
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
FibonacciChick666
  • FibonacciChick666
\[E _{p}=k _{e}\lambda R (x _{p}\int\limits\limits_{0}^{2\pi}\frac{ d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}-R \int\limits\limits_{0}^{2\pi}\frac{ \cos \theta d \theta }{ (x _{p}^{2}-2Rx _{p}\cos \theta +R ^{2})^{3/2} })\] We can combine the two integrals:\[ E _{p}=k _{e}\lambda R \int\limits\limits_{0}^{2\pi}\frac{ x_p-Rcos( \theta) d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}\]
FibonacciChick666
  • FibonacciChick666
Then long division
FibonacciChick666
  • FibonacciChick666
or complex integration
FibonacciChick666
  • FibonacciChick666
complex would be easier
ivancsc1996
  • ivancsc1996
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.
FibonacciChick666
  • FibonacciChick666
integration by parts would work too you would need two degrees of integration then notice they look the same
FibonacciChick666
  • FibonacciChick666
good luck I'll take a look later gtg to class
anonymous
  • anonymous
are you sure you didn't screw up deriving this integral?
ivancsc1996
  • ivancsc1996
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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