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ivancsc1996
 one year ago
Best ResponseYou've already chosen the best response.1\[E _{p}=k _{e}\lambda R (x _{p}\int\limits\limits_{0}^{2\pi}\frac{ d \theta }{ (x _{p}^{2}2Rx _{p} \cos \theta +R ^{2})^{3/2}}R \int\limits\limits_{0}^{2\pi}\frac{ \cos \theta d \theta }{ (x _{p}^{2}2Rx _{p}\cos \theta +R ^{2})^{3/2} })\]

fabsam14
 one year ago
Best ResponseYou've already chosen the best response.0dw:1375125061000:dw

ivancsc1996
 one year ago
Best ResponseYou've already chosen the best response.1I think it may have something to do with partial fractions. I need somebody to guide me through it

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1is this physics?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1ok so, can I assume that the only variable is theta?

ivancsc1996
 one year ago
Best ResponseYou've already chosen the best response.1Yes, everything is constant except theta and cos theta.

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1ok give me a second to think

ivancsc1996
 one year ago
Best ResponseYou've already chosen the best response.1Take 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
 one year ago
Best ResponseYou've already chosen the best response.1\[E _{p}=k _{e}\lambda R (x _{p}\int\limits\limits_{0}^{2\pi}\frac{ d \theta }{ (x _{p}^{2}2Rx _{p} \cos \theta +R ^{2})^{3/2}}R \int\limits\limits_{0}^{2\pi}\frac{ \cos \theta d \theta }{ (x _{p}^{2}2Rx _{p}\cos \theta +R ^{2})^{3/2} })\] We can combine the two integrals:\[ E _{p}=k _{e}\lambda R \int\limits\limits_{0}^{2\pi}\frac{ x_pRcos( \theta) d \theta }{ (x _{p}^{2}2Rx _{p} \cos \theta +R ^{2})^{3/2}}\]

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1Then long division

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1or complex integration

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1complex would be easier

ivancsc1996
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1integration by parts would work too you would need two degrees of integration then notice they look the same

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1good luck I'll take a look later gtg to class

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0are you sure you didn't screw up deriving this integral?

ivancsc1996
 one year ago
Best ResponseYou've already chosen the best response.1Jajaja. 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 xaxis. 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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