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

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

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

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

abb0t Group TitleBest ResponseYou've already chosen the best response.0
OH YEAH RIGHT! LOL
 one year ago

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

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

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

ivancsc1996 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

FibonacciChick666 Group TitleBest 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:\[ E _{p}=k _{e}\lambda R \int\limits\limits_{0}^{2\pi}\frac{ x_pRcos( \theta) d \theta }{ (x _{p}^{2}2Rx _{p} \cos \theta +R ^{2})^{3/2}}\]
 one year ago

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

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

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

ivancsc1996 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

FibonacciChick666 Group TitleBest ResponseYou've already chosen the best response.1
integration by parts would work too you would need two degrees of integration then notice they look the same
 one year ago

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

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

ivancsc1996 Group TitleBest ResponseYou've already chosen the best response.1
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 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.
 one year ago
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