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ivancsc1996
 3 years ago
Anybody can solve this?
ivancsc1996
 3 years ago
Anybody can solve this?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1375125061000:dw

ivancsc1996
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1is this physics?

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

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

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

ivancsc1996
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1Then long division

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

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

ivancsc1996
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1good luck I'll take a look later gtg to class

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you sure you didn't screw up deriving this integral?

ivancsc1996
 3 years 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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