A community for students.
Here's the question you clicked on:
 0 viewing
ivancsc1996
 2 years ago
Anybody can solve this?
ivancsc1996
 2 years ago
Anybody can solve this?

This Question is Closed

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

ivancsc1996
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1is this physics?

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

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

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

ivancsc1996
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1Then long division

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

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

ivancsc1996
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1good luck I'll take a look later gtg to class

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

ivancsc1996
 2 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.