ivancsc1996 2 years ago Anybody can solve this?

1. ivancsc1996

$E _{p}=k _{e}\lambda R (x _{p}\int\limits\limits_{0}^{2\pi}\frac{ d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}-R \int\limits\limits_{0}^{2\pi}\frac{ \cos \theta d \theta }{ (x _{p}^{2}-2Rx _{p}\cos \theta +R ^{2})^{3/2} })$

2. fabsam14

|dw:1375125061000:dw|

3. ivancsc1996

I think it may have something to do with partial fractions. I need somebody to guide me through it

4. FibonacciChick666

is this physics?

5. ivancsc1996

Yes

6. abb0t

OH YEAH RIGHT! LOL

7. FibonacciChick666

ok so, can I assume that the only variable is theta?

8. ivancsc1996

Yes, everything is constant except theta and cos theta.

9. FibonacciChick666

ok give me a second to think

10. 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

11. 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:$E _{p}=k _{e}\lambda R \int\limits\limits_{0}^{2\pi}\frac{ x_p-Rcos( \theta) d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}$

12. FibonacciChick666

Then long division

13. FibonacciChick666

or complex integration

14. FibonacciChick666

complex would be easier

15. 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.

16. FibonacciChick666

integration by parts would work too you would need two degrees of integration then notice they look the same

17. FibonacciChick666

good luck I'll take a look later gtg to class

18. oldrin.bataku

are you sure you didn't screw up deriving this integral?

19. 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.