Question

Anybody can solve this?

Answer 1

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

Answer 2

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

Answer 3

is this physics?

Answer 4

Yes

Answer 5

OH YEAH RIGHT! LOL

Answer 6

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

Answer 7

Yes, everything is constant except theta and cos theta.

Answer 8

ok give me a second to think

Answer 9

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

Answer 10

\[E _{p}=k _{e}\lambda R (x _{p}\int\limits\limits_{0}^{2\pi}\frac{ d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}-R \int\limits\limits_{0}^{2\pi}\frac{ \cos \theta d \theta }{ (x _{p}^{2}-2Rx _{p}\cos \theta +R ^{2})^{3/2} })\]
We can combine the two integrals:\[ E _{p}=k _{e}\lambda R \int\limits\limits_{0}^{2\pi}\frac{ x_p-Rcos( \theta) d \theta }{ (x _{p}^{2}-2Rx _{p} \cos \theta +R ^{2})^{3/2}}\]

Answer 11

Then long division

Answer 12

or complex integration

Answer 13

complex would be easier

Answer 14

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.

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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.


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.