The Electric Field Produced by a Finite Charged Wire XD

[vorcil]

http://img22.imageshack.us/img22/1958/physicsl.jpg

My attempt,
After all the integrals I've got the final equation

(1/4piE0) * |q| / ( d * sqrt ( d^2 + (L/2)^2) )

i'm not too sure how to express it how the question asks,

please help!

the bit I'm confused on, is the ( d * sqrt ( d^2 + (L/2)^2) )

can it also be seen as (d * d) + (l/2)
or is it d * (d+(l/2))

bah

 

[vorcil]

d * (sqrt ( d^2 + (l/2)^2) ) = d*(L/2)?

 

[vorcil]

d * (sqrt ( d^2 + (l/2)^2) ) = d*(L/2)?

no because that's the thing if it becomes infinitely long and I'm dealing with a finite length

 

[Cyosis]

Not sure what you're trying to do here. The question simply asks you to give the answer in terms of L, lambda, d and k. You're missing a k and a lambda, use the definition of k and lambda to get them into your expression.

 

[vorcil]

Not sure what you're trying to do here. The question simply asks you to give the answer in terms of L, lambda, d and k. You're missing a k and a lambda, use the definition of k and lambda to get them into your expression.

yes I am not a total retard as i was able to make the integral to the final equation.
anyway lol, I just don't know how to convert the

|q| / ( d * sqrt ( d^2 + (L/2)^2) )


part of the equation, in terms of lambda and d

-

 

[Cyosis]

What is the definition of k and what is the definition of lambda?

 

[vorcil]

What is the definition of k and what is the definition of lambda?

k=1/4piEo
lambda=q/L

i don't know how to make lambda from q / d*sqrt(d^2+(L/2)^2)

 

[Cyosis]

So if \lambda=q/l then q=...? Note that the l you're using here is not the same l as in your problem, but a general symbol for length. Perhaps it is wise to show us your integration.