# The Electric Field Produced by a Finite Charged Wire XD

1. Jul 27, 2009

### vorcil

http://img22.imageshack.us/img22/1958/physicsl.jpg [Broken]

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

Last edited by a moderator: May 4, 2017
2. Jul 27, 2009

### vorcil

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

3. Jul 27, 2009

### vorcil

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

4. Jul 27, 2009

### 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.

5. Jul 27, 2009

### vorcil

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

-

6. Jul 27, 2009

### Cyosis

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

7. Jul 27, 2009

### vorcil

k=1/4piEo
lambda=q/L

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

8. Jul 27, 2009

### 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.

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