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I know it has something to do with electric flux

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

how do I make the electic flux symbol?

Google is your best friend.

I'm looking at a Flux through a Piecewise-continues closed surface

I'll draw it

|dw:1359307997058:dw|

E=+(200 N/C)k
through the region z>0 and by E=-200 N/C k through the region z<0

|dw:1359308120623:dw|

Who does the E on top penetrate the surface and is still parallel to the surface?

ok, I have an integral question though

\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}\]

Oh interesting

ok

Do they explain how to do this ?

ok it starts off with \[dE=dE_rr\] how do I make an r hat?

and the E has an arrow on top of it

|dw:1359310750008:dw|

ok. so now you find E due to dq ?
then integrate over all the dq's in the rod ?

Sorry I'm typing it all...might take me a couple of mins

\[dE_r=\frac{kdq}{r^2}\]
and
\[cos\theta=\frac{-x_s}{r}\]

|dw:1359311776834:dw|

yes, I think I follow it so far.

now we need an equation ?

\[dE_x=\frac{k\;dq}{r^2}cos\theta=\frac{k\;cos\theta\;\lambda dx_s}{r^2}\]

\[tan\theta=\frac{y_p}{|x_s|}=\frac{y_p}{-x_s}\]

so \[x_s=-\frac{y_p}{tan\theta}=-y_pcot\theta\]

\[sin\theta=\frac{y_p}{r},\;\textrm{so}\;r=\frac{y_p}{sin\theta}\]

\[dx_s=-y_p\frac{dcot\theta}{d\theta}=y_pcsc^2\theta d\theta\]

ok here comes a big integral...

that is some kind of ugly

because they keep reusing that formula over and over in the examples that follow

yes, but it seems they could have gone about it in a more straight-forward manner,

don't you mean Ey (not Ez) ?

|dw:1359314142343:dw|

|dw:1359314391724:dw|

sweet jesus...

Yeah this is Calculating E from Coulomb's Law

Jennifer whats the name of your text?

physics for scientist and engineers by tipler and mosca

volume II

how would you have done it phi? I anything simpler than this please...

Ahhhhaaa!!!! yeah that makes sense

I feel ....sad....I wish I understood this better

so is all that hassle to find that resultant vector?

I find this stuff very confusing.

so is all that hassle to find that resultant vector?
yes.

in this equation y is constant (we put our point at 0,y and the rod is at y=0)

oh I see, so x is the length of the rod...

how is y a constant?

why is it plus?

it probably should be minus

no plus seems right...I guess

|dw:1359315838647:dw|
can you show me?

|dw:1359315883222:dw|