Flux through a piecewise-continuous closed surface

- anonymous

Flux through a piecewise-continuous closed surface

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

I know it has something to do with electric flux

- geerky42

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

- anonymous

how do I make the electic flux symbol?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- geerky42

Google is your best friend.

- anonymous

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

- anonymous

I'll draw it

- anonymous

|dw:1359307997058:dw|

- anonymous

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

- anonymous

|dw:1359308120623:dw|

- anonymous

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

- anonymous

@phi

- anonymous

@hartnn

- anonymous

ok, I have an integral question though

- anonymous

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

- phi

The flux is the part of E normal to the surface. So to answer your question
how does the E on top penetrate the surface and is still parallel to the surface? It doesn't
the flux through the sides is 0

- anonymous

Oh interesting

- anonymous

can you help me understand an example in my book. It's a different problem. It's about electric field due to a line charge of finite length. It doesn't involve gauss...do you have a min?

- phi

ok

- anonymous

A thin rod of length L and charge Q is uniformly charged, so it has a linear density of \[\lambda=\frac Ql\] Find the electric field at point P, where P is an arbitrarily positioned point

- phi

Do they explain how to do this ?

- anonymous

Yes, they draw it out and everything. I can write their explanation and we can talk about it. I kinda get it, but I need to be able to reproduce this on my own

- anonymous

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

- anonymous

and the E has an arrow on top of it

- anonymous

|dw:1359310750008:dw|

- phi

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

- anonymous

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

- anonymous

\[\vec{dE}=dE_r \hat{r}\]
\[dE_x=dE_r\hat r \cdot \hat i=dE_rcos\theta\]
\[dE_y=dE_r\hat r \cdot \hat j=dE_rsin\theta\]

- anonymous

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

- anonymous

|dw:1359311776834:dw|

- phi

yes, I think I follow it so far.

- phi

now we need an equation ?

- anonymous

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

- anonymous

\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}=k\lambda\int_{x_1}^{x_2}\frac{cos\theta \;dx_s}{r^2}\]

- anonymous

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

- anonymous

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

- anonymous

I'm beyond LOST...but it continues...I'll just continue typing because they have another integral coming up...

- anonymous

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

- anonymous

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

- anonymous

ok here comes a big integral...

- anonymous

\[\int_{x_1}^{x_2}\frac{cos\theta dx_s}{r^2}=\int_{\theta_1}^{\theta_2}\frac{cos\theta y_p csc^2\theta d\theta}{y^2p/sin^2\theta}=\frac1{y_p}\int_{\theta_1}^{\theta_2}cos\theta d\theta\]

- anonymous

\[E_x=k\lambda\frac{1}{y_p}\int_{\theta_1}^{\theta_2} cos\theta d\theta=\frac{k\lambda}{y_p}(sin\theta_2 - sin\theta_1)=\frac{k\lambda}{y_p}\left(\frac{y_p}{r_2}-\frac{y_p}{r_1}\right)=k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]

- phi

that is some kind of ugly

- anonymous

yep ...tell me about it. I guess the most important thing is the equation at the end
\[k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]

- anonymous

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

- phi

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

- anonymous

Tell me about it!!!!!
this is what they wrote
\[E_z=\frac{k\lambda}{R}(sin\theta_2-sin\theta_1)=k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]
and
\[E_R=-\frac{k\lambda}{R}(cos\theta_2-cos\theta_1)=-k\lambda\left(\frac{cot\theta_2}{r_2}-\frac{cat\theta_1}{r_1}\right)\]

- anonymous

they have a drawing to go with the above statement, but I don't feel like drawing it. It's similar to the drawing I drew earlier, but it has z and R axis I guess

- phi

don't you mean Ey (not Ez) ?

- anonymous

|dw:1359314142343:dw|

- anonymous

|dw:1359314391724:dw|

- anonymous

sweet jesus...

- anonymous

Yeah this is Calculating E from Coulomb's Law

- phi

For a single charge, the E field at some point is q (the charge)/ r^2 (the distance squared between the charge and the point, all times a unit vector that points from the charge to the point
(scaled by a factor)
\[ E= k \frac{q}{r^2} \mathbf{e} \]
For more than one charge you have to integrate the contributions from all the charges.

- phi

The hard part is setting up the integral. Instead of keeping it a function of x (which is what I would try), they made everything a function of theta

- anonymous

What I have a hard time picturing is the fact that it causes a single arrow electric field. I would expect it go look like this|dw:1359314691161:dw|
I don't know how to explain it...I would expect several arrows

- anonymous

Jennifer whats the name of your text?

- anonymous

physics for scientist and engineers by tipler and mosca

- anonymous

volume II

- anonymous

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

- phi

What I have a hard time picturing is the fact that it causes a single arrow electric field. I would expect it go look like this
all the individual contributions to E could be thought of like that. But they sum up to produce one vector. Just add up all the vectors to get a resultant vector

- anonymous

Ahhhhaaa!!!! yeah that makes sense

- anonymous

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

- anonymous

so is all that hassle to find that resultant vector?

- phi

I find this stuff very confusing.

- phi

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

- phi

If we change the single charge equation
\[E= k \frac{q}{r^2} \mathbf{e}\]
to an integral over all charges
\[E_x=\int\limits_{x_1}^{x_2}k\lambda \frac{x dx}{(x^2+y^2)^{\frac{3}{2}}}\]

- phi

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

- anonymous

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

- anonymous

how is y a constant?

- phi

the rod is at y=0 and the point we are interested in does not move. it is always at (0,y)
as we integrate along the rod, r^2 changes (that is in x^2+y^2),

- anonymous

why is it plus?

- phi

I skipped a step. the vector from the charge at (x,0) pointing at our point is (-x,y)
to make it unit length divide by sqrt(x^2 +y^2)
that makes the denominator (originally r^2 or x^2+y^2 become (x^2+y^2)^(3/2)

- phi

it probably should be minus

- anonymous

no plus seems right...I guess

- phi

the direction vector e pointing from the charge to our point is (-x,y)
the x component of E is the x component of the e vector times all the rest of the stuff,
so we do want -x dx rather than x dx (if that is what you are asking )

- anonymous

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

- phi

|dw:1359315883222:dw|