- anonymous

Flux through a piecewise-continuous closed surface

- jamiebookeater

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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

- anonymous

- 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