## JenniferSmart1 Flux through a piecewise-continuous closed surface one year ago one year ago

1. JenniferSmart1

I know it has something to do with electric flux

2. geerky42
3. JenniferSmart1

how do I make the electic flux symbol?

4. geerky42

5. JenniferSmart1

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

6. JenniferSmart1

I'll draw it

7. JenniferSmart1

|dw:1359307997058:dw|

8. JenniferSmart1

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

9. JenniferSmart1

|dw:1359308120623:dw|

10. JenniferSmart1

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

11. JenniferSmart1

@phi

12. JenniferSmart1

@hartnn

13. JenniferSmart1

ok, I have an integral question though

14. JenniferSmart1

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

15. 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

16. JenniferSmart1

Oh interesting

17. JenniferSmart1

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?

18. phi

ok

19. JenniferSmart1

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

20. phi

Do they explain how to do this ?

21. JenniferSmart1

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

22. JenniferSmart1

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

23. JenniferSmart1

and the E has an arrow on top of it

24. JenniferSmart1

|dw:1359310750008:dw|

25. phi

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

26. JenniferSmart1

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

27. JenniferSmart1

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

28. JenniferSmart1

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

29. JenniferSmart1

|dw:1359311776834:dw|

30. phi

yes, I think I follow it so far.

31. phi

now we need an equation ?

32. JenniferSmart1

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

33. JenniferSmart1

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

34. JenniferSmart1

$tan\theta=\frac{y_p}{|x_s|}=\frac{y_p}{-x_s}$

35. JenniferSmart1

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

36. JenniferSmart1

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

37. JenniferSmart1

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

38. JenniferSmart1

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

39. JenniferSmart1

ok here comes a big integral...

40. JenniferSmart1

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

41. JenniferSmart1

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

42. phi

that is some kind of ugly

43. JenniferSmart1

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

44. JenniferSmart1

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

45. phi

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

46. JenniferSmart1

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 $E_R=-\frac{k\lambda}{R}(cos\theta_2-cos\theta_1)=-k\lambda\left(\frac{cot\theta_2}{r_2}-\frac{cat\theta_1}{r_1}\right)$

47. JenniferSmart1

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

48. phi

don't you mean Ey (not Ez) ?

49. JenniferSmart1

|dw:1359314142343:dw|

50. JenniferSmart1

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