Flux in Through/ Out of Caps Purcell

In summary: When you do the appropriate changing of variables, you implicitly change the reference point for the flux integral.
  • #1
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"Flux in Through/ Out of Caps" Purcell

Homework Statement


I'm working problem 5.11 in Purcell's E&M book. It's about relativistic particles, and the eventual condition about angles at which the E-field lines are directed. The ultimate goal is to prove that tan([itex]\varphi[/itex])=[itex]\gamma[/itex]tan([itex]\theta[/itex]).

Homework Equations


The Attempt at a Solution


I have solved the integral for the "inner cap's" flux, and got that Flux1=2[itex]\pi[/itex]Q(1-cos[itex]\theta[/itex]).

The "outer cap" flux is Flux2=2[itex]\pi[/itex]Q(1-[itex]\gamma[/itex]cot([itex]\varphi[/itex]). (This integral might not be correct, though...)

I'm missing a sin([itex]\theta[/itex]) somewhere in the first flux, but I don't know where it could be.
 
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  • #2


Your flux through the inner cap looks correct (there is no missing sinθ)

Your expression through the outer cap does not look correct. You would need to show some details of your calculation to spot the error.
 
  • #3


Ok, for the second integral, they provide the surface area element as 2[itex]\pi[/itex]r[itex]^{2}[/itex]sin([itex]\theta[/itex]). In the text, they also talk about the electric field of a moving charge as having the value:

E=(Q(1-[itex]\beta[/itex][itex]^{2}[/itex]))/(r[itex]^{2}[/itex](1-[itex]\beta[/itex][itex]^{2}[/itex]sin([itex]\varphi[/itex])).

Although now that I think about it, these lines would all be perpendicular to the circle centered at the point where the charge would have been had it kept moving, not centered at where it stopped... so when I tried to integrate this in the surface integral, I wouldn't get the right thing I think.

Anyway, when I did evaluate the integral ∫E[itex]\bullet[/itex]dA, I ended up needing to integrate 2[itex]\pi[/itex]Q(1-[itex]\beta[/itex]2) ∫(sin[itex]\varphi[/itex])/(1-[itex]\beta[/itex]2sin2([itex]\varphi[/itex])) d[itex]\varphi[/itex]

I used the substitution x= -cos([itex]\varphi[/itex]) so that dx= sin([itex]\varphi[/itex]) d[itex]\varphi[/itex], meaning I then had to integrate 2[itex]\pi[/itex]Q(1-[itex]\beta[/itex]2)∫ dx/(1-[itex]\beta[/itex]2+[itex]\beta[/itex]2x2)(3/2) between -1 and -cos([itex]\varphi[/itex])

Doing this, I realize one mistake I made at first, because I end up getting the final answer of 2[itex]\pi[/itex]Q(1-(cos[itex]\varphi[/itex])/sqrt(1-[itex]\beta[/itex][itex]^{2}[/itex]sin2([itex]\varphi[/itex]))).

But when I set these two fluxes equal, I still can't get the tangents in the right places. How do I get the extra factor involving sine-- I feel like it might have to do with my point about the flux being perpendicular to a circle centered at a different point than the "inner cap" circles.

Thanks for the help!
 
  • #4


schaefera said:
I end up getting the final answer of 2[itex]\pi[/itex]Q(1-(cos[itex]\varphi[/itex])/sqrt(1-[itex]\beta[/itex][itex]^{2}[/itex]sin2([itex]\varphi[/itex]))).

Good. I believe that's what you should get.

But when I set these two fluxes equal, I still can't get the tangents in the right places. How do I get the extra factor involving sine-- I feel like it might have to do with my point about the flux being perpendicular to a circle centered at a different point than the "inner cap" circles.

Setting the fluxes equal to one another will yield an expression for cosθ in terms of cos[itex]\varphi[/itex] and sin[itex]\varphi[/itex]. Use this to derive an expression for sinθ in terms of sin[itex]\varphi[/itex]. Then try setting up an expression for tanθ and see if it will simplify.
 
  • #5


Ok thank you! I will work on that. Does that mean we don't need to worry about measuring r from where the charge would have been? Also, why should phi be larger than theta? The picture makes it look like theta is larger unless we measure phi from such a location as where the charge would've been.
 
  • #6


schaefera said:
Does that mean we don't need to worry about measuring r from where the charge would have been?

For the spherical cap associated with phi, r is the distance measured from where the charge would have been if it hadn't stopped at t = 0. It has the same meaning as r' in equation 12.
Also, why should phi be larger than theta? The picture makes it look like theta is larger unless we measure phi from such a location as where the charge would've been.

Phi should be larger than theta for the case being studied (where the charge suddenly stops moving). This is also what figure 5.18 shows. Note that segment CD in the figure makes a greater angle to the x-axis than segment AB.
 
  • #7


Aha, thank you! I thought it would have to be so for the angle to be larger. That means that in doing what the problem calls the appropriate changing of variables from Eq 12 we implicitly moved the origin for that flux integral?

Thank you so much for the help! Much appreciated!
 
  • #8


schaefera said:
That means that in doing what the problem calls the appropriate changing of variables from Eq 12 we implicitly moved the origin for that flux integral?

Yes.
 

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"Flux in Through/ Out of Caps Purcell" is a concept in fluid dynamics that describes the rate of flow through a specific area or region. It is often used in the study of fluid dynamics to understand the movement of fluids through different structures or systems.

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"Flux in Through/ Out of Caps Purcell" is an important concept in scientific research because it allows scientists to understand and quantify the movement of fluids in various systems. This understanding can be applied to a wide range of fields, from engineering to biology, and can help improve the design and functionality of various systems and structures.

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