Electrostatic Potential problem, help!

[mooeypoo]

Hey guys,

 

I have a hw question I'm completely stuck with.. I know which formula I am supposed to use, and I know how, in theory, this should be solved, but I'm having trouble seeting up my vectors.

 

This is a problem I'm constantly having.. I know how to generally solve questions like these, as long as the vectors are 'easy' (on the origin, or stuff like that). The moment the vectors are a bit more complicated, I confuse myself to oblivion.

 

So.. help me out, please, and if you can, explain *how* you do this, so I can finally get rid of my uberconfusion over these type of questions...

 

 

A line charge [math]\lambda[/math] per unit length extends along the negative z-axis from the origin to z=-L.

Find the electrostatic potential V(x,y,z) at a general field point above the xy-plane.

 

So, I know that the electrostatic potential can be found by:

 

[math]V® = \left( \frac{1}{4\pi \epsilon } \right) \int \frac{\rho ®}{r} d\tau [/math]

 

And I know that:

 

[math]

\rho ® = \lambda dx

[/math]

 

But.. but... my r. meh. I'm getting confused setting up the vector r (distance from the arbitrary point to the line charge).

 

Help me... don't solve the equation, please, just help me see how to set up the r vector, so I can continue on my own.

 

Thanks!

 

~moo

[Klaynos]

r = ai + bj + ck

 

So you want to integrate over the vector each part dr (=dzk) between 0 and L

[timo]

So, I know that the electrostatic potential can be found by:

 

[math]V® = \left( \frac{1}{4\pi \epsilon } \right) \int \frac{\rho ®}{r} d\tau [/math]

[math] V(\vec r) = C \int_{\vec x \in V} dV \frac{\rho(\vec x)}{\| \vec r - \vec x \|} [/math]. Your two rs in the integrand refer to different locations (the position of the generating charge and the distance from the position). The equation but not be sensible, otherwise. Of course this relatively general expression simplifies to a 1D integral in your case; I did not rewrite it that way because you might want to figure it out yourself (assuming the notation in your integral was the problem).

 

EDIT: Oh, sorry: Your expression can make sense if you shift your coordinate system appropriately, i.e. if you look at V(\vec 0) for a charge distribution in a line from (-x,-y,-z) to (-x,-y-L,-z) (not exactly sure about the coordinates; due to my currently crappy INet connection I cannot see the original problem).

Edited October 12, 2008 by timo

[mooeypoo]

all hail overcomplicating myself.

 

Help, seriously, I'm sure I'm stressing out over nothing and hence doing things wrong for just confusion, but.. I'm trying..

 

[math]\overrightarrow{x} = a\hat{x}+b\hat{y}+c\hat{z}[/math]

and

[math]\overrightarrow{r} = -z\hat{z}[/math]

 

So.. uhm..

 

[math]

|r-x|=\sqrt{a^2+b^2+c^2-z^2}

[/math]

 

Which would make my integral:

 

[math]

V® = \left( \frac{1}{4\pi \epsilon } \right) \int \frac{\rho ®}{r} d\tau =

\left( \frac{1}{4\pi \epsilon } \right) \int_{-L}^{0}\frac{\lambda dz}{\sqrt{a^2+b^2+c^2-z^2}}

[/math]

 

Is that right??

[timo]

- the TeX code for vectors is \vec, e.g. \vec a -> [math]\vec a[/math].

- Using your expressions for [math]\vec x[/math] and [math]\vec r[/math] then [math]\| \vec r - \vec x \| = \sqrt{ (0-a)^2 + (0-b)^2 + (-z -c)^2} = \sqrt{ a^2 + b^2 + (z + c)^2}[/math]. Not the term you wrote.

- I'd recommend being consistent with signs and use [math] \vec r = +z \hat z[/math] (this will alter the expression for [math]\| \vec r - \vec x \|[/math]!). Your integration range (from -L to 0) indicates that you accidentially switched to it somewhere along the way, anyways.

- The [math]\frac{\rho ®}{r} [/math] integrand still looks strange to me, to say the least. What is r supposed to stand for (in terms of physics)? Is it an R³ vector or a real value?

 

I think your confusion might stem from not having completely understood the expression for the electrostatic potential.

[mooeypoo]

- the TeX code for vectors is \vec, e.g. \vec a -> [math]\vec a[/math].

Thanks, I was looking for that..

- Using your expressions for [math]\vec x[/math] and [math]\vec r[/math] then [math]\| \vec r - \vec x \| = \sqrt{ (0-a)^2 + (0-b)^2 + (-z -c)^2} = \sqrt{ a^2 + b^2 + (z + c)^2}[/math]. Not the term you wrote.

Egh, yeah, I see .. I confused myself too much, and neglected the simple fact that first you add/subtract first to get the distance vector... okay.. i think I got that one now.

- I'd recommend being consistent with signs and use [math] \vec r = +z \hat z[/math] (this will alter the expression for [math]\| \vec r - \vec x \|[/math]!). Your integration range (from -L to 0) indicates that you accidentially switched to it somewhere along the way, anyways.

I didn't think about that (that I do the minus in the integral), that's a good point.. okay.

 

- The [math]\frac{\rho ®}{r} [/math] integrand still looks strange to me, to say the least. What is r supposed to stand for (in terms of physics)? Is it an R³ vector or a real value?

 

I think your confusion might stem from not having completely understood the expression for the electrostatic potential.

I agree. The book has the worst notation EVER, using variable squiggly forms of "r" and the professor is writing on the board with even WEIRDER notation, along with the fact he's insisting on using cursive (which I am having huuuuge difficulties with).

 

So I am left with trying to figure this out on my own, hence my confusion. Please bear with me here, I'm sure that you're right - when I understand exactly what the integral means, I will have no more problems.

 

I thought I did, though so.. I am not sure now. The book is using squigglies and bolded version of r to differentiate between r the vector, r the distance and |r-r0|. I'm desperately trying to make sense of those but they're REALLY confusing.

 

As far as I understood,

 

[math]V® = \left( \frac{1}{4\pi \epsilon } \right) \int \frac{\rho (r')}{r} d\tau ' [/math]

 

(the notation r' is, in the book, bolded, and written exactly like that - with the 'prime' notation. so is the d\tau)

 

r is the distance to the source (the magnitude of the vector \vec{r} to the source point).

r' is the vector of the point or points of integration.

d\tau is integration in all space. In my case, it's a line integral.

 

So.. putting it this way,

 

[math]r= \sqrt{a^2+b^2+(c-z)^2}[/math]

 

[math]r'= z\hat{k}[/math]

 

[math]\rho (r') = \lambda dz [/math]

 

And so the integral should be:

 

[math]\frac{1}{4\pi \epsilon} \int_{-L}^{0} \frac{\lambda}{\sqrt{a^2+b^2+(c-z)^2}} \hat{z} dz [/math]

 

Does that seem better? I have a feeling I might've missed another z on the top, but I'm not sure, since it seems that my rho is set up correctly, but i"m ... not sure anymore.

[timo]

That is definitely better. A few comments/corrections, though:

- [math] \lambda \, dz[/math] is not a charge density. If you consider dz being a small line segment (the typical view physicists take when dealing with differentials) then it is a small charge. Technically, the density would be [math] \rho(x,y,z) = \lambda \delta (x) \delta (y) \Theta(-z) \Theta (L+z)[/math]. If you don't know the delta-distribution and the theta-function then just ignore this comment (it is just a complicated way of saying "zero everywhere except on the line segment from the origin to (0,0,-L) where it equals lambda"). The point is merely about units, anyways.

- Where does the [math] \hat z [/math] come from under the integral? Both, r' and r are mere real values. They should not suddenly become a vector.

[mooeypoo]

That is definitely better. A few comments/corrections, though:

- [math] \lambda \, dz[/math] is not a charge density. If you consider dz being a small line segment (the typical view physicists take when dealing with differentials) then it is a small charge. Technically, the density would be [math] \rho(x,y,z) = \lambda \delta (x) \delta (y) \Theta(-z) \Theta (L+z)[/math]. If you don't know the delta-distribution and the theta-function then just ignore this comment (it is just a complicated way of saying "zero everywhere except on the line segment from the origin to (0,0,-L) where it equals lambda"). The point is merely about units, anyways.

- Where does the [math] \hat z [/math] come from under the integral? Both, r' and r are mere real values. They should not suddenly become a vector.

 

I thought that the z vector notation came from my \rho definition, but I don't think that was right. In any case, I do know dirac delta (errr a bit.. we learned that in some depth a few lessons ago) but I 'simplified' it automatically (the book and the professor seem to do that all the time, I figured it's the way to do it) to apply to a line charge.

 

If we were talking about a volume (like a charged sphere), then my \rho considerations would have been accross all space (I had a few questions like that with \rho function that isn't linear, hence we were supposed to integrate to get the Qenclosed and the Electrical field, etc).

 

I must say I am now completely stuck in the integration... Is this integral even possible to solve? Or am I just letting the bunch of numbers in there confuse me..?

[timo]

To be honest: Yes, it is solvable in the sense that there is an analytic expression for it - I did it with Mathematica . Math handbooks might have it as a standard integral in their tables but I couldn't check that as my math handbook lies in my office. The charge density never is a vector, regardless whether it is charge per volume, per area or per 1D-volume.

[mooeypoo]

Hmm.. I'll have to try again when my head's clear, it's possible I am just overworking myself and missing an obvious.

 

If you don't mind -- could you just verify that my *set up* is good? That's all I'm concerned with, since those are my problems. If I set it up well, for sure, then I think I understand and will be able to do it again.

 

I just need to practice now, and I hope I am not perpetuating a mistake ..

 

Thanks!!!!!!

 

~moo

[timo]

As said: The [math]\hat z[/math] in the integrand is wrong. The rest seems ok.

[mooeypoo]

I had an epiphany. I think...

 

i used a sort of double substitution for this integral, and I hope it's "legal". I don't see anythign wrong with it, since I merely called my constants in a different name, and whatever isn't a constant I did a proper U substition (with differetiating for du, etc).

 

But... just want to make sure.

 

So, I start with:

 

[math]

\frac{1}{4\pi \epsilon} \int_{-L}^{0} \frac{\lambda}{\sqrt{a^2+b^2+(c-z)^2}} dz =

[/math]

 

and then do a double substitution; first:

 

[math]u= c-z[/math]

[math]du=dz[/math]

 

second, just for 'convinience' to see the formula, I renamed my constants:

 

[math]m^2 = a^2+b^2[/math]

 

And so I have:

 

[math]

\frac{\lambda}{4\pi \epsilon} \int_{-L}^{0} \frac{1}{\sqrt{m^2+u^2}} dz =

[/math]

 

Which is a formula (given in our hw, too), and solved like so:

 

 

[math]

- \int \frac{1}{\sqrt{m^2+u^2}} dz = - ln \left(u + \sqrt{m^2+u^2} \right)

[/math]

 

So, replacing back the constant and u-substitution, I have:

 

[math]

= \frac{\lambda}{4\pi \epsilon} \left( -ln(c-z+\sqrt{a^2+b^2+(c-z)^2}) \right) |_{-L}^{0} =

[/math]

 

[math]

= \left( \frac{\lambda}{4\pi\epsilon} \right) \ln \left( \frac{c-z+\sqrt{a^2+b^2+(c+L)^2}}{c-z+\sqrt{a^2+b^2+c^2}} \right)

[/math]

 

Ha!

 

Well, it seems like my substitutions are okay, but.. there's no actual problem in substituting m^2=a^2+b^2 , specifically since I'm not sqrting it or doing anything to it (and hence risking losing signs), right? it's just "calling it" a different name..?

 

 

Anyhoo, I think it's solved fine.. if you spot an error, do tell.

 

Thanks again!!

 

~moo

 

Just an update -- the professor solved this question in class today, and my solution is, indeed, true (including the 'trick' I've used, yay! ).

 

Thanks a lot!

 

~moo

Edited October 14, 2008 by mooeypoo