How to Calculate Electric Potential Along the Axis of a Charged Tube?

[maherelharake]

Homework Statement

You need to calculate the electric potential for a thin, long tube with a uniform surface charge density (σ). This tube has a length L and a radius R. Find the electric potential along the tube axis for points outside the tube.

Homework Equations

The Attempt at a Solution

I am having trouble getting this problem. If I am understanding it correctly, we are supposed to calculated V either above or below the tube, as long as it is along the tube's axis. I am not sure how to set up my problem. Should I set the origin of my axes between L/2 and -L/2? I am not sure where to start. Thanks.

 

[Mindscrape]

I would just find E using Gauss's law, usually if a problem specifies that a tube is long then you are free to ignore the ends of the tube (i.e. treat as if it were infinite). Then use V=-∫E*dl

 

[maherelharake]

But it says that it has a length of L, so it isn't infinite (I think)? Also, I think he wants E along the axis of the tube, so wouldn't the ends be the only parts that contribute. This is where I'm confused.

 

[maherelharake]

My teacher told us today in class to calculate V directly using the formula for it. He told me not to use the integral of E dl.

 

[gabbagabbahey]

That seems like a good place to start. I'd use cylindrical coordinates $s$, $\phi$ and $z$, with the z-axis coinciding with the axis of the tube. Your field point is then at $\textbf{r}=z\mathbf{\hat{z}}$ with $|z|>\frac{L}{2}$.

 

[maherelharake]

Ok so I agree with everything you said. I tried to use the equation v=k Integral[(sigma*Pi*R^2)/z]dr. When i convert to cylindrical coords, everything that was an 'r' term, becomes an 's' term. Does it then become a tripe integral where I integrate with respect to S, phi, and z?

 

[gabbagabbahey]

Ok so I agree with everything you said. I tried to use the equation v=k Integral[(sigma*Pi*R^2)/z]dr. When i convert to cylindrical coords, everything that was an 'r' term, becomes an 's' term. Does it then become a tripe integral where I integrate with respect to S, phi, and z?

The general form for a volume charge distribution $\rho(\textbf{r})$ is

$$V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\textbf{r}')d^3r'}{|\textbf{r}-\textbf{r}'|}$$

which is a triple integral (volume integral). In the case of surface charge distributions $\sigma(\textbf{r})$, the integral simplifies to a double integral (surface integral):

$$V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\sigma(\textbf{r}')dA'}{|\textbf{r}-\textbf{r}'|}$$

and for linear charge distributions $\lambda$, the integral simplifies to a single integral (line or path integral):

$$V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\lambda(\textbf{r}')dl'}{|\textbf{r}-\textbf{r}'|}$$


So, for the surface charge distribuion in this problem, you will want to use the double integral form...

 

[maherelharake]

Oh yeah that's right. I keep wanting to use a volume charge, even though the problem gave us a surface charge.

So the denominator of that integrand is just 'z' and dA becomes dtheta dr?

If that is correct, then I'm not really accounting for the non-endcaps part of the cylinder, am I?

 

[gabbagabbahey]

So the denominator of that integrand is just 'z'

No, the field point (the point(s) at which you want to determine the potential) is at $\textbf{r}=z\mathbf{\hat{z}}$, but what about the source points (the points where there is charge that creates the potential)? Where are they? What does that make $\textbf{r}'$ ( their position(s) )? And $|\textbf{r}-\textbf{r}'|$ (the distance from each source point to the field point)?

and dA becomes dtheta dr?

Close, but you're missing something (what are the units of area? what are the units of your expression?)

 

[maherelharake]

Would the source points be (R, Phi, Z) where Z is -L/2<z<L/2?

And would dA become r*dtheta*dr since the r*dr factor give the proper units?

 

[gabbagabbahey]

Would the source points be (R, Phi, Z) where Z is -L/2<z<L/2?

And would dA become r*dtheta*dr since the r*dr factor give the proper units?

Right, in unit vector notation, that means $\textbf{r}'=R\mathbf{\hat{s}}+z'\mathbf{\hat{z}}$ (where I use $s$ as the radial coordinate, so you don't confuse it with the distance from the origin $r=|\textbf{r}|$ )...if you are wondering where the $\phi$ dependence is, just draw a picture...you will see that the direction of $\mathbf{\hat{s}}$ depends on $\phi$.

So, what does that make $|\textbf{r}-\textbf{r}'|$? What does that make your integral?

 

[maherelharake]

Ok before I start the integral, I want to make sure that value for the denominator is correct. Would that value become (z-z', -Phi, -R)?
I tried to draw a picture to better visualize this, but I think the Phi is still throwing me off a little bit.

 

[gabbagabbahey]

Ok before I start the integral, I want to make sure that value for the denominator is correct. Would that value become (z-z', -Phi, -R)?
I tried to draw a picture to better visualize this, but I think the Phi is still throwing me off a little bit.

If the phi is throwing you off, Google "position vector in cylindrical coordinates", you'll probably find some useful sites.

I'm not sure what you mean by the ordered triplet (z-z', -Phi, -R)? An ordered triplet usually represents a vector, and in cylindrical coordinates, it is usually in the order $(s,\phi,z)$. The value in the denominator shouldn't be a vector, it should be the magnitude of the vector $\textbf{r}-\textbf{r}'$ (the distance from the source point to the field point...distance is a scalar quantity, not a vector).

 

[maherelharake]

Oh ok I think I got the phi now.

I am trying to set it up as the correct vector, before I take the magnitude. Is the difference in the s-components [0-R]? And for the Phi compononet is it [0-Phi] (this is the most confusing one to me), and for the z-components [z-z']?

 

[gabbagabbahey]

Oh ok I think I got the phi now.

I am trying to set it up as the correct vector, before I take the magnitude. Is the difference in the s-components [0-R]? And for the Phi compononet is it [0-Phi] (this is the most confusing one to me), and for the z-components [z-z']?

Well, there is no phi comonent for a position vector (some vector fields, like an electric/magnetic field will have phi components, but porsition vectors do not), but your other two are correct.

If $\textbf{r}=z\mathbf{\hat{z}}$ and $\textbf{r}'=R\mathbf{\hat{s}}+z'\mathbf{\hat{z}}$, then you should immediately be able to say $\textbf{r}-textbf{r}'=-R\mathbf{\hat{s}}+(z-z')\mathbf{\hat{z}}$... this is simple vector addition/subtraction in unit-vector notation.

So, what is the magnitude of that vector (usually called the separation vector) then?

 

[maherelharake]

Would it be sqrt[(R^2) + (z^2) -2zz' + (z'^2)]?

And I have to go to campus now. I will try to respond when I am over there, but if I can't, I will definitely respond ASAP this evening.

 

[gabbagabbahey]

Would it be sqrt[(R^2) + (z^2) -2zz' + (z'^2)]?

And I have to go to campus now. I will try to respond when I am over there, but if I can't, I will definitely respond ASAP this evening.

Yup...now just calculate your integral

 

[maherelharake]

Would all those values in the integral be constant except for 'r'? Wouldn't k, sigma, R, z, and z' all be constant? If that's true, then it just becomes the integral of 'r' dtheta dr where 'r' goes from 0 to R, and theta goes from 0 to 2Pi.

 

[gabbagabbahey]

Would all those values in the integral be constant except for 'r'? Wouldn't k, sigma, R, z, and z' all be constant? If that's true, then it just becomes the integral of 'r' dtheta dr where 'r' goes from 0 to R, and theta goes from 0 to 2Pi.

Why would $z'$ be constant...it's the z-coordinate of a given source point, and there are source points all along the cylinder from $z'=-L/2$ to $z'=+L/2$...since $z'$ and $\phi'$ vary over the surface of the cylinder, you need to determine the area subtended when you vary each by a differential amount, you should find $dA'=Rd\phi'dz'$.

 

[maherelharake]

Ok I understand now why z' isn't constant. I should have recognized that.

I am a bit confused on your value of dA'. If I use that value, does it incorporate the surface charge on the endcaps? Charge is located all along the endcaps, not just at R distance from the axis.

I guess I am still a bit confused on how you got dA'.

 

[gabbagabbahey]

Ok I understand now why z' isn't constant. I should have recognized that.

I am a bit confused on your value of dA'. If I use that value, does it incorporate the surface charge on the endcaps? Charge is located all along the endcaps, not just at R distance from the axis.

I guess I am still a bit confused on how you got dA'.

my dA' is for the curved part of the tube. If the tube has encaps, then you will have to divide your surface integral into 3 parts...one for the curved surface and one for each endcap. The dA' for the endcaps will be s'ds'dphi' since s' and phi' vary over them.

 

[maherelharake]

Ok so for the integral involving the curved part of the tube, do I need a computer to solve it?

And for the integrals involving the endcaps, is it the exact same except for that value of dA'? Z' would be constant for the endcaps wouldn't it?

 

[gabbagabbahey]

Ok so for the integral involving the curved part of the tube, do I need a computer to solve it?

You shouldn't, just start with the substitution $u=z-z'$ and go from there (post your work if you get stuck)

And for the integrals involving the endcaps, is it the exact same except for that value of dA'? Z' would be constant for the endcaps wouldn't it?

Yes and Yes.

 

[maherelharake]

Here is where I got stuck. He told us we can use a computer, but I would still like to refresh my memory a little bit.


http://i51.tinypic.com/2ii8hlg.jpg

 

[gabbagabbahey]

Here is where I got stuck. He told us we can use a computer, but I would still like to refresh my memory a little bit.


http://i51.tinypic.com/2ii8hlg.jpg

You need to change your limits of integration when doing the u-sub, and you are missing a negative sign $du=-dz'$.

Anyways,

$$\int \frac{du}{\sqrt{R^2+u^2}}=\ln\left(2\sqrt{R^2+u^2}+2u\right)+\text{constant}$$

This is a fairly common integral that you can lookup in a table of integrals. If you want to calculate it out yourself for practise, I think you need to use the substitution $v=\sqrt{R^2+u^2}$ and then use partial fraction decomposition

 

[maherelharake]

Yeah I found that in a table in my old Calculus book, but for some reason there wasn't a factor of 2.

Ok so this is where I am now, I am currently working on plugging in the values.

 

[maherelharake]

I have also attached what I hope is the correct V's for the endcaps.

 

[maherelharake]

I tried to do that integral by hand, and I used a substitution of u=Rtan(theta).
I ended up with an integral of sec(theta) d(theta) which integrated to ln[sec(Theta)+tan(theta)] which transformed back into ln{sqrt(R^2 + u^2)/z + u/z)

 

[gabbagabbahey]

I have also attached what I hope is the correct V's for the endcaps.

For the endcaps, $s'\neq R$ (except at the very edge of the endcap)... your denominator should be $$\sqrt{(s')^2+(z\pm\frac{L}{2})^2}$$...do you see why?

 

[maherelharake]

Ohh yes of course. I see that now. Ok so I think I am done with the endcaps. Now for the integration of the other one, did you see my comment about the way I did it by hand? Are they somehow equivalent (using some rearranging that I am not able to figure out right now)?

 

[gabbagabbahey]

I tried to do that integral by hand, and I used a substitution of u=Rtan(theta).
I ended up with an integral of sec(theta) d(theta) which integrated to ln[sec(Theta)+tan(theta)] which transformed back into ln{sqrt(R^2 + u^2)/z + u/z)

Don't you mean $$\ln\left(\frac{\sqrt{R^2+u^2}+u}{R}\right)$$? If so, keep in mind that [tex]\ln\left(\frac{f(u)}{R}\right)=\ln(f(u))-\ln(R)[/itex], and $\ln(R)$ is just some constant that can be absorbed into the integration constant (same thing with the factor of 2 in the result I posted earlier)

 

[gabbagabbahey]

Ohh yes of course. I see that now. Ok so I think I am done with the endcaps.

Are you sure? Changing the denominator of the integrand means changing the way you integrate...

 

[maherelharake]

Curved part:
Oh yes that's what I meant. Sorry. So then I just plug in the bounds and it should give me the correct potential?

Endcaps: Ohh of course. Now I can use substitution and set u=s'^2 +(z-z')2

 

[maherelharake]

Here is my work for the endcaps. I believe that both endcaps should give the same result after substitution

 

[maherelharake]

Ok I have attached (what I hope is) my final draft for this problem. One attachment is for the end caps, and one is for the curved part. I am going to sleep, and these are due tomorrow morning. If you have any more pointers, I will read them before I go to campus. Thank you for your help.