Help me find the Potential V from a situation with charge

[RJLiberator]

Homework Statement

A disc of radius R lying in the xy-plane is composed of an inner disc of radius R/2 carrying a uniform surface charge density +σ and an anulus (inner and outer radii of R/2 and R) carrying a uniform surface charge density -σ. Assume that the inner disc and the annulus are electrically insulated from each other by an insulating strip of neglibible thickness placed at their intersection at R/2.

a) Calculate the potential V at the point (0,0,z) along the z-axis using infinity as the reference point. Show by making a Taylor expansion i n1/z that z dependence of your answer makes sense in the z->∞ limit.

Homework Equations

$V = \frac{1}{4 \pi ε_0} \int \frac{σ}{r} dτ$
3. The Attempt at a Solution

So this is a multi-part problem and I can get parts b,c, and d if I can figure this potential out.

When I try I get this:
V(0,0,z) = \frac{1}{4 \pi ε_0}\int_0^z \frac{σ}{z} dz

Which evaluates to $V(0,0,z)=\frac{1}{4 \pi ε_0}\frac{σ}{z^2}$

But this doesn't make sense to me. I mean, I have the differences in charge densities to consider, and so on. So I ask for hints on what I am doing wrong here and how I can understand how to use this equation better. Am I not understanding what r is suppose to be in this case? How can I consider the -σ and +σ.

 

[Charles Link]

Your integral needs to be an integral over $dA=2 \pi r \, dr$ and your denominator of $z$ is incorrect. Your denominator needs to be the distance $s=\sqrt{r^2+z^2}$. Hopefully this input is helpful and didn't violate the forum rules of giving you too much information.

 

[RJLiberator]

No-- That makes a lot of sense.
I had on my paper "Q" instead of σ which made me not think it was an area integral. :(.
I see why it must be dA and spherical.

With that being said, here is where I am going now.

V(0,0,z) = \frac{1}{4 \pi \epsilon_0)} \int_0^z \frac{σ } {\sqrt{r^2+z^2}} dA

I have this evaluating out to be
\frac{σ}{4 \epsilon_0}(\sqrt{2}-1)z

But this must be wrong as the question tells me to take the taylor expansion around 1/z.

Debugging my answer, I must have done something wrong with the sigma charge or the bounds of integration.

The bounds of integration goes from infinity = 0 to z, I am pretty sure, no?

 

[Charles Link]

You need to read the problem more carefully. $\sigma=+\sigma_o$ from $r= 0$ to $r= R/2$, and $\sigma=-\sigma_o$ from $r=R/2$ to $r=R$. There should not be any $z$ on the limits of your integrals. For the purposes of the integration, $z$ is a constant.

 

[Charles Link]

And a follow-on : For the second part=the Taylor series in $1/z$, you can guess what it needs to be: i.e. Compute the net charge $Q$ on the disc. That should give you the necessary $Q$ for the potential when you get far from the disc.

 

[RJLiberator]

You need to read the problem more carefully. $\sigma=+\sigma_o$ from $r= 0$ to $r= R/2$, and $\sigma=-\sigma_o$ from $r=R/2$ to $r=R$. There should not be any $z$ on the limits of your integrals. For the purposes of the integration, $z$ is a constant.

I'm struggling to create this integration.

Typically, I would split the integral from 0 to R/2 and R/2 to R, however, my interpretation of this is that I am integrating from infinity to R.

I guess I could take the integral from infinity to R/2 and R/2 to 0. Does this make sense to you?

 

[Charles Link]

They confused you in the original problem statement. By taking infinity as the reference point, that is where the potential $V=0$. For a point charge $q$ at distance $s$, the potential $V$ is $V=\frac{q}{4 \pi \epsilon_o s }$. (Notice as $s \rightarrow +\infty$ , $V=0$). $\\$ To get the total potential $V$ (at location $(0,0,z)$) from a distribution of charge, all you need to do is sum up all of the individual potentials across the charge distribution. You do this by integrating them, i.e. integrating across the charge distribution. $\\$ Additional item: The potential $V(s)=\frac{q}{(4 \pi \epsilon_o s )}$ comes from $V(s)= \int\limits_{s}^{+\infty} \frac{q}{4 \pi \epsilon_o r^2} \, dr =\frac{q}{4 \pi \epsilon_o s}$, where we're integrating the electric field function to get the potential. We have this result, so that we don't need to do any further integrations to $+\infty$.

 

[RJLiberator]

Ok, that explanation (i think) helps me understand what I need to do.

integration :

V = \frac{-σ}{4 \pi \epsilon_0} \int_{R/2}^R \frac{ 2 \pi r}{\sqrt{r^2+z^2}}dr+\frac{σ}{4 \pi \epsilon_0} \int_0^{R/2} \frac{ 2 \pi r}{\sqrt{r^2+z^2}}dr

This makes sense to me as we take in the understanding of the change in charge and evaluate their distributions correctly.
The problem is, I get a very messy result featuring multiple parts.

Result:

V= \frac{-σ}{\epsilon_0} (\sqrt{\frac{R^2}{4}+z^2}) + \frac{-σ}{\epsilon_0} \sqrt{R^2+z^2}-\frac{zσ}{2 \epsilon_0}

This doesn't make sense when consider taylor expansion around 1/z.

 

[Charles Link]

Your first term there should have a plus sign, and your second term needs a 2 in the denominator. $\\$ Meanwhile, to do the Taylor series, pull a $z$ out of the first two terms. That will leave you with e.g. $\sqrt{1+R^2/(4z^2)}$, etc. It does work=I checked it. $R^2/(4z^2)=\Delta$ as $z \rightarrow +\infty$. You Taylor expand $\sqrt{1+\Delta}$, etc.

 

[Charles Link]

See my edited version above: You need a 2 in the denominator of the 2nd term. Notice in the Taylor expansion, the 3 terms that are first power in $z$ cancel.

 

[Charles Link]

It remains to compute the total charge $Q$ on the disc, and your Taylor expansion should give you $V(0,0,z)=\frac{Q}{4 \pi \epsilon_o z}$ for large $z$, which is equivalent to small $1/z$.

 

[RJLiberator]

I see where I made some typos, long latex for me, sorry!

I have been able to pull the z out. I think the potential is correct.

I'm having trouble understanding what to do with the Taylor Expansion.

If I taylor expand just the sqrt terms, then I get https://www.wolframalpha.com/input/?i=taylor+expansion+of+sqrt(1+R^2/(4x^2))

Which doesn't compute to what we need.

I feel good about the Potential, though.

 

[Charles Link]

Taylor expansion is simple: $\sqrt{1+\Delta} =1+\frac{\Delta}{2} +...$. To show this, let $f(x)=\sqrt{1+x }$ and expand about $x=0$. $f(x)=f(0)+f'(0)(x-0)+...$. $\\$ $\sqrt{1+R^2/(4z^2)}=1+R^2/(8z^2)+...$ $\\$ And remember, this gets multiplied by the $z$ that was factored out... The $1$ out front with a $z$ on it is going to get canceled by the other two terms with their $z$'s.

 

[RJLiberator]

Ok, so this makes sense.

After doing the Taylor expansion of the two sqrt's, I get z in the denominator of the terms, then I get 1-1/2-1/2 = 0 so the entire potential is 0 when z goes to infinity which makes crystal clear sense. Beautiful? :D

 

[Charles Link]

Ok, so this makes sense.

After doing the Taylor expansion of the two sqrt's, I get z in the denominator of the terms, then I get 1-1/2-1/2 = 0 so the entire potential is 0 when z goes to infinity which makes crystal clear sense. Beautiful? :D

You are partly correct=Collect the $1/z$ terms: They should add to give $Q/(4 \pi \epsilon_o z )$ where $Q$ is the net charge which is not zero because the outer ring has larger area=the total charge will be negative.

 

[Charles Link]

Your first term gives $R^2/(8z)$, and your second term gives $-R^2/(4z)$ . These add to give $-R^2/(8z)$.

 

[RJLiberator]

You are getting the Q/4pi term from letting sigma = the charge density over the area, correct?

How are you collecting the 1/z terms? I have the taylor expansion going like $1 + \frac{R^2}{8z^2} + \frac{R^2}{16z^2} +...$ and when z goes to infinity, these go to 0.

so it would be the charge density*0.

 

[Charles Link]

You are getting the Q/4pi term from letting sigma = the charge density over the area, correct?

How are you collecting the 1/z terms? I have the taylor expansion going like $1 + \frac{R^2}{8z^2} + \frac{R^2}{16z^2} +...$ and when z goes to infinity, these go to 0.

so it would be the charge density*0.

They get multiplied by the $z$ that was factored out so they become $1/z$ terms. Yes, you need to compute them. Note again, you got the minus sign wrong in the first term and you need a 2 in the denominator of the second term.

 

[Charles Link]

From post 16, $V=-\frac{\sigma R^2}{8 \epsilon_o z}=-\frac{\sigma \pi R^2}{2} \frac{1}{4 \pi \epsilon_o z}$. It remains to show that $Q_{total}=-\sigma \pi R^2/2$.

 

[RJLiberator]

I'm sorry for my swiftness and lack of coherency towards the end of this thread. Electromagnetism has buried me tonight. Let me reply in full:

They get multiplied by the z" role="presentation">z that was factored out so they become 1/z" role="presentation">1/z terms. Yes, you need to compute them. Note again, you got the minus sign wrong in the first term and you need a 2 in the denominator of the second term.

Yes, absolutely, but when z goes to infinity they are still 1/infinity whether its 1/z^2 or 1/z.

Let me show you what I have, perhaps there in lies the discrepancy. After computer 2 taylor expansions for the major sqrt signs I get:

\frac{z \sigma}{ \epsilon_0} [1 + \frac{R^2}{8z^2}+\frac{R^2}{16z^2}+\frac{R^2}{32z^2}+...-\frac{1}{2} (1+\frac{R^2}{2z^2}+\frac{R^2}{4z^2}+...)-\frac{1}{2}]

From here, we can cancel one z in the denominator everywhere and then we are left with.

\frac{ \sigma}{ \epsilon_0} [z + \frac{R^2}{8z}+\frac{R^2}{16z}+\frac{R^2}{32z}+...-\frac{1}{2} (z+\frac{R^2}{2z}+\frac{R^2}{4z}+...)-\frac{z}{2}]

But, still, as z goes to infinity, the 1/z terms go to 0 and the remainder is z-z/2-z/2 which is 0. No?

Thank you for your generous help on this problem. Without you, I'd have nowhere to start on this.

 

[Charles Link]

Keep the $1/z$ terms and throw away $1/z^3$ terms, etc. They want you to compute it keeping the first order in $1/z$.

 

[Charles Link]

And the result you get should be simply $V(0,0,z)=\frac{Q_{total}}{4 \pi \epsilon_o z }$. Compute $Q_{total}$ and you will see the disc is not electrically neutral.

 

[Charles Link]

$Q_{total}=\int\limits_{0}^{R/2} \sigma \, 2 \pi r \, dr +\int\limits_{R/2}^{R} (-\sigma) \, 2 \pi r \, dr$. Of course, you know how to do this, but it's getting late, so I'm trying to get you to the result. :) From the result of post 19, it should be the case that $Q_{total} =-\sigma \pi R^2/2$. Hopefully this answer is in agreement, and yes, I get that it computes correctly. :)

 

[RJLiberator]

Mate, why are we computing the Q here for potential? That integral gives us $Q_total = \frac{- \sigma \pi R^2}{2}$

If I just focus on the 1/z terms then I find in the taylor expansion as we exclude those terms we get
V = \frac {-R^2 \sigma}{8z \epsilon_0}

 

[Charles Link]

Mate, why are we computing the Q here for potential? That integral gives us $Q_total = \frac{- \sigma \pi R^2}{2}$

If I just focus on the 1/z terms then I find in the taylor expansion as we exclude those terms we get
V = \frac {-R^2 \sigma}{8z \epsilon_0}

Exactly. And the question they asked is "does your result make sense when you consider the total charge of the disc, which from far away looks like a point charge.

 

[Charles Link]

Also, see post 19 again. You have the correct answer=simply do the algebra now that I did in post 19. :) Or alternatively, compute $V=\frac{Q_{total}}{4 \pi \epsilon_o z } =-\frac{\sigma R^2}{8 \epsilon_o z}$.

 

[RJLiberator]

Oh man, I need to go on a walk and digest this.
I am starting to see where you led me. Part b says to calculate the E field which should be pretty easy as I just take the negative gradient of the potential.

I thank you for all your kind help again. *bow*

 

[RJLiberator]

BOOM. I understand it now. Thank you for your help.
Walked out for a moment, re did some of the calculations. I see how it behaves like a point charge from z far away in the taylor expansion. Marvelous.

But man oh man Electromagnetism kicks my ass!