Voltage Lab Conductive Paper Questions

[Albertgauss]

Good Day

Hi all,

I am doing a project with the standard lab about measuring equipotential surfaces with conductive paper, a voltmeter, and a power supply. I have a couple of questions about this project.

I attach a jpeg of the equations for voltage calculations in the file “Equations.jpeg”. My confusion is centered around making V = 0 as r --> ∞ (as you move away from the charge distributions) for the configurations of this project.

Question 1: In the slides “Dipole.jpeg”, “LineOfCharge.jpeg”, “ParLines.jpeg”, “PointCharge.jpeg”, do I need to paint a conductive stripe (attached to the negative to terminal of the power supply) on the edges of my conductive paper, so that V --> 0 as r --> ∞?

Note that in the jpeg “PointCharge.jpeg”, there is already a silver ring for this purpose. Should I move that ring closer to the edge of the paper? Could I replace the ring with a rectangular border on the edge of the paper? The ring in that paper does not seem far away enough from the point charge for “r --> ∞”.

I have not seen any such border on any Parallel lines for this project, yet, though. Usually one line is at the (negative terminal of power supply and COM of voltmeter) and the other line will be at V ~ 20 volts and connected to the positive terminal of the power supply. Many labs just have the bars at opposite ends of the paper and only focus on voltages in between the lines; for those labs with the bars centered in the middle of the paper, no border was ever present.

Question 2. Should I use the expression of the voltage due to a finite wire, even though it is complicated, for a single line of charge (but possibly more realistic)? Or can I use the voltage due to an infinite wire? If the later, do I need a border (yellow) at which the negative terminal of the power supply and COM of the voltmeter would be attached, as in the jpeg “LineOfCharge.jpeg”?

Question 3. I have never seen any calculations for the potential to distinguish between two dimensions or three-dimensions? Does the voltage expression change between two and three dimensions? I have never seen any calculations for the voltage due to these configurations expressly developed for one case or the other but would be glad to know where to find them. My equations above are what I understand about this last question.

 

[kuruman]

You ask a lot of questions about what you should do, and you provide a lot of details about the designs on the conductive paper, but you have not told us what the goal of your project is. Are you trying to go beyond drawing the equipotentials specific to an electrode configuration? If so, where are you headed?

 

[Albertgauss]

I'm trying to correctly measure the voltages around the paper for the various configurations I presented and so verify the equations I presented in the Equations.jpeg. (the voltage of a || plate is not included in the Equations.jpeg but I know it is linear with distance in between the plates) ) I would measure those voltages of with the + lead of a voltmeter with respect to the V = 0 (ground) of each situation. But I'm stuck because I don't know how to make sure V --> 0 as r --> 0, or if I have the electrical grounds set up properly . I put forth what I think I should do, but I am not sure.

 

[zoki85]

(the voltage of a || plate is not included in the Equations.jpeg but I know it is linear with distance in between the plates)

That's correct only for infinitely long & wide plates

 

[Albertgauss]

Is there a more practical expression for something finite like conductive paint painted on conductive paper in these labs and projects?

 

[ZapperZ]

Summary:: Assorted charge distributions and voltages measured on classic conductive paper lab or project. Some questions. Two dimensions? Three Dimensions? V --> 0 as r --> 0 questions.

Good Day

Hi all,

I am doing a project with the standard lab about measuring equipotential surfaces with conductive paper, a voltmeter, and a power supply. I have a couple of questions about this project.

I attach a jpeg of the equations for voltage calculations in the file “Equations.jpeg”. My confusion is centered around making V = 0 as r --> ∞ (as you move away from the charge distributions) for the configurations of this project.

View attachment 256771

Question 1: In the slides “Dipole.jpeg”, “LineOfCharge.jpeg”, “ParLines.jpeg”, “PointCharge.jpeg”, do I need to paint a conductive stripe (attached to the negative to terminal of the power supply) on the edges of my conductive paper, so that V --> 0 as r --> ∞?

Note that in the jpeg “PointCharge.jpeg”, there is already a silver ring for this purpose. Should I move that ring closer to the edge of the paper? Could I replace the ring with a rectangular border on the edge of the paper? The ring in that paper does not seem far away enough from the point charge for “r --> ∞”.

View attachment 256774

I have not seen any such border on any Parallel lines for this project, yet, though. Usually one line is at the (negative terminal of power supply and COM of voltmeter) and the other line will be at V ~ 20 volts and connected to the positive terminal of the power supply. Many labs just have the bars at opposite ends of the paper and only focus on voltages in between the lines; for those labs with the bars centered in the middle of the paper, no border was ever present.

View attachment 256773

Question 2. Should I use the expression of the voltage due to a finite wire, even though it is complicated, for a single line of charge (but possibly more realistic)? Or can I use the voltage due to an infinite wire? If the later, do I need a border (yellow) at which the negative terminal of the power supply and COM of the voltmeter would be attached, as in the jpeg “LineOfCharge.jpeg”?

View attachment 256772

Question 3. I have never seen any calculations for the potential to distinguish between two dimensions or three-dimensions? Does the voltage expression change between two and three dimensions? I have never seen any calculations for the voltage due to these configurations expressly developed for one case or the other but would be glad to know where to find them. My equations above are what I understand about this last question.

OK, here's MY question to you: Are you being asked to actually find the dependence on V as a function of r for these various configurations here?

The reason why I ask this is because in the standard labs on this, a student is being asked to just find the equipotential lines for these various configurations, not to investigate the actual V dependence. So is this something that you've been asked to do, or is this just something you WISH to do?

I can see where you are being asked to investigate various configuration to get at the electric field lines, but trying to actually verify, say, the values of V as a function of r from a "point" source on the conductive paper is often rather inaccurate for many reasons.

This type of experiment is good for qualitative demonstration, not so go for quantitative demonstration.

Zz.

 

[Albertgauss]

Yes, you are correct, finding V(r) is something I wish to do. It may be that this kind of lab is only good for qualitative equipotential surfaces. I was hoping to go further and get the actual V dependence on distance from the charges. That was the goal. Obviously, when I did measurements myself, they didn't work so great, so I thought maybe there was something not set up correctly in ensuring V --> 0 as I get farther from the charges. Of course, if V(r) doesn't work for these kind of labs, then V --> 0, r --> 0 is not the problem of the nonconforming data, and is a moot point.

 

[ZapperZ]

Yes, you are correct, finding V(r) is something I wish to do. It may be that this kind of lab is only good for qualitative equipotential surfaces. I was hoping to go further and get the actual V dependence on distance from the charges. That was the goal. Obviously, when I did measurements myself, they didn't work so great, so I thought maybe there was something not set up correctly in ensuring V --> 0 as I get farther from the charges. Of course, if V(r) doesn't work for these kind of labs, then V --> 0, r --> 0 is not the problem of the nonconforming data, and is a moot point.

For a "point" charge, V will approach zero as r gets bigger, but the EXACT relationship will not be very clear that it is as 1/r. This is because the conducting paper will slightly distort the actual field from a point charge.

This is why labs like this will tend to not ask you to do such a thing. Many of us who designed such experiments have already checked it out, and we know it isn't an experiment to verify V dependence on r for textbook cases.

Zz.

 

[Albertgauss]

Oh, I see. Other people have tried this before...with the same results. No problem. But it was worth a try.

 

[kuruman]

For a "point" charge, V will approach zero as r gets bigger, but the EXACT relationship will not be very clear that it is as 1/r. This is because the conducting paper will slightly distort the actual field from a point charge.

Not to mention distortion due to edge effects; the size of the paper is finite after all. Once I tried to see if there is a 1/r dependence with a just a central dot electrode. I had to suspend disbelief to see it to any reasonable accuracy. However what works quite well with these papers is verification of Gauss's Law. You can draw various rectangular boundaries (2-D Gaussian surfaces) with chalk around the electrodes and see whether the sum of voltages across the boundary (using a two-pronged voltage probe) is (a) independent of the boundary if a single electrode is enclosed by two different boundaries and (b) if both electrodes are enclosed by a boundary, the sum of voltages is zero. Make sure that the line joining the two-prongs of the probe is perpendicular to the boundary because you want the perpendicular component of the electric field $E=-\frac{\Delta V}{\Delta x}$.

 

[vanhees71]

Also the expression for the potential of a finite charged rod looks not correct. I guess that's only the correct result in the plane perpendicular to the rod intersecting it in its center. If the rod is directed in $z$ direction from the symmetry under rotations around the $z$ axis, the potential must be a function of $R=\sqrt{y^2+z^2}$ and $z$ (cylinder coordinates $(R,\varphi,z)$), i.e.,
$$V(\vec{r})=V(R,z).$$

 

[Albertgauss]

Okay, I will give Kurunam's idea a try. I will investigate that. @vanhees71, you are correct in what you said. The finite rod calculations were done at the plane perpendicular to the rod intersecting it in its center. I did not do the more general version you mentioned because I could not get the simpler case of the midpoint to work. But, yes, you are correct.

 

[vanhees71]

Maybe the integral for the general case is not so simple or not expressible with standard functions.

 

[kuruman]

Okay, I will give Kurunam's idea a try. I will investigate that. @vanhees71, you are correct in what you said. The finite rod calculations were done at the plane perpendicular to the rod intersecting it in its center. I did not do the more general version you mentioned because I could not get the simpler case of the midpoint to work. But, yes, you are correct.

It's Kuruman. Any way, if you choose to follow my suggestion, you may wish to post your progress here. This is a lab experiment I used to teach and I may have some tips for you.

 

[kuruman]

Maybe the integral for the general case is not so simple or not expressible with standard functions.

Without loss of generality we can consider the point of interest in the $xy$-plane with the finite rod along the $x$-axis. The origin is at the midpoint of the rod of length $L$. Then the potential is $$V=k\lambda\int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{dx'}{\sqrt{(x-x')^2+y^2}}=-k\lambda \ln \left[ (x-x')+\sqrt{(x-x')^2+y^2}\right]_{-\frac{L}{2}}^{\frac{L}{2}}$$At this point we can conventionally put the rod along the $z$ axis and rename the radial distance to $r$ to get the azimuthally symmetric potential$$V(r,z)=k\lambda \ln\left[\frac{ (z+\frac{L}{2})+\sqrt{(z+\frac{L}{2})^2+r^2}}{(z-\frac{L}{2})+\sqrt{(z-\frac{L}{2})^2+r^2}}\right].$$This expression almost reduces to the bottom right expression in post #1 when one sets $z=0$. (The $\frac{L^2}{2}$ terms under the radicals are incorrect.) I am not sure what the expression in the upper right is supposed to express.

 

[vanhees71]

Great, so it's even very simple. I wonder sometimes, why textbooks and/or lectures then only treat a special case...

 

[Albertgauss]

@kuruman, sorry for the misspelled name. But, yes, I will work on the lab you suggested. That sounds like a great idea.

The expression I got in the upper right of the "Equations.jpeg" came about when I searched for proofs about how to make sure V -> 0 as r -> 0. That expression came from a website that posted pages from a book called "Geometry of Vector Calculus" by Tevian Dray. My equation may be wrong, or may be different than what I intended it to do. I attached the webpage below. Furthermore, I thought the expression in the upper right was for three dimensions, and the expression in the lower right was only for two dimensions, since the expression in the lower right was worked out in the xy plane only with no z plane mentioned in that article where I read about it.

 

[kuruman]

@kuruman, sorry for the misspelled name. But, yes, I will work on the lab you suggested. That sounds like a great idea.

The expression I got in the upper right of the "Equations.jpeg" came about when I searched for proofs about how to make sure V -> 0 as r -> 0. That expression came from a website that posted pages from a book called "Geometry of Vector Calculus" by Tevian Dray. My equation may be wrong, or may be different than what I intended it to do. I attached the webpage below. Furthermore, I thought the expression in the upper right was for three dimensions, and the expression in the lower right was only for two dimensions, since the expression in the lower right was worked out in the xy plane only with no z plane mentioned in that article where I read about it.

It is incorrect to expect that $V\rightarrow 0$ when $r\rightarrow 0$. The equation is derived by adding (integrating) elemental contributions to the potential of the form $dV=\dfrac{k\lambda dx'}{|\vec r-\vec r'|}.$ For each of these contributions the potential is taken to be zero at infinity. They all have the same sign, the sign of $\lambda$, so that when you add them all up upon integration, you get zero only at infinity.

My expression in post #15 is the same as equation (1) in your reference with one exception: I assume that the length of the wire is $L$ while the reference assumes that is it $2L$ presumably to get right of the annoying divisions by $2$. Note that this equation is two-dimensional in that the potential depends on only two independent position variables, $r$ and $z$. That's because the charge distribution has azimuthal symmetry. This means that if you put the $z$-axis along the wire, the potential is independent of the polar angle $\theta$. In other words, the potential has the same value on a circle whose plane is perpendicular to the wire, has radius $r$ and is centered at distance $z$ from the origin, here the midpoint of the wire.

Finally, a good way to check expressions involving finite elements for correctness is to see whether they reduce to known expressions asymptotically. When $L<<(z,r)$, the equation in #15 reduces (after a series expansion) to $V(r,z)\approx\dfrac{k\lambda L}{\sqrt{r^2+z^2}}.$ With $Q=\lambda L$, you get the electric potential for a point charge $Q$. This is what the charged wire looks like when you are very far from it (relative to $L$) at distance $\sqrt{r^2+z^2}$ from the origin.