# Electrostatics - finding the work done

• Saitama

## Homework Statement ## The Attempt at a Solution

I don't think the given problem requires me to find an expression of potential energy as a function of the equal side lengths. Also, I am not sure how would I set up the integral even if I want to find the expression for potential energy. There is probably some trick to this question but I am honestly clueless about it.

Any help is appreciated. Thanks!

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Hi voko! :)

I am not sure, I have never done problems in Electrostatics using dimensional analysis, can I have a few hints about how to start?

That starts by enumerating the parameters that describe the problem. Then you figure out how those parameters can be combined so that the units of the sought quantity can be represented from them. The sought quantity here is work. Then you note that there is some similarities between this and that configurations.

That starts by enumerating the parameters that describe the problem. Then you figure out how those parameters can be combined so that the units of the sought quantity can be represented from them. The sought quantity here is work. Then you note that there is some similarities between this and that configurations.

I can think of two, the equal sidelength and the charge of the plate, are these sufficient?

I do not think there are any other parameters possible (assuming that the material of the plate does not oppose folding in any way).

I do not think there are any other parameters possible (assuming that the material of the plate does not oppose folding in any way).

I still don't think I get it. Do I write that potential energy is proportional to some power of sidelength (##a##) and charge of plate (##q##) i.e

$$U\propto a^mq^l$$

Should I just equate the dimensions?

I still don't think I get it. Do I write that potential energy is proportional to some power of sidelength (##a##) and charge of plate (##q##) i.e

$$U\propto a^mq^l$$

Should I just equate the dimensions?

Not exactly sure what Voko has in mind, but I would include the charge density (charge per unit area) in the expression. The one constant through the entire process is this. (EDIT: nope, sorry, it changes)

You can't just equate the dimensions on both sides because your constant of proportionality may not be dimensionless. Instead, try to derive the dimensional relationship given what you know to be the relationship between energy, charge, distance, force and area.

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Hi Curious3141! :)

Instead, try to derive the dimensional relationship given what you know to be the relationship between energy, charge, distance, force and area.

I don't get what you are asking me to do here. Do you ask me to write down the following:
$$E \propto q^ad^bF^cA^d$$
and then equate the dimensions? I still don't think I get it. Do I write that potential energy is proportional to some power of sidelength (##a##) and charge of plate (##q##) i.e

$$U\propto a^mq^l$$

Should I just equate the dimensions?

Gory details: http://en.wikipedia.org/wiki/Buckingham_π_theorem

In simpler cases, such as this one, that does boil down to just equating the dimensions in that formula you wrote.

The one constant through the entire process is this.

Is it not doubled each time the plate is folded?

Hi Curious3141! :)

I don't get what you are asking me to do here. Do you ask me to write down the following:
$$E \propto q^ad^bF^cA^d$$
and then equate the dimensions? Not really. I'm asking you to start by writing down the relationship between energy on one side and force and distance on the other. Then express force in terms of charge and distance (and constants that don't change). Then express charge in terms of charge density and area - and the latter can be expressed in terms of distance (linear dimension). In the final expression you should be able to equate energy on one side to an expression in terms of charge density and linear dimension only (and physical constant(s) that don't change, so they don't matter). Both the charge density and the linear dimension change predictably.

I've pretty much given the game away here actually.

EDIT: since the *charge* is the only constant, voko's method is much simpler, so please follow that. forget about the charge density. Relate the energy to the charge and the linear dimension.

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Is it not doubled each time the plate is folded?

You're right in that it changes. I was initially wrong about it being constant. But it should quadruple, shouldn't it, since the same charge is now distributed over a fourth of the area?

You're right in that it changes. I was initially wrong about it being constant. But it should quadruple, shouldn't it, since the same charge is not distributed over a fourth of the area?

Why fourth? Each time the plate is folded, its area is halved. Am I being silly here?

In simpler cases, such as this one, that does boil down to just equating the dimensions in that formula you wrote.

But the proportionality constant need not be dimensionless as pointed out by Curious. But the proportionality constant need not be dimensionless as pointed out by Curious. We have three parameters. Charge, length, work. Can we create a dimensionless product of their degrees? Can we do that in more than one ways? If the answers are yes and no, in order, then we are good. Any other answers - oops, think again.

Why fourth? Each time the plate is folded, its area is halved. Am I being silly here?

No, I am. You're right. The folding only halves the base, but keeps the altitude of the triangle constant.

Sorry. Must stop trying to help with these late at night.

We have three parameters. Charge, length, work. Can we create a dimensionless product of their degrees?

The product ##q^al^bE^c## is dimensionless if a=b=c=0. Well, that is always the case. Look for the solutions with some of the degrees not zero.

Well, that is always the case. Look for the solutions with some of the degrees not zero.

The dimensions of charge, length and energy are ##A^1T^1##, ##L^1## and ##M^1L^2T^{-2}## respectively. Hence the product is:
$$M^cL^{2c+b}T^{-2c+a}A^a$$
For the product to be dimensionless, ##a=c=0##. So we are left with ##L^b##. ##b## should be also zero for the product to be dimensionless. It looks like we'll have to throw another parameter, such as the potential energy, into the mix. But then, with the potential energy, we can just solve this directly :)

It looks like we'll have to throw another parameter, such as the potential energy, into the mix. But then, with the potential energy, we can just solve this directly :)

I am not sure if I get your point but if I add in potential energy, I would need the potential energy of two configurations, am I right or did I misunderstood your reply?

Observe that the triangle before and after folding is similar. They only change in linear size uniformly, and in charge density, also uniformly. So their potential energies will be exactly the same, except in some scaling factor. So finding one expression, you will find them all.

Observe that the triangle before and after folding is similar. They only change in linear size uniformly, and in charge density, also uniformly. So their potential energies will be exactly the same, except in some scaling factor. So finding one expression, you will find them all.

The product we now have is: ##q^al^bW^cU^d## (W represents work) i.e ##M^{c+d}L^{2c+b+2d}T^{a-2c-2d}A^a##. For the dimensionless product, I have ##a=0##, ##c+d=0## and ##b=0##. I can't find the values for c and d, is this correct now?

It looks like you will have to obtain an expression for the potential energy of the plate to proceed further. This is what I sincerely wanted to avoid by using simple dimensional analysis, but that did not quite work out.

It looks like you will have to obtain an expression for the potential energy of the plate to proceed further. This is what I sincerely wanted to avoid by using simple dimensional analysis, but that did not quite work out.

Ah, can you please provide a few hints about how to set up the integral? I am thinking of placing the triangle in the xy plane with the altitude along the y-axis and base (longest side or hypotenuse) along the x-axis. If I select a small area element, I will have to pair it with every other small element on the triangle but I am not sure how to set up the integral.

I would actually recommend that you position the triangle exactly as shown in the first figure, with the x-axis rightward and the y-axis upward, and the right angle at the origin. You won't actually have to integrate that, so don't worry about having to express the other boundary.

You won't actually have to integrate that, so don't worry about having to express the other boundary.

Sorry, I don't get this. Are you sure we you can't solve this with just dimensional analysis?

To perform a fold, it seems to me that the only quantities that the work can depend on are the initial charge density ##\sigma## and the initial leg length ##a## (as well as Coulomb's constant). That should be enough to get the answer.

But, maybe I'm being naive Hi TSny! :)

Are you sure we you can't solve this with just dimensional analysis?

To perform a fold, it seems to me that the only quantities that the work can depend on are the initial charge density ##\sigma## and the initial leg length ##a## (as well as Coulomb's constant). That should be enough to get the answer.

But, maybe I'm being naive So ##W \propto k\sigma^ma^n##. From dimensional analysis, I get ##m=2## and ##n=-1##. Hence, ##W\propto \sigma^2/a##.

When the plate is folded, the surface charge density gets doubled and the sidelength gets reduced by a factor of ##\sqrt{2}##. Hence
$$\frac{1}{W_2}=\frac{1}{4\sqrt{2}} \Rightarrow W_2=4\sqrt{2} \,J$$
Does this look right?

EDIT: This is incorrect, where did I go wrong? So ##W \propto k\sigma^ma^n##. From dimensional analysis, I get ##m=2## and ##n=-1##.

How did you get ##n = 1##? Remember, ##\sigma## is a charge density, not a charge.

How did you get ##n = 1##? Remember, ##\sigma## is a charge density, not a charge.

Is ##n=3##? Am I right in saying that the charge density doubles after the first fold?

Are you sure we you can't solve this with just dimensional analysis?

To perform a fold, it seems to me that the only quantities that the work can depend on are the initial charge density ##\sigma## and the initial leg length ##a## (as well as Coulomb's constant). That should be enough to get the answer.

You are probably right. There may be a methodological difficulty in justifying the constant, though, unless a deeper analysis is performed.

The potential energy approach should reveal that constant.

Is ##n=3##? Am I right in saying that the charge density doubles after the first fold?

I would say yes to both questions.

I would say yes to both questions.

Thanks a lot TSny and voko! I get ##W_2=\sqrt{2}\,\, J##.

But how do I find the expression for potential energy? I still don't see how to set up the integrals.

And where one can find such problems? Are they present in any text?

For the integrals, let ##\Delta_1## be the first triangle, then ##\Delta_2##, etc.

The potential due to the first triangle is $$\phi_1(\xi, \eta) = \int\limits_{(x,y) \in \Delta_1} k \sigma_1 \frac {dx dy} {\sqrt{(x - \xi)^2 + (y - \eta)^2}}$$ *edit: clarified the potential integral.

The potential energy is $$U_1 = \frac 1 2 \int\limits_{(\xi,\eta) \in \Delta_1} \sigma_1 \phi_1(\xi, \eta) {d\xi d\eta}$$ Now, if we replace ##\sigma_1## with ##\sigma_2## and ##\Delta_1## with ##\Delta_2##, we will get ##U_2##. It is obvious how to replace the charge density. To replace the area of integration, we can map linearly ## \Delta_1 \to \Delta_2 ##, so the integral for ##U_2## will now be taken over ##\Delta_1##, but the change of variables will yield another scaling factor ##s##, so $$U_2 = \frac {\sigma_2^2} {\sigma_1^2} s U_1$$

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