# Electrostatics - finding the work done

1. Feb 19, 2014

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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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2. Feb 19, 2014

### voko

3. Feb 19, 2014

### Saitama

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?

4. Feb 19, 2014

### voko

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.

5. Feb 19, 2014

### Saitama

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

6. Feb 19, 2014

### voko

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

7. Feb 19, 2014

### Saitama

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?

8. Feb 19, 2014

### Curious3141

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.

Last edited: Feb 19, 2014
9. Feb 19, 2014

### Saitama

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?

10. Feb 19, 2014

### voko

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.

Is it not doubled each time the plate is folded?

11. Feb 19, 2014

### Curious3141

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.

Last edited: Feb 19, 2014
12. Feb 19, 2014

### Curious3141

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?

13. Feb 19, 2014

### voko

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

14. Feb 19, 2014

### Saitama

But the proportionality constant need not be dimensionless as pointed out by Curious.

15. Feb 19, 2014

### voko

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.

16. Feb 19, 2014

### Curious3141

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.

17. Feb 19, 2014

### Saitama

The product $q^al^bE^c$ is dimensionless if a=b=c=0.

18. Feb 19, 2014

### voko

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

19. Feb 19, 2014

### Saitama

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.

20. Feb 19, 2014

### voko

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 :)