How Do You Calculate Surface Charge Density Without Knowing Plate Dimensions?

[flyingpig]

Homework Statement

[PLAIN]http://img232.imageshack.us/img232/8974/61038710.png

The Attempt at a Solution

Well $$\sigma$$ = charge/area

But how am I suppose to find the area if I am not even given the dimensions of the plates?

 

[zhermes]

Your right about the definition of surface charge density, but there may be other ways to find it. Can you think of other equations which involve sigma, and the values you have been given?

 

[flyingpig]

I don't understand, I am going to need the area more or less or at least the dimensions of the plates.

 

[zhermes]

That doesn't end up being the case. Check your physics textbook, or this page: http://en.wikipedia.org/wiki/Capacitor for the proper equation.

The answer will be the same regardless of the size of the plates---as long as the charge density is constant. Larger plates would require more charge, but would still have the same charge density.

 

[SammyS]

Use Gauss's Law to find the E field between the plates, the field due to a charge density, σ, on one of the plates. Then find V from:

$$V=-\int_{0}^{0.005}\left|\vec{E}\right|\,dx\ .$$

Do this symbolically, and you will have V in terms of σ. Plug 600 Volts in for V and solve for σ.

 

[flyingpig]

I just thought of this clever way of doing it.

$$\sigma A = q$$

From Gauss's Law

$$\oint \vec{E} \vec{dA} = \frac{\sigma A}{\epsilon_{0}}$$

The As disappear, with the integral

$$E\epsilon_{0} = \sigma$$

E = 600V/m, epsilion 0 is a constant and I got it?

 

[flyingpig]

Did I get it right? Also this sigma is for both plates, does that make sense to have a sigma for two plates that isn't one?

Should I multiply it by two to get the density for one plate?

 

[SammyS]

I just thought of this clever way of doing it.

$$\sigma A = q$$

From Gauss's Law

$$\oint \vec{E} \vec{dA} = \frac{\sigma A}{\epsilon_{0}}$$

The As disappear, with the integral

$$E\epsilon_{0} = \sigma$$

E = 600V/m, epsilion 0 is a constant and I got it?

E = 600V/(0.005 m) = __?__ V/m

Did I get it right? Also this sigma is for both plates, does that make sense to have a sigma for two plates that isn't one?

Should I multiply it by two to get the density for one plate?

No, the other plate has surface charge density, ‒σ. If you use Gauss's law with that plate you should get the same answer. (E will point towards the negative plate.)

 

[flyingpig]

E = 1.2 x 10^5 V/m (forgot lol)

Then σ = E * ε0 = (1.2 x 10^5V/m)(8.85 x 10^-12) = 1.06 x 10^-6C/m^2

There is also a follow up part which I am confident about, but I can't say I am 100% sure that I am right. The question says

"If an electron is 2.00mm from the positive plate, what is the work that must be done to move it to the negative plate"

I did E = 1.2 x 10^5 V/m

|F| = qE = 1.92 x 10^-14N

|F||d| = |1.92 x 10^-14N||3mm| = 5.76 x 10^-17J

Because it is 3mm away from the negative plate. The problem is, THIS JOULES IS RIDICULOUSLY SMALL! I am having second thoughts on my answer...

 

[flyingpig]

Wait why would it have -σ? I thought E = σ/2ε0 of one plate and two plate is E = σ/ε0

 

[SammyS]

Oppositely charged conducting parallel plates are ...

Read the problem you posted.

 

[flyingpig]

Oppositely charged conducting parallel plates are ...

Read the problem you posted.

Oh...right.

Did I get work problem right now? Because it is REALLY small