Am I correct to think that the higher the charge between two plates in a parallel plate capacitor the higher the voltage difference? Is there an equation relating these difference in charge between the plants and the voltage difference between the plates?

Yes, you're correct. For a classical plate capacitor the equation you're looking for is:



where Q = the charge on the plates (+Q on one plate, -Q on the other), V is the voltage between the plates, and C is the capacitance of the device. The capacitance is found from:



where is the permittivity of the material between the plates, A is the area of the plates, and d is the distance between the plates. So for a given capacitor if you increase Q you also increase V.

Thanks a lot

Can this idea be applied to a uniform field?
What I mean is can I extrapolate that if voltage difference between two plates is halfed the charge on the plates is also halved?

Quote:

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Originally Posted by Roddilla

can this idea be applied to a uniform field?
What I mean is can I extrapolate that if voltage difference between two plates is halfed the charge on the plates is also halved?

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Yes. C remains constant, so if you cut V in half you also cut Q in half.

I cannot believe how if for example I have an established voltage between two parallel plates (which form a capacitator) and you change the area you manage to change the charge as well.
If you have a 9V battery connected to a capacitor the charge should be built according to the voltage since more electrons can move if the voltage of the battery is higher. So if I suddenly change the area aren't there the same number of electrons and thus the same charge? I know that there is the equation but I want to go beyond that

In fact it's true - if you increase the area there is more space for more electrons to be packed into one plate (and removed from the other) while maintaining the same charge density. Think of it this way: the higher the voltage the more charge per square inch is possible.

Yes but electrons are induced to move because of voltage difference. If the capacitor is charged and you change the area the pd should remain the same between the plates of the capacitor and electrons still move? I know that the equation states so. The fact that the larger the plates are the more electrons can move makes sense but the first statement which I made confuses a bit the things.

What I mean is if pd across capacitor has already been established why should electrons move even more?

Again: the voltage difference between the plates controls the charge density, not the total charge. Here's why: think of the voltage as a force that causes electrons to be packed together more tightly than "normal." So if you increase the voltage you get more electrons per square inch. This also means that if you increase the area of the plates and maintain a constant voltage, the same number of electrons per square inch times more inches means a greater number of electrons, and hence a greater total charge on the plate.

If the capacitance was not a function of area then we would find that wiring multiple capacitors in parallel would not increase overall capacitance above that of a single capacitor. But we know that in fact capacitors in parallel are additive.

You could think of it as a column of water too.

http://p1cture.it/images/b619c05g32416g34d6f8.png

When the area is small, the electrons are concentrated at a point, but when the area is large, the electrons are spread wider across the surface. There is the same volume of water, but the area is affects is greater, hence the depth of the water must be lower. Remember that electrons repel each other. And since the depth is now lower, there is more space for more electrons (space until the depth is similar to that in the first case, assuming that it was the maximum depth of the water).

Very help full answers indeed.

One last question to conclude: I have know understood how charge density is affected by the area. The smaller the charge density the less repelled the electrons are and the more they can gather. But the total charge remains the same whether you increase the area or not.

But...

The equation is capacitance = charge / voltage

NOT

Charge density / voltage

So according to this equation by changing the area of the capacitance you are changing the charge not the charge density.

So, I am in a bit of doubt...

Still thanks a lot for your help.

By changing the area, we are changing the charge density (the charges are further repelled) and the charge remains constant (volume of water the same). So, with the actual voltage (work done is less to move more electrons to the plate as they are further apart, which means lower repulsive forces), you can get more electrons on the plate, until the charge density becomes the same as in the first case (small plate).

And through so, you are getting more charge.

C = Q/V

V is constant, C has been increased due to increase in area, so Q automatically increases.

I cannot understand what you are saying sorry very much but I take time to understand.


What do you want to exactly mean by saying that total charge doesn't change but charge on plates changes?

No no, initially, the charge on the plates do not change. We're talking about split seconds or maybe even smaller than that, or maybe into abstract...

It's like, if you only increased the area, the charge density decreases, because the area has increased, and the charge has remained constant.

But because you keep a constant potential difference, as soon as the charge density decreases, more charge comes on the plate to restore the initial charge density. Thus, also how the capacitance is increased.

So charge does change?

Well I think I've got it.
When you attach the battery to the circuit containing capacitor a certain charge is created. In this case it is the voltage of the battery which is causing the change in the charge from 0 to Q. The battery acts as a pump to move electrons.

Then you change the area and electrons move once more to restore the charge density. As a result charge increases.

If that's it then thanks a lot for your help/

One final conclusion:

Voltage in practice increases if charge density increases since there is more charge per unit area.
This is why voltage stays the same when you increase the area. Charge increases in such a way so that charge density remains the same so that voltage remains the same. It must remain the same since the battery attached has a fixed voltage.

BIG THANKS FOR YOUR HELP

Am I correct?

Well, that's how I understand it anyway. :)