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Old Jan 22nd 2009, 07:05 AM   #1
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Capacitors basics

hiii.......
I have some problem with some basic questions

1) if we take a || plate capacitor (air inbetween) and connect it to a circuit then how 1 plate get '+'vely charged and other '-'vely charged (i.e why current starts flowing(circuit get completed) even when air gap due to capacitor is present.

2) where charge is stored in capacitor and how?

thanks to have a look
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Old Jan 22nd 2009, 07:48 PM   #2
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In a direct current circuit, there is a brief flow of current even though no electrons pass between the plates. After a certain amount of excess electrons runs onto one plate, their presence repels any other electrons that try to follow them - that's the way I look at it. Such brief currents are called "transients". The charge is stored on the plates. One plate has an excess of electrons.


When a capacitor is in an alternating current circuit, no current jumps across the plates. The voltage changes and this brings more or less electrons to the plates. There is an alternating current in the circuit. Since it has a regular pattern, it is not considered to be a "transient".
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Old Jan 22nd 2009, 10:59 PM   #3
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Originally Posted by tashirosgt View Post
1) In a direct current circuit, there is a brief flow of current even though no electrons pass between the plates.


2) After a certain amount of excess electrons runs onto one plate, their presence repels any other electrons that try to follow them - that's the way I look at it. The charge is stored on the plates. One plate has an excess of electrons.
1) how and why there is a brief flow of current as I mentioned circuit will not be completed due to air gap present.
Ok... Do you mean if we connect an ISOLATED metal sheet with negative terminal of cell then the sheet will get exess electrons.

2) one plate has excess of electrons. But due to induction I thing other plate will acquire equal but opposite charge don't you think so.
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Old Jan 22nd 2009, 11:46 PM   #4
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The excess electrons on the negative plate repel the free electrons on the positive plate. This can happen without electrons moving across the air gap ("action at a distance"). The positive plate is positive because it loses free electrons which are repelled by the charge on the negative plate. The battery is some device that tries to push (or "repel") electrons out of its negative terminal. Think of the first electron that is repelled by the battery terminal. It is also repelled by the electrons in the wire, and vice-versa. It moves a very small amount and its neighbor moves a small amount. The electron nearest the plate is nudged onto the plate. In a very short time the plate becomes so negatively charged that it is able to repel any more electrons fom coming onto it. An electron near the battery terminal is repelled by the battery terminal but also repelled by its neighbors, who are repelled by their neighbors,... etc. till we get to the electron that is trying to move on the plate and can't.

Yes, I think the plates will have an equal and opposite charge.

The best analogy I can make for the transient current is to imagine a garden hose with a balloon over the end. When you turn on the faucet there is a flow of water even though no water shoots out into the air. If the ballon is extremely strong, it could fill up to a certain amount and then the flow of water would stop when the pressure in the ballon equaled the pressure from the faucet. (Of course the average balloon would break first, just as capacitors can arc between the plates.)

If we have two electrons near each other and one moves toward the other, it is an interesting question of when the other electron feels an increased repulsion. I think the current theory is that the repulsion is not felt instantaneously. Instead the increased repulsion is transmitted at the speed of light.
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Old Jan 23rd 2009, 03:40 AM   #5
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Ok... Lets suppose that a metal plate whose area is equal to the area of New York is connected to the negative terminal of the dry cell, then will the dry cell get totally exhausted? (as electrons will get accumulated on it)
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Old Jan 23rd 2009, 07:53 AM   #6
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That's an interesting question! We need to consider more than the area of the plate (we need two plates). Your general idea is that a battery could be exhausted by charging a very large capacitor and that is true.

If a capacitor is charged by a battery and the battery is disconnected, a charge remains on the capacitor. In working on electrical equipment, you have to be careful about putting your hands around large capacitors even after the power has been shut off. You can get a shock from them. When you get such a shock, you are getting hit by a transient current. It's the discharge current instead of the charging current.

By the way, I found a site that has the equation for the transient current when a capacitor is charged.or discharged It is Capacitor Discharging.

The definition of "current" is that it is the rate of change of charge with respect to time. Batteries are usually rated in "amp hours". To answer your question quantitatively, we would have to figure out the total amp hours used to charge a large capacitor. Doing this for the transient would involve calculus. Have you studied calculus?.
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Old Jan 23rd 2009, 09:15 AM   #7
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Originally Posted by tashirosgt View Post
1) We need to consider more than the area of the plate (we need two plates).

2) Have you studied calculus?.

1) incase of two plates battery will surely get discharged (as per what you hav taught me so far) I meant what if there is only one plate (will baterry still get discharged)(this will help me to conclude why current flow even though air gap due to capacitor is present).

2)yes,I have studied calculus.
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Old Jan 23rd 2009, 09:47 AM   #8
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The ability of the capacitor to hold a charge is measured in units of "capacitance". I don't know the formula for figuring out the capacitance of a capacitor with one plate. Did you mean that it would have one big plate on one side and only a wire on the other side? Or do you mean a wire from one terminal of the battery to a big plate with no other wire? (In that case the other terminal of the battery would play the role of the other wire.) I don't think single plate arrangements have much capacitance relative to two-plate arrangements. So I don't think a single metal pie plate attached to a terminal of a battery would hold much charge. (About a plate with the area of New York, I don't know.)


The formula for the capacitance of a capacitor with two plates (of the same area) is given on the page ACEE Lesson Page

To understand a capacitor with one big plate and one very small plate (the end of a wire?) in a theoretical sense might be challenging. We would have to look at the electric field that such a setup generates. My intuition is this: An electron has an easier time crowding on to a plate if its presence pushes away the electrons on the opposite plate since that causes the opposite plate to have some positive charge. If there is no opposite plate, the electron has nobody to push away.
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Old Jan 23rd 2009, 10:28 AM   #9
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Ok... The setup that I meant consist of a battery,a wire and a metal plate.wire connects the negative terminal of the battery to the metal plate.(note that other terminal of battery remains as such its not connected to plate)
now will this plate have any negative charge accumulation?
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Old Jan 23rd 2009, 04:18 PM   #10
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It won't have much charge accumulation because the capacitance of that arrangement is very small. I think it would have some small charge accumulation, but I don't know if ordinary instruments could measure it.

A similar situation happens in cars. When you work on a car, you generally disconnect the negative terminal of the battery. But suppose you leave it connected and disconnect the positive terminal instead. The negative terminal of the battery is connected to the frame of the car, to he engine block and almost all metal in the car. That metal has a large area. But I wouldn't be afraid that a spark was going to jump out from the frame of the car and shock me. (However, if I was leaning on the frame and touched the positive terminal of the battery, I could get a shock. This is why the safe procedure is to disconnect the negative terminal.)
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