Why is capacitance inversely related to the distance of the conductor plates

[yhelothar]

Say if there is an x amount of charge in a capacitor. If the distance of the conducting plates is 1mm, its potential is greater than if the distance is 2mm.

|| > | | in voltage

A capacitor can only hold charge up till the charge creates a potential equal to the potential of the voltage source. Thus this would mean if the distance is smaller, it would take less charge for this to occur. Thus it'd be storing less charge per volt?

But the equation shows that capacitors store more charge per volt if the distance is smaller.

Where is my reasoning incorrect?

 

[bobdole369]

The closer the plates are, the more effect the electric fields have on one another, therefore the more electrons can be "stored" in the plate.

The voltage limit is of course the dielectric's breakdown voltage. And the "charge limit" which is in coulombs - is basically volts x amps x time (or another way to say - y amount of amperes will flow). The capacity is measured in farads which is directly proportional to the coulombs that can be stored. Your voltage isn't measured by the amount of charge. Voltage isn't stored at all.

 

[yhelothar]

OK I'm confused. Why is it that a stronger electric field would mean more electrons can be stored in the plate? It seems that the opposite would occur. A stronger electric field would mean less charge is required to produce a voltage equal to the voltage source.

The amount of electrons a plate can store depends on the strength of the voltage source. The voltage source drives a current into the capacitor. As the capacitor builds up charge, it produces a voltage in the polar direction, thus negating the the voltage source. A capacitor can only gain charge until the voltage potential of the capacitor is equal to the capacitor of the voltage source, as at this point, no net current would be flowing into the capacitor.

Why is voltage not related to the amount of charge? As charge flows into the negative terminal of the capacitor, a potential builds up due to the lack of charge in the positive terminal. By looking at the time constant graphs, it is clear that the more charge that flows into the capacitor, the greater the potential in the battery.

The capacity measured in farads, is a measurement of charge per unit of voltage. The amount of charge that a capacitor can hold is thus directly proportional to the voltage source.

http://electronics.howstuffworks.com/capacitor1.htm

 

[TecHNooB]

If I said the E field for an infinite sheet of charge was the same everywhere regardless of where you were (ignoring direction), would that blow your mind? Cuz it blows my mind :O

What's messing you up is that maybe you're imagining single point charges when thinking about the E field. You need to imagine more! The charges contribute radial E fields so when you integrate from plate 1 to plate 2, not all the E fields are pointing in the direction of integration so their contribution is less. Bringing your test point close to 1 plate will cause some charges to contribute more E field due to lesser distance but others to contribute less because the angle of the E field vector becomes closer to 90 with respect to your direction of integration. I also confused myself when thinking about halving distances vs 1/r^2 relationship (would have 4 times the E field for half the distance).

 

[exdeath]

To keep it simple...

The relationship between the plates is attraction, the - are attractive to the +.

Forget capacitors for a minute and realize that a natural law in electromagnetism is that opposite charges attract and the force of attraction follows an inverse square function. Halving the distance quadruples the attraction force.

In a capacitor you also have a repulsion force of like charges piling up on a single plate preventing charge from accumulating normally.

The closer the opposite charged plates are together, but never touching, the stronger the attraction (due to inverse square), such that more charge on the same plate can pile up and "overcome" the repulsion of like charges on their own plate due to attraction of charge on the opposite plate.

One electron is not going to follow a path to a plate holding a million other like charge electrons. If you move the plates slightly closer, the force of attraction is higher between unlike charges and provides the motivation for that million+1 electron. Thus stored charge capacity is greater at the same potential.

It might help if you start with basic Coulomb's / Gauss's law and integrate the charge over two opposite charged plates of area pi*r*r separated by a distance d and work your way to deriving the formula for a classic parallel plate capacitor and see where the inverse square of separation distance comes in and contributes to the capacitance.

Capacitance is a measurement of charge, not voltage. Voltage is what "moves" electrons, but the actual electrons are what are stored over time at a given voltage. The potential can only be several volts, but the charge can be anything from a few picofarads to several farads, and that varies as a function of plate geometry even when voltage is held constant.

Start with a 1 Farad 1 Volt capacitor. This stores 1 Coulomb. Half the distance between the plates and you get a 4 Farad 1 volt capacitor that stores 4 Coulombs. Double the voltage as well to 2 volts and now you have a 4 Farad 2 volt capacitor that stores 8 Coulombs.

Also understand how the unit of Farad is defined and derived. http://en.wikipedia.org/wiki/Farad

Or to keep it simple again, think of a AAA battery vs a D battery. Same 1.5 volts but there is much more "charge" in a D cell.

 

[exdeath]

Oh this should work for you

From http://electronics.howstuffworks.com/capacitor2.htm

Farad
A capacitor's storage potential, or capacitance, is measured in units called farads. A 1-farad capacitor can store one coulomb (coo-lomb) of charge at 1 volt. A coulomb is 6.25e18 (6.25 * 10^18, or 6.25 billion billion) electrons. One amp represents a rate of electron flow of 1 coulomb of electrons per second, so a 1-farad capacitor can hold 1 amp-second of electrons at 1 volt.

 

[TuxDave]

Say if there is an x amount of charge in a capacitor. If the distance of the conducting plates is 1mm, its potential is greater than if the distance is 2mm.

This is where you're wrong. If they contain an equal amount of charge, using Gauss' Law they will have equal E field and therefore the one that is further apart will have a greater voltage since you have to pass through more distance in the same E field.