Capacitance Question

[abbarue]

I have a very strange question that's been on my mind lots lately.
Best way to ask it is with an example.

Say I have a 100 microfarad capacitor charged with 10 volts,
and then was able to bring the capacitor plates closer together,
to say half the distance apart.
What would happen to that 10 volt charge?
Would the voltage increase to say 20 volts?
Would it then have 200 microfarads of charge stored there?

My logical understanding of capacitance tells me something
has to increase, either voltage or stored charge or both.

Thanks.

 

[bananasiong]

As I know, capacitor is charged by the current, and the voltage of the capacitor is approaching the supplied voltage as time increases.
If it is fully charged when its capacitance is 100 uF, I don't think that the voltage will increase to 20 V when the capacitance is increased to 200 uF, maybe it will decrease.
BTW, 200 uF is capacitance but not charge.

 

[Leftyretro]

The voltage potential will stay the same. Charge potential and charge quantity are two different things.

Moving the plates closer will increase the capacitance and if still hooked to a voltage source will increase it's total charge quantity, but the charged voltage potential will remain that of the source voltage.

Now with some slight of hand tricks with multitudes of diodes and capacitors it possible to create a higher voltage then the charging source, hence voltage doubler and tripler circuits, however that has nothing to do with the fundamentals physics of a given capacitor.

I don't have the various capacitor formulas at hand but I'm sure an analysis of such formulas will bring light upon the question.

Lefty

 

[ronsimpson]

If you have a 100uf capacitor with a charge on it (10 volts). You decrease the spacing of the plates until the capacitor is now 200uF. What is the voltage? What is the energy stored?

Energy stored on a capacitor is one half capacitance voltage squared:
E=(1/2)CV^2

When you change the spacing on a capacitor the energy stored remains the same.
E=E
(1/2)CV^2=E=(1/2)CV^2
(100uf X 10^2)/2=(200uf X 7^2)/2
I think the voltage will drop to 7.07 volts.

 

[bananasiong]

Besides,
V = Q/C
As the amount of charge remain the same and the capacitance increases, the voltage will be decreased.
Does this make any sense?

 

[ronsimpson]

Almost
There is a 2 and square root missing.
If the capacitor gets bigger then the voltage must get smaller in order for the power remains constant.

 

[abbarue]

I thought that as the plates get closer together the static attraction
between them increases, thus the voltage would increase accordingly.
I was hoping I had stumbled upon a new source of energy.
Thankyou for the speedy reply.

 

[Pommie]

I thought that as the plates get closer together the static attraction between them increases, thus the voltage would increase accordingly. I was hoping I had stumbled upon a new source of energy.
Thankyou for the speedy reply.

Think of it this way, as the static attraction becomes greater there is less outward force to push the charge around a circuit and therefore less voltage.

Mike.

 

[johnnitwenty]

is there anyone think about another question, the energy wille decrease when the capacitance increase.according equality as this:
E=(1/2)CU^2
U=Q/C
then E=(1/2)(Q^2)/C
Q is constant at here,C increase,therefor E decrease.
where the energy has gone.

 

[bananasiong]

is there anyone think about another question, the energy wille decrease when the capacitance increase.according equality as this:
E=(1/2)CU^2
U=Q/C
then E=(1/2)(Q^2)/C
Q is constant at here,C increase,therefor E decrease.
where the energy has gone.

It is not energy but voltage (edited). The charge is still the same.
E is inverse proportional to C, that's why if C increases, E is reduced.

I=V/R, I is proportional to V. So when V increases, I will increase. Are you going to ask, 'where the current come from?'?

 

[johnnitwenty]

If you have a 100uf capacitor with a charge on it (10 volts). You decrease the spacing of the plates until the capacitor is now 200uF. What is the voltage? What is the energy stored?

Energy stored on a capacitor is one half capacitance voltage squared:
E=(1/2)CV^2

When you change the spacing on a capacitor the energy stored remains the same.
E=E
(1/2)CV^2=E=(1/2)CV^2
(100uf X 10^2)/2=(200uf X 7^2)/2
I think the voltage will drop to 7.07 volts.

what i refered to is this,energy is NOT equal at here.

 

[bananasiong]

what i refered to is this,energy is NOT equal at here.

Why shouldn't the energy be equal? Only the capacitance is increased right?

 

[johnnitwenty]

think over this question,you might know what i means.

 

[Nigel Goodwin]

Think of it like this:

You have a rubber bucket, that holds one gallon.

You pour one pint of water in it.

You then stretch the bucket until it can hold two gallons.

How much water does the two gallon bucket hold? - obviously still the original one pint.

Why make things difficult and complicated?.

 

[ljcox]

If you have a 100uf capacitor with a charge on it (10 volts). You decrease the spacing of the plates until the capacitor is now 200uF. What is the voltage? What is the energy stored?

Energy stored on a capacitor is one half capacitance voltage squared:
E=(1/2)CV^2

When you change the spacing on a capacitor the energy stored remains the same.
E=E
(1/2)CV^2=E=(1/2)CV^2
(100uf X 10^2)/2=(200uf X 7^2)/2
I think the voltage will drop to 7.07 volts.

No, the stored energy will change. Only the charge remains constant.

What everyone seems to be overlooking is that there is a force of attraction between the plates. Therefore, mechanical energy (positive or negative) is required to move the plates one way or the other. In this case, the plates are being moved closer together, so energy is removed from the system.

E1 = 0.5 C1 V1^2, E2 = 0.5 C2 V2^2.

So E2/E1 = C2/C1 (V2/V1)^2

In this example we know that C2/C1 = 2 and V2/V1 = 0.5 (since Q is constant)

Therefore E2/E1 = 2 (0.5)^2 = 0.5, ie. half the energy has been removed by the mechanical effort.

QED

 

[johnnitwenty]

Think of it like this:

You have a rubber bucket, that holds one gallon.

You pour one pint of water in it.

You then stretch the bucket until it can hold two gallons.

How much water does the two gallon bucket hold? - obviously still the original one pint.

Why make things difficult and complicated?.

yes, there still has one pint of water,so that the Q is equal,but the E is'n equal.

 

[johnnitwenty]

No, the stored energy will change. Only the charge remains constant.

What everyone seems to be overlooking is that there is a force of attraction between the plates. Therefore, mechanical energy (positive or negative) is required to move the plates one way or the other. In this case, the plates are being moved closer together, so energy is removed from the system.

E1 = 0.5 C1 V1^2, E2 = 0.5 C2 V2^2.

So E2/E1 = C2/C1 (V2/V1)^2

In this example we know that C2/C1 = 2 and V2/V1 = 0.5 (since Q is constant)

Therefore E2/E1 = 2 (0.5)^2 = 0.5, ie. half the energy has been removed by the mechanical effort.

QED

i have found a discussion about this question on another forum,energy stored in a capacitor will change when the capacitor plates move,electromagnetic wave should be produced.not only the exchange of energy between machanical energy and electric field,but also between electromagnetic wave
and electric field. the capacitance is changed by parallel connect with another capacitor ,no machanical energy is appended.

 

[ronsimpson]

I thought the question was theoretical. Energy cannot be destroyed.
0.5 C1 V1^2 = E =0.5 C2 V2^2
If C magically changes by moving the plates then V^2 must change.

LJCOX brings up an interesting point. Theory may not equal reality. It takes energy to move the plates.

Lets look at the question another way. We have talked about moving the plates together. In school (1,000 years ago) we made a capacitor. Charged it. Then greatly increased the spacing. The C became very small. The V became very large. This question was asked in ELE101 and proved in the lab. From my shocking experience the voltage goes up when the places are separated. We never thought about taking energy to move the plates.

1,000 years ago ELE 101 was called ELE CI and ELE 201 was called ELE CCI.

 

[Pommie]

Isn't this like the energy required to pull two magnets apart, the only energy that increases is the potential energy available that is released when they rejoin. The interesting thing here is, when the plates are moved together, potential energy is lost, if we then move them apart then we regain that energy. If we discharge them before we move them apart then we have lost some energy somewhere. Or, did we just dissipate that energy when we moved them together, a bit like lowering a weight.

Mike.

 

[ronsimpson]

Parallel Plate Capacitor Capacitor Charge, Plate Separation, and Voltage

The question is asked here:
https://www.csun.edu/~gsl05670/labs/cap_plate_sep_volt.htm
The answer is homework.