Capacitance of a parallel plate capacitor varies when the distance between the two parallel plates

Consider a parallel plate capacitor, with distance between plates [itex] = d_1 [/itex]. As we know the voltage between them [itex] V = Ed [/itex]. The electric field of two parallel plates is perpendicular to the surface and of the same intensity no matter where we are between the surfaces (accurate for small d's). Now, We approached the two parallel plates to each other so as the distance is reduced to half. [itex] d_2 = \frac{1}{2} d_1[/itex]. So now, the voltage between the plates is reduced to half too. This means that to transfer a positive charge from the negative plate to the positive plate you require half of the work now.

There is the contradiction: How is the work required is sliced in half to transfer a charge from a plate where the fields acting on it stayed the same to another plate where the fields on it stayed the same too??

The voltage on the plates will change when the capacitance is changed by changing plate separation.
Capacitance is defined as; C = Q / V, if the charge is fixed then V must change with 1/C.

No information is being given whether the capacitor is isolated or still connected to a supply.

The voltage on the plates will change when the capacitance is changed by changing plate separation.
Capacitance is defined as; C = Q / V, if the charge is fixed then V must change with 1/C.

How to solve this contradiction.

No information is being given whether the capacitor is isolated or still connected to a supply.

The electric field gradient may not have changed, but the distance you must move the charge between plates has been reduced. When you halve the plate separation you double the capacitance. With a fixed charge the voltage halves, C = Q / V.

Energy before is E = ½ C * V2;

The electric field gradient may not have changed, but the distance you must move the charge between plates has been reduced. When you halve the plate separation you double the capacitance. With a fixed charge the voltage halves, C = Q / V.

Energy before is E = ½ C * V2;

Energy after is E = ½ * 2*C* (V/2)2 = ¼ * C * V2
So energy was released when you allowed the plates to be moved closer together.

Hope this helps op with the concept..

Thank you for all these explanations, they make perfect sense from the overall point of view. However, They must also make sense from the point of view of the voltage source. from this point of view, nothing is changed. Assuming I am battery, I must move an electron from the negative plate (with the same profile of electric fields in both cases) to the positive plate (with the same profile of electric fields in both cases), so logically, I must expend the same energy on this electron in both cases.

Whereas from the explanations you mentioned above, distance is halved means voltage is halved means I must expend half the amount of energy to move an electron from - to +. Therein lies the Contradiction.

How is the work required is sliced in half to transfer a charge from a plate where the fields acting on it stayed the same to another plate where the fields on it stayed the same too?? ........... Assuming I am battery, I must move an electron from the negative plate (with the same profile of electric fields in both cases) to the positive plate (with the same profile of electric fields in both cases), so logically, I must expend the same energy on this electron in both cases.

Whereas from the explanations you mentioned above, distance is halved means voltage is halved means I must expend half the amount of energy to move an electron from - to +. Therein lies the Contradiction.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html

old jim

They must also make sense from the point of view of the voltage source. from this point of view, nothing is changed. Assuming I am battery, I must move an electron from the negative plate (with the same profile of electric fields in both cases) to the positive plate (with the same profile of electric fields in both cases), so logically, I must expend the same energy on this electron in both cases.

If you think there is a "paradox" here because "the two situations are the same", the mistake is that they are NOT the same.

You are missing the basic difference between the two situations. If the capacitor is isolated, the number of electrons on each plate stays constant when you move the plates, and the voltage between the plates changes as the capacitance changes. If the capacitor is corrected to a battery, the voltage between the plates stays constant (equal to the battery voltage) but the number of electrons on the plates changes when the distance changes. Current flows into or out of the battery, while the plates are moving.

If you think there is a "paradox" here because "the two situations are the same", the mistake is that they are NOT the same.

The question is: why is the capacitance halved? If we look at it from purely field-energy point of view we can't (at least me) see the difference

Seems to me the answer lies in the definitions firstly of capacitance, which is coulombs per volt, Q/V;

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html

secondly voltage, which is joules per coulomb;

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html

logically when you halve the distance you double the capacitance (because C=εA/D ) ; and since you chose to keep numerator of your capacitance Q/V ratio constant( by expressing charge in electrons), in order to double capacitance you must halve your denominator. Good questions. It is necessary to make your mental model agree with the math, else you have a flawed mental model. These things are good to ponder in your morning shower , or a boring office meeting . EDIT i dont understand this statement "(with the same profile of electric fields in both cases)" maybe that's the flaw in your mental model ? What profile? Edit again - Baluncore said same thing way up in post #2 ....

old jim

If we are to move an electron from + to - through the wire, we need to apply a force equal and opposite the electric field force at every point along the wire, and to tick the electron with infinitesimal force just to get it moving. We need half the work in 2 to do the same thing as in 1. So [itex]∫_LEq\,dx[/itex] is halved, [itex]q[/itex] stayed the same, [itex]L[/itex] (length of the wires) stayed the same, I suppose [itex]E[/itex] (along the path of motion or at least at some segment) decreased, how and where?

If we are to move an electron from + to - through the wire, we need to apply a force equal and opposite the electric field force at every point along the wire,

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html

Conductor at Equilibrium For a conductor at equilibrium: 1. The net electric charge of a conductor resides entirely on its surface. (The mutual repulsion of like charges from Coulomb's Law demands that the charges be as far apart as possible, hence on the surface of the conductor.)

2. The electric field inside the conductor is zero. (Any net electric field in the conductor would cause charge to move since it is abundant and mobile. This violates the condition of equilibrium: net force = 0.)

Its energy is deposited in the dielectric, maybe aligning the polar molecules of an oil or plastic dielectric or maybe just 'warping the fabric of space'* according to ε0 .

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c1

Work and Voltage: Constant Electric Field The case of a constant electric field, as between charged parallel plate conductors, is a good example of the relationship between work and voltage.

The electric field is by definition the force per unit charge, so that multiplying the field times the plate separation gives the work per unit charge, which is by definition the change in voltage.

Assuming I am battery, I must move an electron from the negative plate (with the same profile of electric fields in both cases) to the positive plate (with the same profile of electric fields in both cases), so logically, I must expend the same energy on this electron in both cases.
Whereas from the explanations you mentioned above, distance is halved means voltage is halved...

We never connected a battery.

If a charged capacitor is disconnected, then the charge is fixed. If you then halve the distance between the plates, the capacitance will be doubled. Energy will be released when the plates are moved together because the plates attract each other. C = Q / V; Where charge is fixed, doubling the capacitance must halve the voltage. The force on the electron is proportional to the electric field gradient, V/m. The total distance moved by the electron is the distance between the plates, m. The energy difference for the electron is therefore E = m * V / m = V. That is why, in particle physics, one unit of energy is the electron volt = eV. Half the distance implies half the voltage = half the electron energy. Which is exactly what you can expect since half the energy was released when you allowed the plates to move half way together.

Remember from earlier that E = ½ * C *V2, but that both C and V2 changed when the plates moved together.

If you can't decide whether you connected a battery or not, then I'm not going to waste time trying to answer your questions.

If a charged capacitor is disconnected, then the charge is fixed. If you then halve the distance between the plates, the capacitance will be doubled. Energy will be released when the plates are moved together because the plates attract each other. C = Q / V; Where charge is fixed, doubling the capacitance must halve the voltage. The force on the electron is proportional to the electric field gradient, V/m. The total distance moved by the electron is the distance between the plates, m. The energy difference for the electron is therefore E = m * V / m = V. That is why, in particle physics, one unit of energy is the electron volt = eV. Half the distance implies half the voltage = half the electron energy. Which is exactly what you can expect since half the energy was released when you allowed the plates to move half way together.

Remember from earlier that E = ½ * C *V2, but that both C and V2 changed when the plates moved together.

well now i get confused, because my mental model of a wire says you can't have an electric field inside one.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html

My mental model of a capacitor has the energy stored not on the plates but in the dielectric. So to move your electron(more correctly one electron's worth of charge) from negative plate to positive plate, it flows from negative plate to right hand dot of your open circuit at constant potential, ie no increase of energy then through your "open circuit" where it picks up some energy, then from left hand dot to positive plate again at constant potential.

Its energy is deposited in the dielectric, maybe aligning the polar molecules of an oil or plastic dielectric or maybe just 'warping the fabric of space'* according to ε0 .

*(which is to me one of life's great mysteries - why does space have dielectric and magnetic properties?)

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c1

Might that be a contributor to your perceived paradox?

I still didn't get where the electric field we are working against exists from the explanation so I want to propose another slightly modified example that answers the same question.

In the Image attached, we have two tubes of glass each filled with charged metal and they are in space. As we know, the electric field is only between them and negligible in other places. If we moved an electron worth of charge from - to + we have execute some work on it. In case 1 we have to execute double the amount of work as in case 2.

I see that the electric field on the plates are the same, so where is the increase in electric field force that we have to work against? Can you describe this to me on the picture?

What I fail to understand is if we move the electron along a path "not" between the plates but "around" the plates. Since there is no field around the plates, where is the force countered responsible for the work done on the electron (or the potential energy increase of the electron)?

If the electron starts away from the capacitor and circulates once around the capacitor, in a closed path, it will see no net change in energy. If it is not in the capacitor's field, then it is irrelevant to the capacitor analysis.

In the Image attached, we have two tubes of glass each filled with charged metal and they are in space.

Power = voltage * current, half the voltage, for the same electron current, implies half the power.

Thank you Jim and Baluncore