Electric field and distance in Parallel-Plate Capacitor

In summary, when the distance between the plates of a parallel plate capacitor is reduced to half, the voltage between them is also reduced to half. This is because the capacitance is doubled, resulting in a decrease in voltage with a fixed charge. This decrease in voltage leads to a release of energy when the plates are brought closer together, as the opposite charges on the plates attract each other. This may seem contradictory from the perspective of the voltage source, as the amount of work required to move an electron from the negative to the positive plate should theoretically be the same in both cases. However, this contradiction can be resolved by understanding the definitions of capacitance and voltage, and making sure that our mental model aligns with the mathematical equations.
  • #1
Amerez
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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??
 
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  • #2
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.
 
  • #3
No information is being given whether the capacitor is isolated or still connected to a supply.
 
  • #4
Baluncore said:
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.

Well, the question is does this happen? We have same electric field profiles before and after so this shouldn't happen.
How to solve this contradiction.
 
  • #5
grzz said:
No information is being given whether the capacitor is isolated or still connected to a supply.

It doesn't make a difference as to the contradiction, let it be we isolated the capacitor for simplicity.
 
  • #6
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.
 
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  • #7
Baluncore said:
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.

right on.
It was explained to me in beginning electronics class thus:
The opposite charges on the plates attract one another. So when you let them move closer together you get mechanical work out, Force X Distance. If instead you pulled them apart you would do mechanical work on them and that would show up as increased voltage.
At that point we boys had not yet been exposed to integrals so teacher didn't ask us to calculate it.
Hope this helps op with the concept..
 
  • #8
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.
 
  • #9
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.

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 don't 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
 
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  • #10
Amerez said:
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.

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.
 
  • #11
AlephZero said:
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.

We never connected a battery.
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
 
  • #12
jim hardy said:
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 don't 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

Indeed This was how I passed my last 2 office meetings :D

Same profile of electric field: same magnitude, shape and direction of electric fields.

Image attached:
What I don't understand is the the energy perspective, [itex] W = ∫_LEq\,dx [/itex]
[itex]E[/itex]: electric field, [itex]q[/itex]:charge, [itex]L[/itex]:distance the force was applied.
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?
Also note that the electric field between the plates stayed the same.
 

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  • #13
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,
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

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.)

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
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.

ewor.gif


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.

ewor2.gif

ewor3.gif

Might that be a contributor to your perceived paradox?
 
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  • #14
Amerez said:
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...

Amerez said:
We never connected a battery.

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.
 
  • #15
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.
 
  • #16
AlephZero said:
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.

I explained when we connected a battery. no need for negativity. Thanks
 
  • #17
Baluncore said:
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.

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)?
 
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  • #18
:frown:
jim hardy said:
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 know there is no electric field inside a conductor, I thought I could say it "move it against an electric field (of zero when we don't have one)" equally well, sorry for the confusion

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.
 
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  • #19
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?
 

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  • #20
Amerez said:
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)?
The charged capacitor analysis cannot have it both ways. Either the capacitor is connected to a charge transfer circuit such as a resistor or a battery, or it is open circuit with a fixed charge. If one electron moves from one plate to the other, then the plates are connected by that electron's path, and the charge is changing.

If an electron starts at one of the charged plates and travels to the other plate by any route whatsoever, the electron will change it's energy. The force on the electron will be countered by the electric field and therefore, in turn, by the physical capacitor's plates.

Since the oppositely charged plates attract, the dielectric insulator that keeps the plates apart counters the force due to electric charge attracting the plates. Moving an electron between the plates changes the charge and therefore the force of the plates on that dielectric structure.

Any conductor connected to a plate is part of that plate. There really is an electric field outside the inner volume. The simple assumption, (for the purpose of analysis), that the field is entirely within the restricted volume between the physical plates, is totally invalidated by considering the real electric field that must be outside the plates.

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.
 
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  • #21
Amerez said:
In the Image attached, we have two tubes of glass each filled with charged metal and they are in space.
If there is charge difference between the tubes then there must be an electric field across the two dielectric glass envelopes and the space between them.

What stops the electrically attracted glass tubes from accelerating towards each other, then crashing, breaking the glass and neutralising the charge?

You appear to be hypothesising a capacitor with charge, but without an electric field.
How are you going to move your test electron through the glass dielectric?

C = Q / V. A charge difference implies a voltage difference V = Q / C, where C is the capacitance between the different charges.

If the charges are the same in the two cases you present, and only the separation is different, then the voltages will be different because the capacitance is different.

Power = voltage * current, half the voltage, for the same electron current, implies half the power.
 
  • #23
Thank you Jim and Baluncore
 

1. What is an electric field?

The electric field is a physical quantity that describes the influence that an electric charge has on other charges in its vicinity. It is a vector quantity, meaning it has both magnitude and direction.

2. What is a parallel-plate capacitor?

A parallel-plate capacitor is a type of capacitor consisting of two parallel conducting plates separated by a small distance. It is used to store electrical energy by creating an electric field between the plates.

3. How is the electric field strength calculated in a parallel-plate capacitor?

The electric field strength in a parallel-plate capacitor can be calculated by dividing the voltage difference between the plates by the distance between them. It is also affected by the material between the plates and the size of the plates.

4. How does the distance between the plates affect the electric field in a parallel-plate capacitor?

The electric field in a parallel-plate capacitor is inversely proportional to the distance between the plates. This means that as the distance between the plates increases, the electric field strength decreases.

5. What is the significance of the electric field in a parallel-plate capacitor?

The electric field in a parallel-plate capacitor is important because it determines the amount of energy that can be stored in the capacitor. It also plays a role in the capacitance of the capacitor and the behavior of the charges on the plates.

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