# Electric field and distance in Parallel-Plate Capacitor

1. Jul 6, 2014

### Amerez

Consider a parallel plate capacitor, with distance between plates $= d_1$. As we know the voltage between them $V = Ed$. 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. $d_2 = \frac{1}{2} d_1$. 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??

Last edited: Jul 6, 2014
2. Jul 6, 2014

### Baluncore

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. Jul 6, 2014

### grzz

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

4. Jul 6, 2014

### Amerez

Well, the question is does this happen? We have same electric field profiles before and after so this shouldn't happen.

5. Jul 6, 2014

### Amerez

It doesn't make a difference as to the contradiction, let it be we isolated the capacitor for simplicity.

6. Jul 6, 2014

### Baluncore

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.

Last edited: Jul 6, 2014
7. Jul 6, 2014

### jim hardy

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. Jul 6, 2014

### Amerez

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. Jul 6, 2014

### jim hardy

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

Last edited: Jul 6, 2014
10. Jul 6, 2014

### AlephZero

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. Jul 8, 2014

### Amerez

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. Jul 8, 2014

### Amerez

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, $W = ∫_LEq\,dx$
$E$: electric field, $q$:charge, $L$: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 $∫_LEq\,dx$ is halved, $q$ stayed the same, $L$ (length of the wires) stayed the same, I suppose $E$ (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.

#### Attached Files:

• ###### ScannedDocument001.jpg
File size:
10.7 KB
Views:
117
13. Jul 8, 2014

### jim hardy

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

Last edited: Jul 8, 2014
14. Jul 8, 2014

### AlephZero

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. Jul 8, 2014

### Baluncore

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. Jul 10, 2014

### Amerez

I explained when we connected a battery. no need for negativity. Thanks

17. Jul 10, 2014

### Amerez

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)?

Last edited: Jul 10, 2014
18. Jul 10, 2014

### Amerez

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.

Last edited: Jul 10, 2014
19. Jul 10, 2014

### Amerez

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?

#### Attached Files:

• ###### fields.jpg
File size:
28.1 KB
Views:
105
20. Jul 10, 2014

### Baluncore

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.