# B Why/how does distance affect capacitance?

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1. Aug 6, 2017

### Joshua Sejin Yoon

As distance between two capacitor plates decreases, capacitance increases - given that the dielectric and area of the capacitor plates remain the same.

So, why does this occur?

2. Aug 6, 2017

### stevendaryl

Staff Emeritus
3. Aug 6, 2017

### sophiecentaur

Capacitance is the amount of charge imbalance (Q) for a given Potential difference. When the plates are closer together, the same PD, applied across the plates will bring opposite charges on the two plates closer together. They will attract each other more (reduced spacing) and so even more charges will be attracted to the faces (it takes only a minuscule field to make the charges in the metal move towards the face. So, by the way Capacitance is defined (more charge, same PD) , the Capacitance is increased.
If you put a dielectric between the plates, the molecules in faces of the the dielectric are much nearer to each plate and the molecules inside the dielectric will polarise easily. So, again, the actual charge imbalance on the plates will be greater for a given PD. Another way I look upon a dielectric is that it's a bit like putting a conductor in between that's just a tiny bit narrower than the original gap. That is producing a capacitor with a much smaller gap and, hence, a greater Capacitance. The only problem is that such a narrow gap between metal plates could flash across with only a tiny PD. A dielectric will only allow Polarisation and not a flow of charge so no 'short circuit'.

4. Aug 6, 2017

### Joshua Sejin Yoon

Does that mean that capacitance is:
- inversely proportional to Electric field strength (E)
- inversely proportional to Distance (d)
- proportional to Charge (Q)
- proportional to Area (A)

And because distance affects capacitance and distance is proportional to the Electric field strength, is capacitance affected by the Electric field strength?

Last edited: Aug 6, 2017
5. Aug 6, 2017

### sophiecentaur

The Capacitance is a description of the Quality of an object. It depends on the dimensions and materials used and not the conditions it is exposed to. A mathematical formula is only a formula. It does however, tell you how to Calculate the Capacitance, if you measure the Charge and PD.

PS. There are materials for which C is not constant for all values of charge and PD, in the same way that there are materials with a Resistance that is not the same for all currents (i.e. Non-Ohmic) But that is definitely not Introductory Level.

6. Aug 6, 2017

### Joshua Sejin Yoon

If I, for example, have paper as the dielectric, and the distance between the two capacitor plates were decreased (so there are less "molecules" of paper), why and how does this increase capacitance? Shouldn't the plates hold more charge if there are more polarised molecules in the dielectric, as the pull on the nucleus will be greater (due to all of the electrons), and thus the atom's electrons will be pulled towards the nucleus with greater force, allowing more charges on the capacitor plates?

7. Aug 6, 2017

### sophiecentaur

The best explanation for this is that there is no shortage of polarisable molecules in the dielectric and, of course, the applied Electric field (Volts per metre) would be Increased with a narrower gap.

8. Aug 6, 2017

### stevendaryl

Staff Emeritus

Capacitance $C$ is defined as $\frac{Q}{V}$.

The voltage between two parallel plates, one with a charge $+Q$ and one with a charge $-Q$ is computed as:

$V = E d$

where $E$ is the electric field, and $d$ is the distance between the plates. The electric field, in turn, for two parallel plates is computed by:

$E = \frac{\sigma}{\epsilon}$ where $\sigma$ is the charge density (charge per unit area).

So putting it all together,

$C = \frac{Q}{V} = \frac{Q}{E d} = \frac{Q}{\frac{\sigma}{\epsilon} d} = \frac{Q}{\frac{Q}{A} \frac{d}{\epsilon}} = \frac{\epsilon A}{d}$

where $A$ is the area of one plate, and $d$ is the distance between the plates. So the charge drops out, and the only quantities that $C$ depends on are the area of the plates, the distance between them, and the dielectric constant.

9. Aug 6, 2017

### pixel

Just to clarify, this derivation is for large i.e. infinite parallel plates.

10. Aug 7, 2017

### jartsa

Let's say a charged capacitor A is two metal plates 1 cm apart. And a non-charged capacitor B is two metal plates 0.1 cm apart.

When we let the side of capacitor B touch the side of capacitor A, what happens?

Well, opposite charges like to be close to each other, so the charges go from the capacitor A to the capacitor B. We say capacitor B has larger capacitance than capacitor A, when charges prefer to be go from A to B.

Or maybe this is better:

Negative charges like to be far from each other, so some negative charges go from the negatively charged plate of A to the originally neutral plate of B. And positive charges like to be far from each other, so some positive charges go from the positively charged plate of A to the originally neutral plate of B. Seen from A the charged capacitor B looks quite neutral, because B contains equal amount of positive and negative charges almost at the same place. So almost all charges go from A to B.

11. Aug 7, 2017

A slightly different way of looking at it:
1. Charge the capacitor and disconnect it from the supply.
2. Pull the plates apart by a method that does not allow charge to leak away. In order to do this work must be done in order to overcome the force of attraction between the plates. This results in an increase in V between the plates and since C=Q/V, C decreases.

12. Aug 7, 2017

### Joshua Sejin Yoon

I am having trouble understanding why the voltage increases as the force of attraction is overcome.
Could you please elaborate as to why this occurs?

13. Aug 7, 2017

### sophiecentaur

The plates are attracted to each other. You exert a force against this and, moving away, you do Work. Like when you stretch a spring. SO the Energy per unit charge increases - that's Potential Difference.

14. Aug 7, 2017

### Joshua Sejin Yoon

Could this relate to when capacitor plates are charged at a certain distance (x), then discharged, and charged at another distance (y), then discharged, and the capacitance at those two points differing? If so, how?

15. Aug 7, 2017

### sophiecentaur

The Capacitance depends just on the geometry and the dielectric - not on the charge. to get the same charge on the plates with different separation you need different voltages.BUT putting the same PD across the plates at different distances will change the Charge.
That's what follows from Q=CV.

16. Aug 7, 2017

### Joshua Sejin Yoon

Just to clarify one more thing: Does the distance between the two capacitor plates, therefore (assuming PD is same), affects the charge on each plate?

17. Aug 7, 2017

### sophiecentaur

Q=CV is the way to put it. C is the Capacitance. The "charge on each plate" will be equal and opposite if the whole thing started off at Earth Potential.

18. Aug 7, 2017

### Joshua Sejin Yoon

Does this mean that electric fields affect capacitance, and distance affects electric fields (assuming voltage is the same)?

19. Aug 7, 2017

### jbriggs444

Electric fields do not affect capacitance. This has been explained already. A particular capacitor has the same capacitance whether it is fully charged, half charged or fully discharged. Capacitance is the ratio of charge to potential difference for that capacitor.

If you take a fully charged capacitor and separate the two plates (doing work as @sophiecentaur suggests) while not permitting any charge to flow you will have left the electric field strength in the gap unchanged. The potential difference between the two plates is given by field strength times separation distance and will have increased. The charge on each plate will have remained unchanged. The ratio ($\frac{charge}{potential\ difference}$) will have decreased with the increase in potential difference. Capacitance is that ratio.

20. Aug 7, 2017