Capacitor Tutorial 3  Physics of Capacitors
If we pump electrons onto the negative plate, electrons are repelled from the negative plate. Since positives do not move, a positive charge is induced. The higher the potential difference, the more charge is crowded onto the negative plate and the more electrons repelled from the positive plate. Therefore charge is stored. The plates have a certain capacitance.
Capacitance is defined as:
Since we have a
positive and negative plate, we have an electric
field. Capacitance is
measured in units called farads (F) of which the definition is:
1
Farad is the capacitance of a conductor, which has potential difference of 1
volt when it carries a charge of 1 coulomb.
So
we can write from this definition:
Capacitance
(F) = Charge (C)
Voltage (V)
In code, this is written:
C = Q
V
A
1 farad capacitor is actually a very big capacitor indeed so instead we use microfarads (mF)
where 1 mF
= 1 ×
10^{6} F. Smaller
capacitors are measured in nanofarads
(nF), 10^{9 }F, or picofarads
(pF), 1 ×
10^{12} F. A working
voltage is also given. If the
capacitor exceeds this voltage, the insulating layer will break down and the
component shorts out. The working
voltage can be as low as 16 volts, or as high as 1000 V.
The voltage rises as we charge up a
capacitor, and falls as the capacitor discharges.
The current falls from a high value as the capacitor charges up, and
falls as it discharges. We can see this in the graph below.
The behaviour of a capacitor is a bit like a rushhour commuter train on its journey into a major city. At the remote terminus station, it is easy for passengers to find a seat. However, as the train calls at each station up the line, it gets harder for passengers to find seats. Then they have to stand as the train gets more crowded. At the last station before the big city, the train may be so crowded that no more passengers can get on.
Why does a capacitor allow AC to flow?
If connected to a d.c. circuit, the bulb flashes, then goes out.
In
an a.c. circuit, the bulb remains on.
We can say that a capacitor blocks
d.c., but allows a.c. to flow.
This circuit can be used to measure the value of a capacitor:
The reed switch is operated from a 400 Hz supply.
It operates on the forward half cycle (why?), to charge up the capacitor.
No current flows on the reverse half cycle so the reed switch flies back to discharge the capacitor.
We can use
I
= Q/t to work out the charge going onto the plates.
We also know that f = 1/t,
so we can combine the two relationships to give
I
= Qf, therefore
Q = I/f
Since
C = Q/V, we can now write
C
= I/fV
Worked Example
A capacitor is
connected to a 12volt power supply by a reed switch operating at 400 Hz.
The ammeter reads 45 mA. What
is the capacitance of the capacitor? 
Answer C = 0.045 A × (400 Hz × 12.0 V) = 9.38 × 10^{6} F = 9.38 mF
You can give your answers in farads or microfarads, as long as you are consistent. 
Directly on the area of the plates
Inversely on the gap between them.
C
µ A/d
C = kA/d
This means that we can increase the capacitance by increasing the area of the plates or by reducing the gap between the plates. However this will limit the voltage we can apply across the plates, as the dielectric may break down and sparks jump across the gap.
The proportionality constant, k, is called the permittivity, and is given the physics code e ("epsilon"; epsilon is a Greek letter ‘e’). We can therefore rewrite the formula as:
C = e A/d
If the plates are in a vacuum, the proportionality constant is called the permittivity of free space. It is a fundamental constant which is involved in a number of physics phenomena, such as electric field. It is given the physics code e_{0} (“epsilon nought”). It has the numerical value 8.85 × 10^{12} C^{2} N^{1} m^{2}. If the plates are in air, the value of the permittivity is very close to that of a vacuum.
Worked Example A 1 farad capacitor has
its plates separated by 1 mm of air. What
is the area of its plates? 
Answer Formula first:
C = e_{0} A/d
therefore:
A = Cd/e_{0}
Put the numbers in:
A = (1 F × 1 × 10^{3} m) × 8.85 × 10^{12} C^{2} N^{1} m^{2}
= 1.12 × 10^{8} m^{2} This would give us a capacitor with plates 10 km × 10 km, which is rather impractical. A farad is a very big unit, and we are much more likely to use microfarads (mF) or nanofarads (nF).

When we add an insulating material between the plates, we have added a dielectric. This changes the capacitance by a factor called the relative permittivity (or the dielectric constant). The relative permittivity is given the physics code e_{r}. The definition of the relative permittivity is given below:
In physics code we write:
The relative permittivity or the dielectric constant tells us how much the capacitance of a capacitor is increased relative a vacuum or air. It is given the physics code e_{r} ("epsilonr"), and has no units. Our relationship gets modified to:
A simple parallel plate capacitor has plates that are 30 cm × 30 cm, and are separated by a dielectric material of relative permittivity 2.56 that is 1.5 mm thick. Calculate the capacitance. 
How does a dielectric increase capacitance?
The insulating gap between the plates of a capacitor is called the dielectric. The reference dielectric is a vacuum, but air gives a value that is very similar. We can use a dielectric other than air. Some insulating materials do not affect the capacitance of the capacitor at all, but there are others, for example polythene or waxed paper that make the capacitance rise quite a lot. This happens because the molecules become polarised, which means that the electrons move slightly towards the positive plate, leaving a deficiency of electrons, hence a positive charge, at the other end. We see this:
The presence of the polarised molecules alters the electric field between the plates. Electric field goes from positive to negative.
The field between the plates goes from right to left.
The polarised molecules make a field that goes from left to right.
The overall field is reduced, therefore more electrons can crowd onto the plates, thereby increasing the charge that can be held.
The table shows some typical values of dielectric constant. Water is not a good practical dielectric as the impurities dissolved in it make it conduct.
Dielectric

Dielectric
constant 
Vacuum 
1.00000 
Air 
1.0005 
Polythene 
2.3 
Perspex 
2.6 
Waxed paper 
2.7 
Mica 
7 
Water (pure) 
80 
Barium titanate 
1200 
Very few capacitors consist of flat plates that we have looked at so far. Instead, they consist of two layers of aluminium foil alternating between two layers of dielectric. The whole lot is rolled up like a Swiss roll to make a compact shape.
Non electrolytic capacitors have a mica or polyester dielectric. The value of the capacitors made in this way is quite low, up to about 10 mF.
Electrolytic capacitors are capable of holding a much bigger charge. The aluminium metal plates are either side of a sheet of paper soaked in aluminium borate. When the capacitor is charged up, there is a chemical reaction that deposits an aluminium oxide layer on the positive plate. This acts as the dielectric. The electrolyte soaked paper acts as the negative plate.
The electrolyte itself acts as the negative plate
The aluminium oxide layer is the dielectric.
The dielectric layer is very thin (10 ^{–4 }m), which results in a very large capacitance. This can be as much as 100 000 mF.
New techniques have produced capacitors of capacitance as much as 10 F.
More recently, supercapacitors have been made that have capacitances of as much as 500 F.
In the case of an ultracapacitor, a capacitance of 5000 F is possible.
You can buy a 500 F super capacitor for Ł25. The working voltage is 3 V.
You can find out more about capacitors on my other website
HERE.
From Tutorial 1 we saw that capacitance
is
defined as:
The
charge required to cause unit potential difference in a conductor.
Potential difference is energy per unit charge. For a point charge, potential is energy per unit charge, so we can derive an expression for the capacitance of a point charge.
Consider an isolated point charge, +Q. A small test charge, +q, is brought to a distance r from the isolated charge as shown below:
From Fields Tutorial 5, we know that the work done in bringing in the test charge q is given by:
This work done is the potential energy. The potential is the energy per unit charge:
V = E/q
Therefore:
We also know that for a capacitor:
We combine these equations to give:
It doesn't take a genius to see that the Q terms cancel out:
Then we turn it upside down to give:
An isolated charge of 50 mC is 50.0 cm from a test charge. What is the capacitance due to the charge? 
Note that the capacitance is dependent on distance, not the charge.