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.
Capacitance
(F) = Charge (C)
Voltage (V)
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.
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
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?
C = 0.045 A × (400 Hz × 12.0 V) = 9.38 × 10^{6} F = 9.38 mF
Capacitance of a Capacitor
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 is called the permittivity of free space. It is given the physics code e_{0} (“epsilon nought”; epsilon is a Greek letter ‘e’).
It has the numerical value 8.85 ×
10^{12} C^{2} N^{1} m^{2}.
We can therefore rewrite the formula as:
C = e_{0} A/d.
A 1 farad capacitor has
its plates separated by 1 mm of air. What
is the area of its plates?
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}
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 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:
C = e_{0}e_{r} A/d
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 
Practical
Capacitors
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.
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.
Why does a capacitor discharge exponentially?
Think about a capacitor that is fully charged. Let's suppose we remove 1 % of its charge. Its charge goes down to 99 %. The new charge is 99 % of the original
Now let's remove another 1 % of the charge that is left, i.e. the 99 % of the original. 1 % of 99 % = 0.99 %. New charge is now 98.01 %. And so on...
Now suppose we lose that fraction in each unit time. We can say that the capacitor discharges a constant fraction of what is left in each unit time. The charge lost in each unit of time can be worked out as:
Charge loss per unit time = change in charge ÷ time interval
This is the rate of change of charge which is also the current. So we can write this as an equation:
From Ohm's Law (V = IR) we can also write:
The minus sign tells us that there is a discharge going on.
From basic capacitance we know that Q = CV. Therefore we can write:
The term RC is the time constant, which we have seen in the previous tutorial. Remember that ohms × farads = seconds. We can rewrite the expression as:
In mathematics one kind of differential equation is one in which a quantity, x, decays with a constant fraction for each unit time. An alternative is to say that the rate of decrease in x is proportional to the amount of x that remains. The differential equation is in the form:
The constant, k, is the fraction of the the quantity x that decays every second.
This is a differential equation in which 1/RC is the constant.
The solution to this can be derived using the calculus process of integration. This gives the equation:
We can take natural logarithms to rewrite this as:
If we plot ln V against time, we get a straight line graph of the form y = mx + c.
Notice the following:
Time constant is 1 × gradient;
The yaxis intercept is ln Vo (the starting voltage);
The xaxis intercept is the time taken for the voltage to fall to 1 volt (as ln 1 = 0).