Capacitors are short term chargestores, a bit like an electrical spring. They are used widely in electronic circuits. At its simplest it consists of two metal plates separated by a layer of insulating material called a dielectric.
The picture shows some different capacitors:
The two components on the right are electrolytic capacitors. The others are nonelectrolytic capacitors.
The symbol for a capacitor is shown below:
There
are two types of capacitor, electrolytic
and nonelectrolytic.
We won’t worry at the moment what these terms mean, other than to say:
Electrolytic
capacitors hold much more charge;
Electrolytic
capacitors have to be connected with the correct polarity, otherwise they
can explode.
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:
The
charge required to cause unit potential difference in a conductor.
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:
[Q  charge in coulombs (C);
C – capacitance in farads (F);
V
 potential difference in volts (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),
1 × 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.
Write down what is meant by the following terms:

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.
If
we connect a capacitor in series with a bulb:
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.
The capacitor does NOT conduct electricity. The "flow" of a.c. is due to the charge and discharge of the capacitor.
Question 2 
Why does a capacitor appear to allow ac to flow, but not dc? 

Question 3 
What is the charge held by a 470 microfarad capacitor charged to a p.d. of 8.5 V? 
Measuring Capacitance
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, 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,
Þ
Q = I/f
Since
C
= Q/V, we can now write
C = I/fV
Question 4 
Why does the circuit only operate on the forward halfcycle? 

Question 5 
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? 
When we
charge up a capacitor, we make a certain amount of charge move through a certain
voltage. We are doing a job of work
on the charge to build up the electric field in the capacitor.
Thus we can get the capacitor to do a job of useful work.
1.
Energy = charge × voltage
2. Q = CV
This second relationship tells us that the charge – voltage graph is a straight line:
The
capacitor is charged with charge Q to a
voltage V.
Suppose we discharged the capacitor by a tiny amount of charge,
dQ.
The resulting tiny energy loss (dW)
can be worked out from the first equation:
dW = V × dQ
This is the same as the area of the pink rectangle on the graph.
If
we discharge the capacitor completely, we can see that:
Energy
loss = area of all the little rectangles
=
area of triangle below the graph
=
½ QV
By substitution of
Q
= CV, we can go on to write:
E = ½ CV^{2}
Question 6 
What is the energy held by a 50 000 mF capacitor charged to 12.0 V? 