Download CAPACITANCE: (Boctor, Ch. 8, p. 272)

Unit 8
CAPACITANCE:
The simplest form of a capacitor consists of two conducting metal plates separated by an
insulating material such as air, paper, mica, or ceramic, as shown below.
Charge Q
V
d
+ + + + + +++
++
+
+++++
-- -- - - - - - - - - -
Insulator with a
dielectric constant

Conducting plate with area A
Symbol for a capacitor:
i
C
+ V When a voltage is applied to the capacitor some of the electrons from the top plate will
be attracted to the positive terminal of the voltage source. The same number of electrons
will be repelled from the negative terminal and thus accumulate on the bottom plate.
The result is that the top plate will be deficient in electrons and thus positively charged
while the bottom plate will acquire excess electrons and become negatively charged.
This motion of electric charge takes place through the conducting wires and the source
but not through the insulating material separating the plates.
As further electrons attempt to migrate from the top plate to the positive battery terminal
they are attracted back by the positive charge accumulating on the top plate. Similarly
the electrons repelled from the negative terminal are now also being repelled by the
negative charge accumulating on the bottom plate. The voltage source has to spend
energy in this process of charge separation. Eventually a balance will be established
between the energy supplied by the battery and the attraction and repulsion forces
exhibited by the steady-state charge, Q, accumulating on each of the two plates. This
charging process takes time which means that capacitors cannot be charged instantly.
An increase in the applied voltage separates more charges on the metal plates, resulting
in an increase in the amount of charge, Q, accumulating on each plate. A positive charge
equal to Q accumulates on the top plate and a negative charge, also equal to Q,
accumulates on the bottom plate. The capacitor is said to store the electric charge Q
when the voltage V is applied to it, or that it is charged to the voltage V. The amount of
charge is directly proportional to the capacitor's voltage V.
Thus Q  V
or Q = CV
1
or C =
Q
V
Unit 8
The constant of proportionality C is a measure of the ability of the device to store charge
per unit voltage applied. For the same applied voltage the capacitor with the greater
capacitance C will store the larger amount of charge.
Definition of capacitance: the amount of charge stored on each plate of a capacitor per
unit voltage applied between the plates.
Capital letters such as Q and V are used to refer to quantities which are constant with
respect to time (i.e. dc values) and because capacitors take time to charge the above
equations are often written as follows:
q = C v or C =
q
v
where lowercase v and q are variable time functions.
Thus the unit of capacitance is coulomb/volt which is termed the Farad (F). However
1 F is very large and the more practical units are:
1 F = 10 -6 F
1 nF = 10 -9 F
1 pF = 10 -12 F
where  (pronounced mu) is for micro, n for nano and p for pico.
Dielectric materials:
When charge accumulates on two plates separated by an insulating (or dielectric)
material, an electric field is established in the space (or dielectric material) between the
plates. A dielectric in an electric field can be viewed as a free-space arrangement of
microscopic electric dipoles which are composed of positive and negative charges a
short distance apart. These charges are bound in place, perhaps by being part of the same
atom, and can only shift positions slightly in response to external fields (the positive
charges moving one way and the negative the other, but only as far as the elastic nature
of the bond permits). Each atom becomes distorted by the electric field and becomes
positive on one side and negative on the other.
All dielectric materials, whether solid, liquid or gas, have one characteristic in common,
their ability to store electric energy. This storage takes place by means of a shift in the
relative positions of the internal, bound positive and negative charges against the normal
molecular and atomic forces, as described above. This displacement against a restraining
force is analogous to stretching a spring and represents potential energy. Thus, when an
electric field is applied to the normally randomly oriented dipoles, the action of the field
is to align these molecules, to some extent, in the same direction. Different dielectric
materials respond differently to electric fields and a measure of how well the material
allows the establishment of an electric field is the dielectric constant or permittivity  of
the material where
 = r o
2
Unit 8
o is the permittivity of free space (8.854 pF/m) and r is the relative permittivity of the
dielectric material. Values of r for dielectrics frequently used in capacitors are as
follows: for mica r = 5, for glass r = 7.5, and for ceramic r = 7500. Capacitance is
directly dependent on permittivity.
The larger the area (A) of the plates used, the more charge can be stored on them and
thus the greater the capacitance. The greater the separation of the two plates, the weaker
the electric field between them and hence the less the energy that can be stored in the
dielectric thus reducing the value of capacitance. Capacitance can be expressed as
C=
A
farad (F)
d
This relationship indicates that any required capacitance value, whether fixed or
variable, can be obtained by manipulating the basic physical dimensions A and/or d and
by choosing the appropriate dielectric.
A maximum allowable working voltage is specified for each type of capacitor and is
usually marked on the side of the capacitor. If this voltage is exceeded the dielectric may
breakdown to become a conductor, rendering the capacitor useless.
See any Electric Circuits book for types of practical capacitor.
Ex 1:
(a)
What quantity of electric charge will produce a p.d. of 200 V between the
plates of a capacitor of 5 F?
(b)
Compare the magnitude of two capacitors, one having two circular plates
of diameter 40 mm and placed 0.5 mm apart, the other having two square
plates of side 50 mm and placed 0.75 mm apart. It may be assumed that
the same dielectric material is used between the plates in each case.
[1 mC, 1.33]
Capacitors in series:
When three capacitors are placed in series as shown below, the applied voltage will
provide work to separate a charge Q such that the top plate C1 has a charge of +Q
deposited on it while the bottom plate of C3 will have -Q deposited on it. Since the
charge on each plate of a capacitor must be equal, but with opposite polarities, this
implies that the charge on each of the series-connected capacitors must be the same, Q.
Thus the equations below can be written.
V
C1
V1
C2
V2
C3
V3
Q
V1 = C1
3
Q
V2 = C2
Q
V3 = C3
Unit 8
Applying KVL:
V = V1 + V2 + V3
Now if CT is the equivalent capacitance of three capacitors in series then:
V =
Q
CT
Thus from above
Q
Q
Q
Q
=
+
+
CT
C1
C2
C3
Dividing both side above by Q we can write that for n capacitors in series:
1
1
1
1
1
=
+
+
+ ... +
CT
C1
C2
C3
Cn
For the special case of two capacitors in series we can write that
CT =
C1 C2
C1 + C2
Capacitors in parallel:
When three capacitors are placed in parallel as shown below, the voltage across each of
them will be the same voltage, V. Each capacitor will be charged by a different amount
of charge. Thus the equations below can be written.
C1
V
C2
Q1 = C1 V
C3
Q1
Q2
Q3
Q2 = C2 V
Q3 = C3 V
The energy source provided the total work necessary to separate the total amount of
charge QT which is
QT = Q1 + Q2 + Q3
To replace the parallel connected capacitors by a single equivalent capacitor CT the
same voltage V should do the same amount of work to separate the same amount of
charge QT as in the original network.
QT = CT V
CT V = C1 V + C2 V + C3 V
Dividing both sides by V we can see that for n capacitors in parallel
CT = C1 + C2 + C3 + ... + C n
4
Unit 8
Ohm's law for a Capacitor:
Electric current in any circuit element is defined as the rate of change of charge w.r.t.
time (i.e. the rate of flow of charge). If a charge q flows into one terminal of an
element in a time t (where  refers to small or incremental variations) then
i
q
t
As the charge q does not build up linearly on a capacitor as a function of time t, the
graph of q versus t is a curve whose slope, at any instant in time, is a tangent line at a
point when t tends to zero. Therefore
q dq

dt
t  0 t
i  lim

dq
slope = dt = i
q
t
0
For the capacitor q = Cv when both q and v are functions of time, with C a constant.
Therefore q  Cv or
i
q
v
C
 C  rate of change of voltage
t
t
Hence the capacitor's current is
q
v
dv
 lim
C
C
t
dt
t 0 t
t 0
i  lim

This is Ohm's law for a capacitor.
Energy stored in a Capacitor:
The instantaneous power p supplied to a capacitor and the energy w are defined as
p  iv 
dw
dt
 w   p dt   vi dt
v
v
v2 
dv
C 2
 w   vC
dt  C  v dv  C   
v 0
0
dt
 2 0 2
5


Unit 8
Therefore, the energy w stored in a capacitor is
w
1 2
Cv joules (J)
2
Ex 2:
Three capacitors of 10 F, 25 F, and 50 F are connected (a) in series and (b)
in parallel. Calculate the equivalent capacitance and the energy stored for each
of the cases (a) and (b) above when the capacitors are connected to a 500 V
supply.
[6.25 F, 85 F, 0.781 J, 10.625 J]
Ex 3:
A 50 F capacitor is charged from a 200 V supply and after being disconnected
is immediately connected in parallel with a 30 F capacitor. The 30 F
capacitor is initially uncharged. Calculate:
(a)
the p.d. across the combination;
(b)
the electrostatic energy stored before and after the capacitors are
connected in parallel. Account for any difference in energy stored.
[125 V, 1 J, 0.625 J]
Capacitance of a multi-plate a Capacitor:
If the total number of plates is n, there are n-1 layers of dielectric and n-1 electric fields.
Thus the total capacitance is n-1 times the capacitance between one pair of plates
(assuming of course that the area of all the pairs of plates is the same).
C =  r o
An - 1
F
d
where n is the number of plates, A is the area of one of the plates and d is the spacing
between each of the plates (assuming equal spacing between them all).
Ex 4:
A parallel plate capacitor consists of 11 plates each having an area of 100 cm2
and spaced 0.02 mm apart. If the capacitance is 0.25 F, what is the relative
permittivity of the dielectric? If the capacitor receives a charge of 50 C, what
is the p.d. across the plates?
[5.65, 200 V]
6