2.3 a) Define and use dielectric constant
Dielectric is an insulating material such as
rubber, plastic or waxed paper. Hence no
free electron are available in it.
The effect of a dielectric is to increase the
capacitance. ( increase the charge storage
capacity of the capacitor )
The capacitance increases by a factor κ or εr
which is called the dielectric constant
(relative permittivity) of the material.
50
Dielectric constant, (relative permittivity)
is the ratio between the capacitance of
given capacitor with space between plates
filled with dielectric, C with the
capacitance of same capacitor with plates
in a vacuum, C0.
where ε
C = Capacitance with dielectric.
o 51
C0 = Capacitance with vacuum /
without dielectric
is also known as relative permittivity
where = Permittivity of insulator
(dielectric)
Example: = Permittivity of free space
Dielectric Dielectric constant, εr
Paraffin 2.1
Mica 7.0
52
2.3 b) Describe the effects of dielectric on a parallel plate
capacitor
Consider a parallel plate capacitor in air
( no dielectric ) is charged by a battery to a charge
Q0.
SF 027 Matriculation Physics 53
A dielectric material is inserted between the plates of
an isolated charged capacitor.
Because of the electric field between the plates, the
molecules of the dielectric (whether polar or non-
polar) will tend to become oriented as shown in the
figure below.
54
When the material is in the capacitor, the dipoles
orient themselves with the field, creates a `reverse`
electric filed ( Eind ).
The negative ends are
attracted to the positive
plate and the positive ends
are attracted to the
negative plate.
Because of the end-to-end
orientation, the left
surface of the dielectric
become negatively
charged, and the right
surface become positively
charged.
55
Since the capacitor is isolated- not
connected to an external circuit, no
pathway for charge to leave or to be added
to the plates, thus Q0 remains same.
56
The dipole field partially
cancels the field due to the
plate charges.
the net field ( E ) between the
plates is reduced – the voltage
across the plates ( V = Ed ) also
decreases.
Since the stored charge
remains the same, the
capacitance increases.
57
• From the definition of the capacitance,
and
where Q is constant (capacitor is disconnected
from battery) @ an isolated capacitor, V
decreased.
since
therefore
If the dielectric is inserted and battery
remains connected, the original voltage is
maintained and the battery must supply
more charge and therefore do work and the
energy storage to go up in this situation.
59
The potential difference across the capacitor
is :
But if a dielectric is now inserted between the
capacitor plates , charge (Q) on the plates
that increase by a factor εr to the value Q.
since more charge is now stored at the same
voltage (Q increased, V constant).
Capacitance will increased .
60
Apply the definition of capacitance :
For a parallel plate capacitor with a
dielectric, the capacitance is increased over
its ( air ) value by a factor εr
61
2.3 c) Apply capacitance of air-filled parallel plate capacitor
Consider 2 parallel plates with a small air gaps which
separated by a distance d.
2 plates have equal area A
The plates are very close together – assume the
electric field is uniform between the plates.
62
• The capacitance of a parallel-plate capacitor, C is
proportional to the area of its plates and inversely
proportional to the plate separation.
Parallel-plate capacitor separated by a vacuum
or
Parallel-plate capacitor separated by a dielectric material
= 8.85 x 10-12 C2 N-1 m-2
2.3 d) Determine capacitance with dielectrics
εr : dielectric
constant
Material Dielectric constant, εr Dielectric Strength
(106 V m-1)
Air 1.00059 3
Mylar 3.2 7
Paper 3.7 16
Silicone oil 2.5 15
Water 80 -
Teflon 2.1 60
a) The advantages of inserting the dielectric between the plates of
the capacitor are :
i) Increase in capacitance.
ii) Increase in maximum operating voltage.
iii)Possible mechanical support between the plates, which allows the
plates to be close together without touching. (decreasing d,
increasing C)
Example 6
A parallel plate capacitor consists of 2 plates
each with area 200 cm2 separated by a 0.4 cm
air gap.
(a) compute its capacitance
(b) if the capacitor is connected across a 500 V
source, what are the charge on either
plate ?
66
Solution
A = 200 cm2 = 200x10-4 m2
d = 0.4 cm = 0.4x10-2 m
(a)
(b)
67
Example 7
A parallel–plate capacitor has charge of
magnitude 9x10-6 C on each plate & capacitance
3μF when there is air between the plates. The
plates are separated by 2.0 mm. With the
charge on the plates kept constant, a dielectric
with εr = 5.0 is inserted between the plates.
What is the potential difference & electric field
between the plates of the capacitor before &
after the dielectric has been inserted ?
68
Solution 69
The potential difference before :
The electric field before :
The potential difference after :
The electric field after :
SF 027 Matriculation Physics
Exercise
The dielectric of a parallel-plate capacitor is
to be constructed from teflon that completely
fills the volume between the plates. The area
of each plate is 0.50 m2.
a)What is the thickness of the teflon if the
capacitance is to be 0.10 F?
b) Calculate the charge on the capacitor if it is
connected to a 12 V battery.
(Dielectric constant for teflon is 2.1)
ANS. : 92.9 m; 1.2 C 70
Check Your Understanding
Figure shows a parallel plate capacitor
connected to a battery. A piece of insulator is
then inserted into the capacitor.
insulator
battery capacitor
State if any changes to 71
(a) the potential difference across the
capacitor.
(b) the capacitance of the capacitor
(c) the charge in the capacitor.SF 027 Matriculation Physics
Answer
(a) Since capacitor still connected to
battery, V remain unchanged.
(b) As dielectric is inserted, capacitance
increase by a factor εr .
(c) From:
For V constant,
Charges store in capacitor increase.
72
End of Chapter 4
73