L7/L8
FAD1011 (PHYSICS 3)
2. Capacitor and
Dielectric
2.1 Capacitors
2.2 Capacitors in series and parallel
2.3 Capacitors with dielectrics
Lecturer:
Dr. Salmiah Ibrahim
Centre for Foundation Studies in Science
University of Malaya
Session 2020/2021
2.1 Capacitor and Capacitance
▪ A capacitor is an electrical device, sometimes called
condenser is a device that can store electric charge.
▪ A capacitor consists of two metal plates separated by
an insulator called dielectric.
▪ The insulator can be a vacuum, air or any other
materials such as polystyrene, oil or waxed paper.
▪ In a diagram, a capacitor is represented by this
symbol.
▪ Capacitors have many applications:
– Computer RAM memory and keyboards.
– Electronic flashes for cameras.
– Electric power surge protectors.
– Radios and electronic circuits.
Types of Capacitor
Parallel-Plate Capacitor
•
Cylindrical Capacitor
A typical capacitor consists of a pair of parallel plates of area, A separated by
a small distance, d. Often the two plates are rolled into the form of a cylinder
with paper or other insulator separating the plates.
▪ When used in an electric circuit, the plates are connected
to the positive and negative terminals of a battery or some
other voltage source.
▪ When this connection is made, electrons are pulled off one
of the plates, leaving it with a charge of +Q and
transferred through the battery to the other plate, leaving
it with a charge of –Q.
▪ This charge transfer stops when the potential
difference across the plates equals the potential
difference of the battery.
▪ Thus, a charged capacitor is a device that stores
energy that can be reclaimed when needed for a
specific application.
Capacitance, C
▪ The measure of the extent to which a capacitor can
store charge is called capacitance.
▪ The number of charges stored in capacitor is
proportional to the potential difference between the
plates, Q = CV where C is capacitance of the capacitor.
▪ The capacitance of a capacitor is defined as the ratio
of the magnitude of the charge on either plate to the
magnitude of the potential difference, V between
them.
𝑄
𝐶=
𝑉
…(1)
▪ S.I unit of the capacitance is farad (F).
1F=1
𝐶
𝑉
▪ 1 Farad: capacitance of a capacitor where 1 coulomb of
charge produces a potential difference of 1 volt.
▪ Practical capacitors usually have the capacitance in the
order of F or pF.
▪ For example, if a 3.0 F capacitor is connected to a 12 V
battery, the magnitude of the charge on each plate of
the capacitor is
Q = C V = (3.0 × 10 -6 F) (12 V) = 36 C .
A capacitor in a simple electric
circuit.
Charge stored:
Q = CV
(The charges stored, Q is directly proportional to the potential difference,
V across the conducting plate.)
Relationship between Q and V
Q
Q = CV
Q αV
V
▪ The capacitance C is a constant (not depend on Q or V. its
value depends on the structure and dimensions of the
capacitor itself).
- for a larger A, more charge can be held on each plate and
for a given potential difference, V
- the charge on the plates increases with decreasing plate
separation, d.
Parallel Plate Capacitor
▪ A simple parallel-plate capacitor
consists of two conducting plates of
area A separated by a distance d.
• Charge +Q is placed on one plate
and –Q on the other plate.
• An electric field E is created
between the plates (uniform electric
field)
Cont..
▪ The electric field between the plates is
𝐸=
𝜎
𝜀𝑜
=
𝑄
𝜀𝑜 𝐴
𝑄
𝐴
where σ = is surface
charge density
▪ Because the field between the plates is uniform, the
magnitude of the potential difference between the
plates equals Ed, where d is distance between two
plates, therefore
𝑉 = 𝐸𝑑
𝑄𝑑
=
𝜀𝑜 𝐴
Cont..
▪ Substituting into Eq (1)
𝑄
𝑄
𝐶= =
𝑉 𝑄𝑑/𝜀𝑜 𝐴
Therefore,
𝐶=
𝜀𝑜 𝐴
d
▪ The capacitance of a parallel plate capacitor is proportional
to the area A of its plate and inversely proportional to the
plate separation d.
Example 1 :
Calculating C for a Parallel-plate
Capacitor
A parallel plate capacitor has an area A of 2.00 cm2 and a
plate separation d of 1.00 mm. Find its capacitance.
Solution 1
𝐹𝑟𝑜𝑚 𝐶 =
𝜀𝑜 𝐴
d
𝜀𝑜 𝐴
8.85 × 10−12 C 2 N −1 m−2 2 × 10−4 m2
𝐶=
=
𝑑
1 × 10−3 m
= 𝟏. 𝟕𝟕 × 𝟏𝟎−𝟏𝟐 𝐅 = 𝟏. 𝟕𝟕 𝐩𝐅
Example 2:
2.2 Combinations of Capacitors
(A) Parallel combination
(a)
(b)
(c)
(a) A parallel connection of two capacitors
(b) The circuit diagram for the parallel combination
(c) The potential differences across the capacitors are the same, and
the equivalent capacitance is Ceq = C1 + C2
Cont..
▪ The potential difference across the capacitors
is the same and each is equal to the voltage of
the battery
V = V1 = V2
▪ The charge in each capacitor is
Q1 = C1 V
Q2 = C2 V
▪ The total charge is equal to the sum of the
charges on the capacitors
Qtotal = Q1 + Q2
▪ It can be written as
Q = Ceq V = (C1 + C2 )V
Example 3:
Four Capacitors connected in Parallel
(a) Determine the capacitance of
the single capacitor that is
equivalent to the parallel
combination of capacitors.
(b) find the charge on the 12.0 μF
capacitor
(c) total charge contained in the
configuration
Solution 3:
(a) C e q = C 1 + C 2 + C 3 + C 4
= 4 5 μF
(b) The potential difference across the
12.0 μF capacitor is equal to the
voltage of the battery and so,
Q = C3 ∆V
= (1 2 × 10-6 F) (18.0 V) = 2 1 6 × 10-6 C = 2 1 6 μC
(c) Total charge QT = Ceq ∆V = (4 5 × 10-6 F) (18.0 V)
= 810 × 10-6 C
= 810 μ C
(B) Series combination
(a) A series combination of two capacitors. The charges on the
two capacitors are the same.
(b) The circuit diagram for the series combination
(c) The equivalent capacitance can be calculated from the
1
1
1
relationship
= +
𝐶𝑒𝑞
𝐶1
𝐶2
• The potential difference across each capacitor is split
between two capacitors ∆ V = ∆ V 1 + ∆ V 2
• ∆V =
•
𝑄
𝐶𝑒𝑞
=
𝑄
𝐶1
+
𝑄
𝐶2
Cancelling Q , So
1
𝐶𝑒𝑞
=
1
𝐶1
+
1
𝐶2
• So for three or more capacitors, the equivalent
capacitance is
1
𝐶𝑒𝑞
=
1
𝐶1
+
1
𝐶2
+
1
𝐶3
+⋯
Example 3:
Four capacitors connected in series
Four capacitors are connected in
series with a battery, as shown.
(a) Find the capacitance of the
equivalent capacitor.
(b) Find the charge on the 12 μF
capacitor.
(c) Find the voltage drop across
the 12 μF capacitor.
Solution 3
(a)
1
𝐶𝑒𝑞
1
𝐶𝑒𝑞
=
=
1
1
+
𝐶1
𝐶2
1
3 μF
+
+
1
𝐶3
1
6 μF
+
+
Note
that
the
equivalent capacitance
is less than the
capacitance of any of
the
individual
capacitors
in
the
combination.
1
𝐶4
1
12 μF
+
1
24 μF
Ceq = 1.6 μF
(b) Charge on the equivalent capacitor
Q = C eq V = (1.6 μF) (18.0 V) = 28.8 μC
(c) Voltage drop across the 12 μF capacitor.
Q = CV
V=
𝑄
𝐶
=
28.8 ×10−6 C
12 ×10−6 𝐹
= 2.4 V
Example 4
Find the equivalent capacitance between a and b for the
combination of capacitors as shown below. All capacitances
are in microfarads.