Note Physics 2 SP025

Relationship between uniform E and 1.4b) Use equation for uniform E
potential difference.

The relationship between uniform electric field produced by two parallel plates
separated by a distance d and potential difference is given by :

The negative sign indicates 50
that the value of electric
potential decreases in the
direction of electric field.

Example 8

Two parallel plates, each is connected to +100 V and -100 V respectively. If an
electric field of 4000 V m-1 is produced between the plates, what is the separation
distance between the plates?
Solution

51

Example 9

A uniform electric field exists in the space between two 52
identical parallel charged metal plates. The plates are 1.0
cm apart. An electron is released from rest at the
negatively charged plate. It arrives at the positively
charged plate 2.0 ns later. Determine ;

(a) the electric field strength
(b) the speed of the electron when it arrives at the

positively charged plate.
(given ; me = 9.11  10–31 kg)

  


E

-

 

Solution

(a) From : F = ma and F = qE

From kinematics equation ;

Given that u = 0 (from ret), so ;

53

Substitute into (1) :
(b) Use :

54

CHAPTER 2

CAPACITOR AND DIELECTRIC

(1 HOURS)

SUBTOPICS :
2.1 Capacitance and capacitors in series and parallel
2.2 Charging & Discharging of capacitors
2.3 Capacitors with dielectrics

MODE Face to face Non Face to face
SLT SLT
Lecture
Tutorial 1 1

4 4

1

Learning Outcome:

At the end of this subtopic, students should be
able to:

2.1 Capacitance and capacitors in series

and parallel

(a) Define and use capacitance,



(Lecture : C1, CLO1, PLO1, MQF LOD1)

(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

(b) Derive and determine the effective capacitance of capacitors in

series and parallel.

(Lecture : C2, CLO1, PLO1, MQF LOD1)

(Tutorial : C4, CLO3, PLO4, CTPS3, MQF LOD6)

(c) Derive and use energy stored in a capacitor,


(Lecture : C2, CLO1, PLO1, MQF LOD1)

(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

2

Learning Outcome:

At the end of this subtopic, students should be able to:

2.2 Charging and discharging of capacitors
(a) State physical meaning of time constant and use

(Lecture : C1, CLO1, PLO1, MQF LOD1)
(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)
(b) Sketch and explain the characteristics of Q-t and I-t graph for
charging and discharging of a capacitor.

(Lecture : C2, PLO1, MQF LOD1)
(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)
(c) Use:



i.



ii.
(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

3

Learning Outcome:

At the end of this subtopic, students should be able to:

2.3 Capacitors with dielectrics

(a) Define dielectric constant,


(Lecture : C1, CLO1, PLO1, MQF LOD1)

(b) Describe the effect of dielectric on a parallel plate capacitor.

(Lecture : C2, CLO1, PLO1, MQF LOD1)

(c) Calculate capacitance of air-filled parallel plate capacitor,
(Tutorial : C4, CLO3 PLO4, CTPS3, MQF
LO D6)

(d) Use dielectric consta(nTtu,to rial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

(e) Use capacitance with dielectric,

(Tutorial : C3, CLO3 PLO4, CTPS3, MQF LOD6)

4

Overview:

Capacitors and dielectrics

Capacitance Capacitors Charging Dielectric
and discharging

of capacitor

In series In parallel

5

2.0 Introduction MOE7

Capacitor

-- A device that is capable of storing electric

charges or electric potential energy.

-- come in different shapes & sizes.

-- used in a variety of electric circuits.

6

Slide 6 for instance they are used to tune the frequency of radio receivers, as filters in power supplies, to eliminate sparking in automobile ignition
MOE7 systems & as energy-storing device in electronic flash units

Ministry of Education, Malaysia; 10/09/2004

-- consists of 2 conductors separated by a
small air gap or a thin insulator (called
dielectric)

conductor

Air gap OR Alumimium
dielectric Silver
Or other metals

Paper
Glass
Ceramic
Or other
nonconductive
materials

7

• The conducting plates could be in the
shape of
1. cylindrical
2. spherical
3. parallel plate

• circuit symbol for capacitor :
or

8

2.1 Capacitance & Capacitors in series
and parallel

Capacitance:
is a measure of the ability of a capacitor to
store charge ( or in other word store energy).

is defined as the ratio of the charge on either
plate to the potential difference between
them.

Formula capacitance C  Q MOE2 MOE3
V

where Q : magnitude of charge on either plate

V : potential difference (voltage) across the

two plates 9

Slide 9 One farad is defined as the charge of 1 coulomb stored on each of the conducting plates as a result of a potential difference of 1 volt between
MOE2 the two plates.

MOE3 Ministry of Education, Malaysia; 10/09/2004

As one farad is an extremely large quantity, practical capacitors have sizes in millionths of a farad or smaller.
example : 1 microfarad or 1 picofarad.

Ministry of Education, Malaysia; 10/09/2004

Scalar quantity

SI unit : Farad ( F ) @ C V-1

1 Farad is defined as the charge of 1 coulomb
stored on each of the conducting plates as a
result of a potential difference of 1 volt
between the two plates.

The capacitance for a capacitor does not

change unless it is designed to be a variable

capacitor. 10

rearrange

The charges stored ( Q ) is directly
proportional to the potential difference
( V ) across the conducting plate.
A capacitor with a large capacitance can
hold more charges than one with a
smaller capacitance for the same
potential difference applied across them.

CC
11

Capacitors in series

MOE5

o Regardless of their capacitance, all capacitors in series

contain charges of same magnitude ( )

o Total voltage, is shared by the capacitors.

o

12

Slide 12 when capacitors are wired in series to a battery of voltage V. when the circuit is complete, electrons are transferred onto the plates such that the
MOE5 magnitude of the charge Q on each plate is the same.

Ministry of Education, Malaysia; 11/09/2004

Generally, for number of capacitors connected
in series, the effective capacitance is given by:

equation to find
effective capacitance
for capacitors in
series

Value of is always smaller than the smallest
capacitance in the combination.

13

Capacitors in Parallel

o They have the same voltage, across their plates.
o Charge, and on each capacitors are different.

14

o Total charge,

o Generally, for number of capacitors connected
in parallel, the effective capacitance is given by

equation to find
effective
capacitance for
capacitors in
parallel

o Value of is always larger than the largest
individual capacitance.in the combination.

15

Energy stored in a charged capacitor,

Derivation

When a capacitor stores charge, it also stores

energy

A small amount of work ( ) is done in

bringing a small amount of charge ( ) from

the battery to the capacitor.

and

o The total work W required to increase the ‒
accumulated charge from zero to Q is given
16
by

The work done to charge it, appears as energy
stored in the capacitor,
Energy stored in a capacitor

17

Example LO 2.1

Determine the effective capacitance of the configuration shown in
Figure 2.1.

Figure 2.1
All the capacitors are identical and each has a capacitance of 2 F.

18

Solution : C1  C2  C3  C4  C5  C6  C7  2 F

 Label all the capacitors in the circuit.

C2 C1 C4 C5 C7
C3 C6

 To calculate the effective capacitance, it is easier to solve it
from the end of the circuit (left) to the terminal (right).

 Capacitors C1, C2 and C3 are connecte1d in s1eries,1then 1
C5  1C2 
Cx C4 C6C7 C1x 1C1 1 C3

Cx  2 2  2
Cx  2 μF
3 19

Solution : C1  C2  C3  C4  C5  C6  C7  2 F
 Capacitors Cx and C4 are connected in parallel, then

C5 C7 Cy  C x  C4
Cy C6 Cy  2 2
3
8
Cy  3 μF

 Capacitors Cy, C5 and C6 are connected in series, then
1 111
Cz C7 Cz  Cy  C5  C6

1 1 11
  
Cz 8  2 2
3
8
Cz  11 μF

20

Solution : C1  C2  C3  C4  C5  C6  C7  2 F

 Capacitors Cz and C7 are connected in parallel, then the
effective capacitance Ceff is given by

Ceff Ceff  C8z  C7
Ceff  11  2

Ceff  30 μF
11

OR 2.73 μF

21

Example LO 2.1

A 4 μF and 6 μF capacitor connected in
series are charged by a 240 V power supply.
Calculate
(a) The charge on each capacitor
(b) The potential difference across each

capacitor.
(c) The total energy stored in each capacitor.

22

Sketching Diagram

Solution
The combined capacitance for 2 capacitors
in series :

23

Q on each capacitor connected in series is the
same & is equal to the Q on the combined
capacitor.

Potential difference across 4μF capacitor :

24

Potential difference across 6μF capacitor :
The total energy stored in 4μF capacitor :
The total energy stored in 6μF capacitor :

25

2.2 Charging & Discharging of capacitors

Capacitors can undergoes two different
process mainly known as:

(i) Charging (ii) Discharging

e e

R R

V0 A C V0 A 
B B C



switch,S e switch,S e

26

27

Charging a capacitor through a resistor

e

R

V0 A  C

B  

switch,S e

o When switch S is closed, current I0 immediately begins
to flow through the circuit.

o Electrons will flow out from the negative terminal of the
battery and accumulate on the plate B of the capacitor.

28

o Then electrons will flow into the positive terminal
of the battery through the resistor R , leaving a
positive charges on the plate A.

o As charges accumulate on the capacitor, the
potential difference across it increases and the
current is reduced until eventually the maximum
voltage across the capacitor equals the voltage
supplied by the battery, V0.

o At this time, no further current flows (I = 0)
through the resistor R and the charge Q on the
capacitor thus increases gradually and reaches a
maximum value Q0

29

Q Q Q0 1  e  t  V V  V0 1  e  t 
RC  V0 RC 
Q0 
0.63Q0
0.63V0

0 τ  RC time ,t I 0 τ  RC time ,t

Figure b : the charge on Figure a : the potential
the capacitor increases difference across
with time. capacitor increases with
time.

I0  t
RC
Figure c : the current through I  I 0e

the resistor decreases 0.37 I0
exponentially with time.

0 τ  RC time , t 30

31

Discharging a capacitor through a resistor

e

R

V0 ABC

switch,S e

o When switch S is closed, electrons from plate B begin to
flow through the resistor R and neutralizes positive

charges at plate A.

o Initially, the potential difference (voltage) across the

capacitor is maximum, V0 and then a maximum current

I0 flows through the resistor R. 32

o When part of the positive charges on plate A is
neutralized by the electrons, the voltage across the
capacitor is reduced.

o The process continues until the current through the
resistor is zero.

o At this moment, all the charges at plate A is fully
neutralized and the voltage across the capacitor
becomes zero.

33

QI

Q0 t 0 τ  RC time ,t
RC
0.37Q0 Q  Q0e 0.37 I 0
t
0 I  I 0e  RC

τ  RC time ,t I0

Figure a : the charge on the V Figure b : the current through

capacitor decreases the resistor decreases

exponentially with time. exponentially with time.

V0 V  V0 e t
RC
Figure c : the potential
difference across capacitor 34

decreases exponentially with 0.37V0 τ  RC time ,t

time.

0

Time constant,

The quantity that appears in the exponent for
all equations in the charging and discharging
process is called time constant,

Formula :

scalar quantity

Unit : Second (s)

It is a measure of how quickly the capacitor

charges or discharges. 35

Physical meaning for Time constant,
Charging

Discharging

36

Example LO 2.21

In the RC circuit shown in Figure 2.6, the battery has fully
charged the capacitor.

aS R

V0 b C

Figure 2.6

At time t = 0 s, a switch S is thrown from position a to b. The
battery voltage V0 is 12.0 V and the capacitance C = 3.00 F.
The current I is observed to decrease to 0.45 of its initial value

in 60 s. Determine

a. the value of R.

b. the time constant, 

c. the value of Q, the charge on the capacitor at t = 0.

d. the value of Q at t = 100 s 37

Solution :V0  12.0 V; C  3.00 106 F; I  0.45 I0;t  60 106 s
a. By applying the equation of current for discharging process,
t
I eI e RC
 I  0

0.45 I 0
 R 60106
3.00106
0
Then by taking natural logs on both sides, thus the value of

 R is
ln 0.45  ln  R 60106
3.00106
e

0.45 60 106
3.00 106
 ln   R

R  25 

38

Solution : V0  12.0 V; C  3.00 106 F; I  0.45 I0;t  60 106 s

b. The time constant is given by

τ  RC τ  253.00 106
τ  7.5 105 s
c. By using the equation of charge for discharging process

Q  Q0e  t and Q0  CV0
RC

and the time, t= 0 hence t
RC
Q  CV0e
 Q  3.00 106 12.0
Q  3.6 105 C

d. By using the equation of charge for discharging process
100106
 t   Q  e 7.5105
Q  CV0 e RC 3.00 106 12.0

and the time, t = 100  106 s hence Q  9.49 106 C 39

2.3 Capacitors with dielectrics MOE10
Capacitance of Parallel plate capacitor

Air filled
parallel plate capacitor

where
ε0 : permittivity of free

space (8.85 10-12 F m-1)
A : area of the plate
d : plates separation

40

Slide 40 If we connect the plates to the terminals of a battery only for a moment on 1 of the surfaces a charge of +Q is accumulated, while on the other,
MOE10 there is a cahrge of - Q.

Ministry of Education, Malaysia; 10/09/2004

Parallel plate capacitor with
dielectric

where

ε0 : permittivity of free space
8.85 10-12 F m-1

: dielectric constant

A : area of the plate

d : plates separation 41