CHAPTER 2:

Capacitor and dielectrics

2.1 Capacitance and capacitors series and parallel
At the end of this chapter, you should be able to:
 Define and use capacitance.

Q
C =
V

 Derive and determine the effective capacitance of

capacitors in series and parallel.

 Derive and use energy stored in a capacitor

2
1 1 1Q
U = CV = QV =
2
2 2 2 C
 Introduction

A capacitor is a device which is used to store electric charge.

Effectively, all capacitors consists of a pair of conducing plates
separated by an insulator.
The insulator is called a dielectric and is often air, oil or paper.

Figure 1 Various type of capacitors

2.1 Capacitance & Capacitors in series and
parallel
Capacitance:

The capacitance of a capacitor is defined as the ratio of the

magnitude of the charge on either plate to the potential
difference between them.

Q
C=
V
where Q : charge on one of the plates
V : potential difference across them

Unit of capacitance : C V-1 @ Farad ( F )

Definition of 1 Farad:

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.
➢ By rearranging the equation from the definition of
capacitance, thus
Q = CV
where the capacitance of a capacitor, C is constant then

Q V
(The charges stored, Q is directly proportional to the
potential difference, V across the conducting plates).

➢The greater capacitance of a capacitor, the more

charge is required.

6
Capacitors in series
 Figure shows three capacitors are connected
in series to a battery of voltage, V.

+Q1
C1,V1 −QI
V equivalent to V +Q
+Q2 Ceff,V
C2,V2
−Q2 −Q
+Q3
C3,V3 − Q3

 When the circuit is complete, electrons are transferred

onto the plates such that the magnitude of the charge
Q on each plate is the same.
 The total charge (Q) on the effective capacitor is

Q = Q1 = Q2 = Q3

 The potential difference across each capacitor C1,C2 and C3

are V1,V2 and V3 respectively. Hence
Q1 Q
V1 = =
C1 C1

Q2 Q
V2 = =
C2 C2

Q3 Q
V3 = =
C3 C3
 The total potential difference V is given by

V = V1 + V2 + V3
Q Q Q
V= + +
C1 C2 C3
V 1 1 1 V 1
= + + and =
Q C1 C2 C3 Q Ceff
 Therefore the effective (equivalent) capacitance Ceff for
n capacitors in series is given by

1 1 1 1 1 capacitors in
= + + + ... +
Ceff C1 C2 C3 Cn series!!
 Capacitors in parallel
 Figure shows three capacitors are connected in parallel to
a battery of voltage, V.

V +Q1 +Q2 +Q3

−Q I −Q 2 −Q 3
C1,V1 C2,V2 C3,V3

equivalent to V +Q
Ceff,V
−Q

 The potential difference across each capacitor is the same

as the supply voltage (V).
 The potential difference on the effective
capacitor is

V = V1 = V2 = V3

 The charges stored by each capacitor C1,C2

and C3 are Q1,Q2 and Q3 respectively.

Q1 = C1V1 = C1V ;
Q2 = C2V2 = C2V ;

Q3 = C3V3 = C3V
➢ The total charge Q on the effective capacitor is
given by
Q = Q1 + Q2 + Q3
Q = C1V + C2V + C3V
Q Q
= C1 + C2 + C3 and = Ceff
V V

➢ Therefore the effective (equivalent) capacitance

Ceff for n capacitors in parallel is given by

Ceff = C1 + C2 + C3 + ... + Cn capacitors in

parallel !!
Example 1
What is the total capacitance in a, b and c ?

C 1 = 5 F
C 1 = 5 F
+ - + -
C 1 = 5 F C 3 = 15 F C 3 = 15 F
C 2 = 10 F
+ + - - + - +
+ - -
+- +-+ -
C 2 = 10 F
+ -
+ -
C 3 = 15 F C 2 = 10 F

(a) (b) (c)

(a) In parallel , the total capacitance is given by

C E = C 1 + C 2 + C 3 = 5 F + 10 F + 15 F
= 30 F

(b) In series, the total capacitance is given by

1 1 1 1 1 1 1
= + + = + +
CE C 1 C 2 C 3 5F 10 F 15F
1 11
=
CE 30 F
30
CE = F = 2.7 F
11
(c) The equivalent capacitor for the 2 capacitors
connected in parallel :

C p = C 1 + C 2 = 5 F + 10 F
C p = 15 F

The total capacitance for the parallel & series

connections is CE where

1 1 1 1 1 2
= + = + =
CE Cp C 3 15F 15F 15F
15
CE = F = 7.5 F
2
Example 2 :

Determine the effective capacitance of the

configuration shown in Figure below.

All the capacitors are identical and each has a

capacitance of 2 µF.
Solution : C1 = C2 = C3 = C4 = C5 = C6 = C7 = 2 F

Label all the capacitors.

C1 C5
C2 C4 C7
C3 C6

 Solve from the end of the circuit to the terminal

(this case left to right).

C5
Cx C4 C7
C6
 Identify the connection (series or parallel) C1, C2
and C3 are connected in series, then
1 1 1 1
= + +
C x C1 C2 C3
1 1 1 1 2
= + + C x = μF
Cx 2 2 2 3

C5 C5
Cx C4 C7 Cy C7
C6 C6
• Cx and C4 are connected in parallel, then
C y = C x + C4
2
Cy = + 2
3
8
C y = μF
3

C5
Cy C7 Cz C7
C6
• Cy, C5 and C6 are connected in series, then
1 1 1 1
= + +
C z C y C5 C 6
1 1 1 1 8
= 8 + + Cz = μF
C z (3 ) 2 2 11

Cz C7 Ceff

• Cz and C7 are connected in parallel, then the effective

capacitance Ceff is
Ceff = C z + C7
30
8 Ceff = μF OR 2.73 μF
Ceff = + 2 11
11
Solution :
C1 = 100 F; C2 = 200 F; C3 = 300 F;VAB = 8.0 V

C12
C1
A A
C2
D D
C3 C3
B B

A
Ceff

B
Energy stored in a charged capacitor

When the switch is closed, charges begin to accumulate on

the plates.
A small amount of work ( ∆W ) is done in bringing a small
amount of charge ( ∆Q ) from the battery to the capacitor.

Q W = V Q
since C = :
V
Q
W =   Q
C 
The total work ( W ) required to increase the
accumulated charge from zero to Q is given by

Q Q
 Q
W =  VdQ =   dQ
0
0 C

1 Q2
W=
2 C
We know, Q = CV thus : 1
W = CV 2
2
Q 1
C can also be written as : C= , thus :
V W = QV
2
The work done in charging the capacitor appears as
electric potential energy U stored in the capacitor.
Thus:
Work done, W = Energy, U
Sketching Diagram
240V

C1 C2

4 F 6 F
Solution

The combined capacitance for 2 capacitors in series

:
1 1 1 5
= + = 
CE 4 6 12
C E = 2 .4  F
2.2 Charging and discharging of capacitors
At the end of this chapter, you should be able to:
 State physical meaning of time constant and use τ = RC
 Sketch and explain the characteristics of Q-t and I-t
graph for charging and discharging of a capacitor.

 Use
i) t for discharging
−
Q = Q0 e RC

ii) for charging.

 −
t

Q = Q0 1 − e RC 
 
 
Charging & Discharging of capacitors

Capacitors can undergoes two different process mainly

known as:

(i) Charging Process

(ii) Discharging Process

 = RC
2.2 Charging & Discharging of capacitors
Charging a capacitor

A capacitor in series with a resistor, switch & battery.

e−

e−

When switch S is closed, charges (electrons) begin to flow

setting up a current.
Charging process
Simultaneously
Electrons (e) , e flow into the
Leaving plate B
flow out from (+) terminal of
negatively
the (-) terminal the battery
of the battery charged.
through the
resistor

Until Vc max=V0 As Q increase, Leaving a

Vc max :voltage the V increases positive
across capacitor but the I charges on the
V0:voltage supplied reduced. plate A.

That time, no Charge on the

current flows (I capacitor
= 0) through reaches Q0 =
the resistor R Qmax.
 Charging graphs and formulas

i) Charge :
Q

Q0
t
−
0.63Q0 Q = Q0 (1 − e RC
)

time, t where
0 τ Q0 : maximum charge
Charge on the capacitor R : resistance of the resistor
increases with time. C : capacitance of the capacitor
ii) Current: I

Current through the

resistor decreases I0
exponentially with
time.
0.37 I 0
−
t V0 time, t
I = I 0e RC and I0 = 0 τ
R

where V0 : maximum voltage = supply voltage

I 0 : maximum current
R : resistance of the resistor
C : capacitance of the capacitor
Discharging a capacitor through a resistor

Figure shows a simple circuit for discharging a charged

capacitor. e−
R

A+ + + + + +
V0
B − −− − −−C

switch, S e−

When switch S is closed, electrons from plate B begin to flow

through the resistor R and neutralizes positive charges at
plate A.
Discharging process

Potential difference
is maximum, V0 and At this moment, all
maximum current ,I0 Q at plate A is fully
flows through the neutralized and V=0
resistor R.

Part of the (+)

charges on plate A Until (I=0) through
is neutralized by e, the resistor .
so V reduced.
Disharging graphs and formulas
Q
i) Charge:

t
− Q0
Q = Q0 e RC

where 0.37Q0
Q0 : maximum charge 0 τ time, t
R : resistance of the resistor
Charge on the capacitor
C : capacitance of the capacitor decreases exponentially with
time.
ii) Current: I
t
− τ
I = − I 0e RC
0 time, t
Q0 0.37 I 0
and I0 =
RC Current through the
where
I0 resistor decreases
I 0 : maximum current
exponentially with
Q0 : maximum charge
time.
R : resistance of the resistor
C : capacitance of the capacitor
(-) sign : the current direction is opposite to its direction
when the capacitor was being charged.

For calculation, the negative sign can be ignored.

Time constant, τ

• Measure of how quickly the capacitor charges

 = RC
or discharge
• Product of resistance and capacitance
• Scalar quantity
• Unit: second (s)

For charging process

• Time required for the capacitor to reach (1-e-1)=0.63 or 63% of
its maximum charge (or maximum voltage).
• Time required for the current drops to 1/e = 0.37 or 37% of its
initial value (I0).
For discharging process
• Time required for the charge on the capacitor (or voltage or
current in the resistor) decreases to 1/e = 0.37 or 37% of its
initial value.
 How fast a capacitor to get charged depends on time
constant (the product of RC).
From:
 = RC
R ;  C  ; 

If the capacitance C is larger,is larger, it would take

a longer time to charge the capacitor.

If the resistance R of the resistor is higher, the current

in the circuit is lower, and the capacitor takes a longer
time to get charged.
Graph of charge versus time (Q-t)
Charging Capacitor Discharging Capacitor

Q = Q 0 (1 − e − t / RC ) Q = Q 0 e − t / RC
Graph of Current versus time (I-t)
Charging Capacitor Discharging Capacitor

− t / RC
I = I0e I = I0e − t / RC
2.3 Capacitors with dielectrics
At the end of this chapter you should be able to:
ε
Define dielectric constant. εr =
ε0

Describe the effect of dielectric on a parallel plate capacitor.

Calculate capacitance of air-filled parallel

plate capacitor. 0 A
C=
d
ε
Use dielectric constant εr =
ε0

Use capacitance with dielectric

C =  r C0
Parallel-plate capacitor
A two parallel metallic plates capacitor of equal area A are
separated by a distance d and the space between plates is
vacuum or air as shown.
positive
terminal
A
+Q
V
d
−Q

negative E
terminal

When the capacitor is charged, its plates have charges of

equal magnitudes but opposite signs: + Q and -Q
Potential difference V across the plates is produced.
Since d << A so that the electric field strength E is
uniform between the plates.

The magnitude of the electric field strength within the

plates is given by σ and σ = Q
E=
ε0 A

Q
E=
Aε0

where σ : surface charge density on either plate

Unit : Cm-2
Scalar quantity
 Since Q=CV and V=Ed CV
E=
Aε0
CEd
E=
Aε0

Separated by εA ε0 A Separated
dielectric C= OR C= by vacuum
material d d

where
ε0 : permittivity of free space ( 0 = 8.85  10 −12 C 2 N −1 m −2 )
ε : permittivity of dielectric material
d : distance between the two plates
A : area of each plate
 Dielectric Constant, εr
Definition:
Ratio of the capacitance of a capacitor when a dielectric
is between its plate to the capacitance of a capacitor
when there is free space (vacuum) between its plates.

C
r =
C0
where
εr = dielectric constant ( relative permittivity of a dielectric material)
C = the capacitance of a capacitor when a dielectric is between its plate,
and
C0 = capacitance of a capacitor when there is free space (vacuum)
between its plates.
The capacitance of any capacitor is proportional to he
permittivity of the material between its plates,

C 
=
C0  0
where
ε = the permittivity of the dielectric
ε0 = the permittivity of free space

•Dimensionless, has no units.


r = •For free space, εr =1.
0 •Equation defines the permittivity of a
material.
Parallel Plate Capacitor

2 parallel metallic plates of equal area A are

separated by a distance d.

The plates are very close together – assume the electric field is
uniform between the plates.
The surface charge density on either plate :
Q
=
A
Applying the Gauss’s Law, the value of the electric
field between the plates :
 Q
E= =  (1)
 0 A 0

Since the field between the plates is uniform, the

magnitude of the potential difference between the
plates :
V = E d  (2)
Dielectric
Dielectric is an insulating material such as rubber, plastic or
waxed paper.
Dielectric materials: high electrical resistivity, but an
efficient supporter of electrostatic fields.

Can store energy / charge.

Able to support an electrostatic field while dissipating

minimal energy in the form of heat.

The effect of a dielectric is to increase the capacitance by a

factor of k ( increase the charge storage capacity of the
capacitor )
k is called dielectric constant / relative
permitivity. – dimensionless & is greater than 1.
If the gaps between the capacitor plates is not
vacuum but filled with a dielectric

mica
(dielectrics)

Permittivity of dielectric is given as 

r =
Rearrange:
0
ε = εr ε0
Capacitance, C for parallel plate filled with dielectrics :

 A ( r  o ) A
C= C=
d d
where
ε : dielectric permittivity of the material ( ε always
greater than ε0 )
εr : relative permittivity

Example : εr = 7.0 for mica

εr = 3.7 for paper
CHECK EITHER THE CAPACITOR CONNECTED
TO A BATTERY FIRST.
DISCONNECTED Q remain same
capacitor from the
battery BEFORE
modifications
Capacitor REMAIN V remain same
CONNECTED to the
battery WHILE
modifications