Chapter 2 Capacitor & Dielectrics SE

CHAPTER 4 : CAPACITORS AND DIELECTRICS
( 3 HOURS )

Terminology :

1. Capacitor : Kapasitor.
Penebat.
2. Insulator : Kapasitans.
Sesiri.
3. Capacitance : Selari.
Pengecasan.
4. Series : Penyahcasan.

5. Parallel :

6. Charging :

7. Discharging :

Introduction :

1. Capacitor :
= An arrangement of conductors separated by an insulator (dielectric) used to
store charge.

The dielectric may be air, paper impregnated with oil or wax, plastic film or ceramic.

2. The figure below shows the diagram of a capacitor :

JOTTING
SPACE

3. If the two conductors are being charged, the magnitude of charge at the both
conductor is the same but in different sign (+Q and –Q respectively) and thus,
the potential difference, V between the conductors is produced.

- -Q +Q +

V

4. Capacitor has several shapes, including :
a) Cylindrical capacitor.
b) Concentric spherical capacitor.
c) Parallel plate’s capacitor.
d) Electrolytic capacitor.

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| Chapter 4 | Capacitor & Dielectrics | page 1/10 |

Capacitance, C :

1. Capacitance refers to the property of a conductor or system of conductors that
describes its ability to store electric charge, Q.

Mathematically, capacitance can be defined as :
= the ratio of the charge which has accumulated on the conductor to the generated

electric potential. It is expressed by the formula :

C= Q …..(4.1)
V

Capacitance is a scalar quantity. In the SI system, the unit of capacitance is CV-1
or normally known as Farad, F.

A farad is a very large unit, that is why in practice we use smaller units :
1µF = 10-6F (microfarads)
1nF = 10-9F (nanofarads)
1pF = 10-12F (picofarads)

Parallel-plate capacitors : JOTTING
SPACE
1. The simplest form of capacitor has two parallel rectangular conducting plates
of area A, separated by a dielectric of air of thickness d.

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Based on the Gauss’s law, the magnitude of E at any point is represented as

follow :

σ where σ = Q
E = κεo A

Thus E= Q …..(i)
κεo A

Relationship between electric field, E and potential difference, V :

E= V …..(ii)
d

From equation (i) and (ii), then the capacitance of a parallel-plate capacitor is
shown as follow :

C = κεo A …..(4.2)
d

where εo = Universal constant known as ‘permittivity of free space’
= 8.85x10-12C2N-1m-2

κ = Dielectric constant.

| Chapter 4 | Capacitor & Dielectrics | page 2/10 |

Note : If there is no dielectric material between the capacitor (vacuum), then :

C = εoA since κvacuum = 1.00
d

Example :

The plates of a parallel-plate capacitor are 8.0 mm apart and each has an area of 4.0 cm2.
The plates are in vacuum. If the potential difference across the plates is 2.0 kV, determine
(a) the capacitance of the capacitor (b) the amount of charge on each plate (c) the electric
field strength was produced.

Capacitors connected in Series and Parallel : JOTTING
SPACE
1. Capacitors play a vital role in electronics, therefore it is important it know how
they can be connected. Capacitors may be joined in parallel or in series or
even in their combination.

Note : The type of connection does not depend on the arrangement.

2. Series combination of capacitors :

The figure below shows the diagram of a capacitors connected in series :

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Hint : Two capacitors are being considered connected in series if only one of
each end is joining each other while the other end joins with other
capacitor (or other component).

In a series combination of capacitors, the potential difference between the
conducting surfaces (in all capacitors) is distributed evenly on all the capacitors :

V = V1 + V2 + V3 + … + Vn …..(i)

where V1 = Q1 , V2 = Q2 , V3 = Q3
or generally, C1 C2 C3

Vn = Qn
Cn

The total charge :

QT = Q1 = Q2 = Q3 = … = Qn

| Chapter 4 | Capacitor & Dielectrics | page 3/10 |

This system of capacitors may be substituted with one capacitor of a specified

capacitance known as the ‘equivalent @ effective capacitance, CT’. From
equation (i), we obtain :

QT = Q1 + Q2 + Q3 + … + Qn
CT C1 C2 C3 Cn

Thus 1 = 1 + 1 + 1 +…+ 1 …..(4.3)
CT C1 C2 C3 Cn

JOTTING
SPACE

3. Parallel combination of capacitors :
The figure below shows the diagram of a capacitors connected in parallel :

T

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Hint : Two capacitors are being considered connected in parallel if both their
ends are joining each other.

In a parallel combination of capacitors, the potential difference between the
conducting surfaces (in all capacitors) is the same and equals V :

V = V1 = V2 = V3 = … = Vn

The total charge :

QT = Q1 + Q2 + Q3 + … + Qn …..(i)

where Q1 = C1V
or generally, Q2 = C2V
Q3 = C3V
Qn = CnV

This system of capacitors may be substituted with one capacitor of a specified

capacitance known as the ‘equivalent @ effective capacitance, CT’. From
equation (i), we obtain :

Thus CTV = C1V + C2V + C3V + … + CnV …..(4.4)
CT = C1 + C2 + C3 + … + Cn

| Chapter 4 | Capacitor & Dielectrics | page 4/10 |

4. Combined Capacitors :
Sometimes capacitors are arranged in the form of combination between series and
parallel in the same circuit.

Stages On Solving Question Related To The Combined Capacitors
(Zahidi, 2007)

STAGE N = 1 Example :

Draw and simplifyS Stage 1 : 1
the circuit. Stage 2 : 24
3
Assign all the individual capacitance
as C1, C2, … 5

Identify all the specific basic 123 4
connections (series or parallel).

Calculate each of the equivalent
capacitances involved using
appropriate formula.

5JOTTING
SPACE

STAGE N = N + 1 Stage 3 : 1234
5
Draw the equivalent circuit (simpler
than the previous stage).

Assign all the equivalent
capacitances (e.g. C12, C345, …)

Identify all the
specific basic connections

(series or parallel).

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Do the connections Calculate each of the
can be solved using a No equivalent capacitances

single formula ? involved using appropriate
formula.

Yes

Calculate the
total equivalent capacitance,

Ceq

| Chapter 4 | Capacitor & Dielectrics | page 5/10 |

Example :

Two capacitors, C1 = 2 µF and C2 = 4 µF are connected in series to a 12 V battery.
Calculate (a) equivalent capacitance (b) charge on each capacitor (c) potential difference
across each capacitor.

C1, Q1, C2, Q2, Example 4 :
V1 V2
Four capacitors are connected as shown in figure.
p=o1i5n.t0Cs V3V,.Q3 3, C4, Q4, Calculate (a) the equivalent capacitance between
V4 a and b (b) the charge on each capacitor if Vab

Energy stored in a Charged Capacitor :

1. The energy of a charged capacitor is the work that needs to be performed in JOTTING
order to carry the charge from one conducting surface to the other. SPACE

2. The work needed to transport small additional charge dq :

dW = Vdq where V = q/C
=  q dq
C

If the whole charge Q is transported, the total work is equal to energy stored in the
charged capacitor :

∫U = dW
∫= Q  q dq

C

0

∴ U = Q2 …..(4.5)
2C

3. From the previous lesson, it stated that :
C= Q
V

Therefore the energy stored also can be written as :

U = 1 QV …..(4.6)
2 …..(4.7)

Or U = 1 CV 2
2

| Chapter 4 | Capacitor & Dielectrics | page 6/10 |

Charging and Discharging of Capacitors :

1. Charging of capacitors :
The figure below shows the diagram of a capacitors in charging process :

I C R
V

mA

R

The magnitude of the current which flows in the circuit (across the resistor)
decreases exponentially with time according to the following equation :

−t …..(4.8)

I = Io e RC

The charge on the capacitor increases exponentially with time according to the JOTTING
following equation : SPACE

Q = Qo 1 − −t  …..(4.9)

e RC



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| Chapter 4 | Capacitor & Dielectrics | page 7/10 |

2. Discharging of capacitors :
The figure below shows the diagram of a capacitors in discharging process :

V −− −−
C+ + + + R

R

mA
I

The magnitude of the current which flows in the circuit (across the resistor)
decreases exponentially with time according to the following equation :

−t …..(4.10)

I = Io e RC

The charge on the capacitor decreases exponentially with time according to the JOTTING
following equation : SPACE

−t …..(4.11)

Q = Qoe RC

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3. Time Constant, τ :
= The time taken for the charge and the voltage across the capacitor to reach 63%
of their maximum value when charging or 37% of their maximum value when
discharging.

τ = RC …..(4.12)

Time constant, τ is a scalar quantity. In the SI unit of time constant is second (s).

Note : Time constant, τ plays a vital role to control as well as to determine the
percentage of charge accumulated or remains in the capacitor within a
certain time (a measure of how quickly the capacitor charges or
discharges) since this process can not be observed by our naked eyes.

| Chapter 4 | Capacitor & Dielectrics | page 8/10 |

Example :

A capacitor of 3 µF is charged using a battery of 1.5 V. The charged capacitor is then
discharged through 50 kΩ resistor. (a) Determine the time constant of the discharge circuit
(b) Calculate the time taken for the charge on the capacitor to decrease to 1/100 of its
initial value.

Dielectrics :

1. Between the conducting surfaces of a charged capacitor, we place in plates of
various dielectric materials : a glass plate, an ebonite plate and a plastic plate.

The increase in the capacitance depends on the type of dielectric that is placed
between the surfaces of a capacitor.

2. Dielectric :

= An insulating sheet placed in between the two conducting surfaces of a capacitor
in order to increase the capacitance of the capacitor with non-dimensional factor
called ‘dielectric constant’, κ (or εr).

C = κCo …..(4.13) JOTTING
SPACE
where Co = capacitance without dielectric (or vacuum).

Values for several types of dielectric :

Dielectric Dielectric constant’,
κ.
Vacuum
Air 1.00000
1.00059
Pyrex
Water 5.6
Teflon 80
Paper 2.1
Polystyrene 3.7
2.56

3. When the dielectric is inserted between the plates, the molecules of the dielectric
are polarized. This means that the surface of the dielectric facing the positive plate
becomes negatively charged, and the other surface becomes positively charged.

VV

+_ +_ +_
+_ +_ +_
Eo + _ + _ + _
+_ +_ +_
+_ +_ +_
+_ +_ +_

No dielectric Dielectric inserted

| Chapter 4 | Capacitor & Dielectrics | page 9/10 |

A reverse electric field opposite in direction to the original electric field, Eo is set up. JOTTING
The resultant electric field, E between the plates decreases and hence the SPACE
potential difference decreases.
If the capacitor is still connected to the same battery, more charge flows into the
capacitor until the potential difference is back to V. Since the charge Q on either
plate increases for the same potential difference, the capacitance of the capacitor
increases.

Example :
Consider a parallel plate capacitor of capacitance C = 110 µF, the plate area A = 0.15 m2
and dielectric constant κ = 6.9. (a) Calculate the separation between the plates (b)
Determine the electric field strength formed if a 10 V potential difference is applied across
the capacitor. (c) State two ways that can increase the capacitance without increasing the
area of the capacitor plate.

| Chapter 4 | Capacitor & Dielectrics | page 10/10 |