Chapter 2

BP3 FYSL 1

Chapter 2: Capacitors And Dielectrics

2.1 Capacitance and capacitors in series and parallel

L.O 2.1.1 Define capacitance and use capacitance

Capacitor is a device that is capable of storing electric charges or electric potential energy.It

is consist of two conducting plates separated by a small air gap or a thininsulator (called a

dielectric).The symbol for a capacitor is:

The capacitance of a capacitor is defined as the ratio of the charge on either plate to the

potential difference between them.

It is a scalar quantity

The unit of capacitance is farad (F) OR coulombs per volt (C V1)

It is always a positive quantity

L.O 2.3.1 Calculate capacitance of air-filled parallel plate capacitor

Parallel plate capacitor consists of a pair of parallel plates of area A separated by a small

distance d.

If a voltage is applied to a capacitor (connected to a battery), it quickly becomes charged.One

plate carries a charge +Q and the other carries a charge –Q then the potential difference

between these two parallel plates is V. Since d<<A so that the electric field strength E is

uniform between the plates.The capacitance of a parallel-plate capacitor, C is proportional

to the area of its plates and inversely proportional to the plate separation.

V

QC

where

Q: Charge on one of the plate

V: potential difference across two plates

d

AC 0

where

ε0 : permittivity of free space

A : area of the plate

d : distance between the two plates

)m N C 1085.8( 21212

0

Chapter 2

BP3 FYSL 2

L.O 2.1.2 Derive and determine the effective capacitance of capacitors in series and

parallel.

Capacitor in series:

Capacitor in parallel:

321 QQQQ

321 VVVV

Derivation:

321 VVVV

321 QQQQ

Derivation:

Chapter 2

BP3 FYSL 3

L.O 2.1.3 Derive and use energy stored in a capacitor

When the switch is closed, charges begin accumulate on the plates.A small amount of work

(dW) is done in bringing a small amount of charge (dQ) from the battery to the capacitor.

Example

Question Solution

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.

(Given0 = 8.85 x 10-12

C2 N

-1 m

-2)

In the circuit shown below, calculate the

a) equivalent capacitance

b) charge on each capacitor

c) the potential difference across each capacitor

VdQdW

dQC

QdW

C

QV and

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

QW

dQC

QdW

00

C

QWU

2

2

1 2

2

1CVU QVU

2

1OR OR

Chapter 2

BP3 FYSL 4

Question Solution

In the circuit shown in figure above, C1= 2.00 F,

C2 = 4.00 F and C3 = 9.00 F. The applied

potential difference between points a and b is

Vab = 61.5 V. Calculate

a) the charge on each capacitor

b) the potential difference across each capacitor

c.) the potential difference between points a and d

Determine the equivalent capacitance of the

configuration shown in figure below. All the

capacitors are identical and each has capacitance

of 1 F.

A 2 µF capacitor is charged to 200V using a

battery.

Calculate the

a) charge delivered by the battery

b) energy supplied by the battery.

c) energy stored in the capacitor.

Chapter 2

BP3 FYSL 5

Exercise

Question

An electric field of 2.80 × 105 V m

-1 is desired between two parallel plates each of area

21.0 cm2 and separated by 250 cm of air. Find the charge on each plate.

(Given 0 = 8.85 x 10-12

C2 N

-1 m

-2)

Four capacitors are connected as shown in figure.

Calculate

a) the equivalent capacitance between points a and b.

b) the charge on each capacitor if Vab=15.0 V.

Answer : 5.96 F, 89.5 C on 20 F, 63.2 C on 6 F,

26.3 C on 15 F and on 3 F

A 3.00 µF and a 4.00 µF capacitor are connected in series

and this combination is connected in parallel with a 2.00 µF

capacitor.

a) What is the net capacitance?

b) If 26.0 V is applied across the whole network, calculate the

voltage across each capacitor.

Answer: 3.71 µF, 26.0 V, 14.9 V, 11.1 V

Two capacitors, C1= 3.00 F and C2 = 6.00 F are connected in

series and charged with a 4.00 V battery as shown in figure.

Calculate

a) the total capacitance for the circuit above.

b) the charge on each capacitor.

c) the potential difference across each capacitor.

d) the energy stored in each capacitor.

e) the area of the each plate in capacitor C1 if the distance between two plates is 0.01 mm and

the region between plates is vacuum.

(Given permittivity of free space, 0 = 8.85 x 10-12

F m-1

)

Answer: 2.00 µF, 8.00 µC, V1 = 2.67 V, V2 = 1.33 V, U1 = 1.07 x 10 -5

J, U2 = 5.31 x 10-6

J,

3.39 m2

Chapter 2

BP3 FYSL 6

2.2 Charging and discharging of capacitors

L.O 2.2.1 Define and use time constant

L.O 2.2.2 Sketch and explain the characteristics of Q-t and I-t graph for charging

and discharging of a capacitor

L.O 2.2.3 Use

for discharging and

charging

Charging Discharging

Originally, both plates are neutral

When switch S is closed, current I0

immediately begins to flow through the

circuit.

Electrons will flow out from the negative

terminal of the battery and accumulate on

the plate B of the capacitor.

Then electrons will flow into the positive

terminal of the battery through the resistor R ,

leaving a positive charges on the plate A

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.

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.

When switch S is closed, electrons

from plate B begin to flow through the

resistor R and neutralize positive

charges at plate A.

Initially, the potential difference

(voltage) across the capacitor is

maximum, V0 and then a maximum

current I0 flows through the resistor R.

When part of the positive charges on

plate A is neutralized by the electrons,

the voltage across the capacitor is

reduced.

The process continues until the current

through the resistor is zero.

At this moment, all the charges at plate

A is fully neutralized and the voltage

across the capacitor becomes zero.

Chapter 2

BP3 FYSL 7

Charging Discharging

Note: For calculation of current in

discharging process, ignore the

negative sign in the formula.

The quantity RC that appears in the exponent for all equation is called time constantor

relaxation time of the circuit or mathematically

It is a scalar quantity.

Its unit is second (s).

It is a measure of how quickly the capacitor charges or discharges.

The time constant for a circuit used to charge a

capacitor is defined as the time taken for the

charging current to decrease toe

1of its initial

value.

The time constant for a circuit used to

discharge a capacitor is defined as the

time taken for thecharge (or potential

difference across) the capacitor to

decrease toe

1of its initial value.

RC

t

eQQ 10

RC

t

eVV 10

RC

t

eII

0

RC

t

eQQ

0

RC

t

eVV

0

RC

t

eII

0R

VI 0

0 RC

QI 0

0

RC

–

Chapter 2

BP3 FYSL 8

Example

Question Solution

The figure shows a simple circuit of the

photographic flash used in a camera. The

capacitance of the capacitor is 40.0 µF, and the

resistance of the resistor is 45.0 kΩ.

a) Explain the function of the capacitor in the

application above.

b) Calculate the time required to charge the

capacitor to 65% so that a good flash can be

obtained.

c) Suggest a way to reduce the charging time of a

capacitor.

In the RC circuit shown in figure below, the battery

has fully charged the capacitor.

Then at t = 0 s the switch S is thrown from position

a tob. The battery voltage is 20.0 V and the

capacitance C = 1.02 F. The current Iis observed

to decrease to 0.50 of its initial value in 40 s.

Determine

a)the value of R.

b) the time constant,

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

d) the value of Q on the capacitor at t = 60 s

45.0 kΩ

40.0 µF

Flash bulb

C

R

0V

S

b

a

Chapter 2

BP3 FYSL 9

Exercise

Question

A 20 µF capacitor and a 1.0 kΩ resistor are connected in series. A battery of emf 12 V is

connected across the series combination. Determine

a) the current at the instant when the battery is connected and current starts to flow in the

circuit

b) the current at time t = 10 ms

c) the time taken by the current to

i) decreases to 36.8% of the initial current

ii) decrease to 13.5% of the initial current

iii) become 2.0 mA

Answer: 12 mA, 7.3 mA, 20 ms, 40 ms, 36 ms

A 2 µF capacitor is charged using a battery of emf 1.5 V. The charged capacitor is then

discharged through a 60 kΩ resistor.

a) What is the time constant of the discharge circuit?

b) Find the time taken for the charge on the capacitor to decrease to (i) e

1 and (ii)

100

1 of its

initial value.

Answer: 0.12 s, 0.12 s, 0.553 s

A charged capacitor is connected in series to a switch and a resistor. A voltmeter of high

resistance is connected across the capacitor. The switch is closed, and a set of values for the

voltage V and the time t is obtained. The graph of lnV against t is plotted as follows.

Determine the time constant of the circuit.

Answer: 10 s

lnV

t (s)

2

1

0 10 20

Chapter 2

BP3 FYSL 10

2.3 Capacitors with dielectrics

L.O 2.3.2 Define and use dielectric constant

Dielectric is defined as a non-conducting (insulating)

material placed between the plates of a

capacitor.Since dielectric is an insulating material,

no free electrons are available in it.

The advantages of inserting the dielectric between the plates of the capacitor are

a) increase in capacitance

b) increase in maximum operating voltage/ prevent capacitor breakdown

c) possible mechanical support between the plates, which allows the plates to be close

together without touching, thereby decreasing d and increasing C

The dielectric strength is defined as the electric field strength at which dielectric breakdown

occurs and the material becomes a conductor.

Dielectric constant (also known as relative permittivity) is defined as the ratio between the

permittivity of dielectric material to the permittivity of free space.

It is dimensionless constant (no unit).

Other equations:

0

r

where

ε0 : permittivity of free space

ε : permittivity of dielectric material

Chapter 2

BP3 FYSL 11

L.O 2.3.3 Describe the effect of dielectric on a parallel plate capacitor

Induced electric field is in the opposite direction of the original electric field; hence reduce

the net electric field strength between the plates:

Since

Case 1: Battery disconnected and then

dielectrics is inserted

Case 2: Battery still connected and then

dielectrics is inserted

Capacitance will increase. SinceV

QC ,

when C increase, V decrease, Q remains

unchanged because the circuit is not

connected to the power supply (battery).

Capacitance will increase. The moment you

insert the dielectric, C increase, V decrease

becauseV

QC . Since the circuit is still

connected to the power supply, Q will be

increasing until VC = V again.

dnet EEE

Q constant

V not constant

Q not constant

V constant

Chapter 2

BP3 FYSL 12

L.O 2.3.4 Use capacitance with dielectric

oo

rC

C

and d

AC o

o

oo

d

A

C

d

AC

Example

Question Solution

A vacuum parallel-plate capacitor has plates of area

A = 150 cm2 and separation d = 2 mm. The capacitor

is charged to a potential difference V0 = 2000 V. Then

the battery is disconnected and a dielectric sheet of the

same area A is placed between the plates as shown in

Figure.

In the presence of the dielectric, the potential

difference across the plates is reduced to 500 V.

Determine

a. the initial capacitance of the capacitor,

b. the charge on each plate before the dielectric is

inserted,

c. the capacitance after the dielectric is in place,

d. the relative permittivity,

e. the permittivity of dielectric material,

f. the initial electric field,

g. the electric field after the dielectric is inserted.

(Given 0 = 8.85 1012

C2 N

1 m

2)

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