Chapter 3 Electric Field 1

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Introduction

Electrostatics

presence of stationary charges thatcause electric field.

Charge polarity: +ve orve, depends on number of

electrons.

The force of attraction & repulsion acts directly

between 2 adjacent charges.

2 types of materialsconductor & insulator.


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Most electric fields exist between 2 conductors.

The space between 2 conductors needs to be filledwith an insulator (known as dielectric), otherwise the

charges would move towards one another and

therefore be dissipated.

Fundamental electronic charge is -1.6x10-19 C.


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Coulombs Law

The force between 2 point charges is proportional tothe product of the 2 charges, & inversely

proportional to the square of the distance between

the charges.

Newtonr

QQkF

2

21


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kis a constant with a value of 1/4 with =

0r= absolute permittivity

0= free space permittivity = 8.85 10-12F/m

r= relative permittivity (no unit) If vectors are used for directional force,

Newtonrr

QQ

kF 12221


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Capacitor & Capacitance

Capacitor

an electronic component consists of 2

conducting surfaces separated by a later of insulating

medium/material (diaelectric)

Capacitance the property of a capacitor to store an

electric charge when its plates ate at different

potentials.

Capacitance Symbol: C Unit: Farad (F)


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1Fof capacitance is when there appears a

potential difference of 1Vbetween the plateswhen it is charged by 1Cof electricity.

Note: In practice, capacitance is usually

expressed in F or pF.


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Example 1

A capacitor having a capacitance of 80F is connected

across a 500V d.c. supply. Calculate the charge.

Since Q = CV

Charge = (8010-6)(500) = 0.04C = 40mC


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Capacitors in Parallel

The charge on C1is Q1coulombs and C2is Q2coulombs

Q1= C1V and Q2= C2V

Suppose 2 capacitors havingcapacitance C1& C2farads

are connected in parallel

across a pd of V volts.


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If we replace C1& C2by a single capacitor ofsuch capacitance C farad that the same total

charge of (Q1+ Q2) coulombs would be

produced by the same p.d., then Q1+ Q2= CV.

Thus, the resultant capacitance, C = C1+ C2

farads


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Capacitors in Series

If V1& V2are corresponding pdacross C1& C2, Q = C1V1+ C2V2

So that

If we replace C1& C2by a singlecapacitance Cfarad such that it

would have the same charge Q

coulombs with the same pd of V

volts, then

2

2

1

1

C

QVand

C

QV

C

QV


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But clearly, V = V1+ V2

Therefore, the resultant capacitance is thesum of reciprocal.


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Example 2If 2 capacitors having capacitances of 6F and 10F

respectively are connected in series across a 200Vsupply, find the potential different across & charge on

each capacitor.

Sol:

a) Let V1& V2be the pd across 6F and 10F capacitors.

From eq. above, therefore


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Solution 2

b) Charge, Q1= (610-6)(125) = 750C

Q2= (1010-6)(75) = 750C


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Example 3

For the circuit below, find the voltage across each capacitor.

First, solve for Ceq, which will be

The total charge is: q = Ceqv = 10 10-330 = 0.3C

This is the charge on the 20-mF and 30-mF capacitors because

they are in series with the source. Therefore:

mFCeq 10

20

1

30

1

60

1

1

VC

q

v 151020

3.0

6

1

1

VC

q

v 101030

3.0

6

2

2

Vv 53


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Gausss Law Here, this law concerns electric field.

It defines that, the electric flux through any closed surface is

proportional to the enclosed electric charge.

The left-side is the surface integral denoting the electric flux E

through a closed surface S, while the right side is the total

charge enclosed by Sdivided by the electric constant.

0

QdsE

S


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Electric Field Strength &

Electric Flux Density

We can investigate an electric field by observing its

effect on a charge.

The magnitude of a force experienced by a unit

charge at any point in a field is termed the electricfield strength at that point.


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Electric field strength is also known as electric stress,

and is measured in newtons per unit charge andrepresented by the symbol E.

1Jof work is necessary to raise the potential of 1Cof

charge through 1V.

Therefore, to move a unit charge through an electric

field so that its potential changes by Vvolts requires

Vjoules of work.


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The most simple arrangement for this is the parallel

charged plates.

Assume that the plates are very large & the distance

between the plates, dis very small & there is free

space between them.


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There is a potential difference of V volts between the

plates, therefore the work in transferring 1C ofcharge between the plates is V joules.

But work is the product of force & distance.

Therefore the force experienced by the charges is theelectric field strength,


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The electric flux density Dis given by:

In electrostatics, the ratio of electric flux density in a vacuum

to the electric field strength is termed thepermittivity of free

space(0).

or

2/ mcoulombs

A

QD

A

Cd

V

d

A

Q

E

D

strengthfieldElectric

densityfluxElectric

0

faradsd

A

C

0


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Permittivity of free space(0) | Unit: farad per

metre (F/m)

From carefully conducted tests,

0= 8.85 10-12F/m


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If the space in between the plates is not vacuum & it

is filled with a certain type of material (glass, rubber,

mica, etc.), the expression becomes

When the vacuum (or air) is replaced with a certain

material as diaelectric, the ratio of the capacitanceincreases greatly.

faradsd

AC r

0


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ris termed as the relative permittivity of that

material and it varies for different type of

materials.

is the absolute permittivity and is equal to

0r


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Example 3

A capacitor is made with metal plates and separated by a

sheet of 0.5mm thick glass with a permittivity of 8. The

area of the plate is 500cm2, what is its capacitance?

Given: d = 0.5mm = 0.0005m ;r= 8;

A = 500cm2= 0.05m2


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Sol 3

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Chapter 3 Electric Field 1