Chapter 1 Electrostatics SE

CHAPTER 3 : ELECTROSTATICS
( 5 HOURS )

Terminology :

1. Electrical charge : Cas elektrik.
Pengkuantuman.
2. Quantization : Aruhan.
Konduktor (pengalir)
3. Induction : Penebat.
Hukum Coulomb.
4. Conductor : Medan elektrik.
Keupayaan elektrik.
5. Insulator : Permukaan sekeupayaan.

6. Coulomb’s law :

7. Electric field :

8. Electric potential :

9. Equipotential surface :

Introduction :

1. Electrostatics refers to the study of electric charge at rest.

2. A particle (matter) consists of positive charged nucleus (consists of neutrons and
protons) and surrounded by negative charged electrons.

JOTTING
SPACE

positive charge negative charge
(proton) (electron)

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Electrical Charges :

1. When rubbed, bodies become charged, i.e. they accumulate electric charges of a
particular type.

2. Beside rubbing, bodies can be electrically charge through :

a) Touching : being touched by other charged bodies.

b) Induction : the flow of electric charges in a body under the influence

of charged objects.

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3. Properties of electrical charge, q :

a) There are two types of natural charge :

Positive charge → proton, q = p

Negative charge → electron, q = e.

b) The electrostatic interaction occurs between charged bodies :
Two bodies charged with identical charge repel while two bodies charged
with opposite charges attract one another.

c) Charge is conserved :
The law conservation of charge states that :
“In a system which is electrically isolated from the surrounding
environment, the total sum of electric charges remains constant.”

d) Charge is quantized :

Electric charges that occur in nature are always multiples of an elementary

charge :

Q = Nq

where : e = -1.602x10-19C
p = +1.602x10-19C
JOTTING
In the SI system, the unit of charge is coulomb (C). SPACE
1C = 1A x 1s

4. On the basis of electric conductivity, we may divide substances into conductors
and insulators (dielectrics).

Conductors are characterised by good conductivity as they contain free carriers

of electric current :

In metals → free electrons.

In electrolytes → ions.

Examples : Metals, graphite, electrolytes.

Insulators (dielectrics) are substances in which charges cannot flow. They have
no free electrons.

Examples : Amber, quartz, porcelain, ebonite, glass, silk.

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Coulomb’s law :

1. The Coulomb’s law states that :
“Two point charges Q and q repel OR attract one another with force F, which is
directly proportional to the product of the charges and inversely proportional to the
square of the distance between them.”

F ∝ Qq
r2

Or F = k Qq …..(3.1)
r2

where k = 1 = 9.0x109Nm2C-2
4πε o

εo is a universal constant known as ‘permittivity of free space’ with value of :
εo = 8.85x10-12C2N-1m-2

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2. The resultant force, FT (exerted on a test charge) can be calculated by adding
the vectors of the individual forces (which are produced by point charges).
FT = F1 + F2 + F3 + …+ Fn
JOTTING
Note : The number of individual forces does depend on the number of point SPACE
charge act upon the test charge.

Mathematically, the resultant force (net force or total force), FT is a vector quantity
with magnitude of :

( )FT = (∑ Fx )2 + ∑ Fy 2 …..(3.2)

and direction of : tan-1  ∑ Fy 
∑ Fx
θ = …..(3.3)

Note : The basic unit for force is kgms-2 or normally known as Newton (N).

| Chapter 3 | Electrostatics | page 3/14 |

Steps Taken to Calculate the Resultant Force, FT :
(Zahidi, 2006)

1. Draw the figure (if not provided).

point F3 y
charge
point F3
test charge θ3 F2
charge
point charge F2 x
θ1
3
F1
F1
free body diagram
system of particles

2. Identify the test charge (based on question) and its type (positive or negative).

3. Draw all individual forces (repel / attracting forces) acting upon the test charge

within direction. JOTTING
SPACE
4. Calculate the magnitude for all the individual forces using equation (3.1), i.e. :

Fi = k Qiq
ri 2

5. Draw the ‘free body diagram’.

- The test charge acts as the origin.

- Draw the x and y axis.

- Draw all individual forces acting upon the test charge (the arrow directs

away from the origin)

- Determine the angle, θi for each force (based on x-axis).

6. Construct a table consisting x and y components of forces as below :

F x-component y-component

F1 F1x = + F1cosθ1 = +……………. F1y = + F1sinθ1 = + …………….
F2 F2x = + F2cosθ2 = +……………. F2y = + F2sinθ2 = + …………….
F3 F3x = + F3cosθ3 = +……………. F3y = + F3sinθ3 = + …………….

Fi Fix = + Ficosθi = +……………. Fiy = + Fisinθi = + …………….

∑ Fx = ……………. ∑ Fy = …………….

Note : + sign for rightwards or upwards direction.
- sign for leftwards or downwards direction.

7. Determine the magnitude of the resultant force using equation (3.2), i.e. :

( )FT = (∑ Fx )2 + ∑ Fy 2

8. Determine the direction of =thteanre-1su∑∑ltaFFnyxt force using equation (3.3), i.e. :
θ

Note : The angle θ can be stated either in quadrant (e.g. θ = 25o (quadrant III))
or in range of 360o (e.g. θ = 205o (180o + 25o))

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Example : JOTTING
In a hydrogen atom, the electron is at a distance of 5x10-11 m from the nucleus which SPACE
consists of one proton. Calculate (a) the force between the two charges (b) the ratio of the
electrostatic force to the gravitational force between the charges. [Ans : (a) F = 9.2x10-8N
(b) FE/Fg = 2.3x1039]

Example :
Three charges each of +3 µC are fixed along the x-axis at x = 0, 0.30 m and 0.50 m.
Calculate the resultant force on the charge in the middle due to the other two charges. [Ans
: FT = 1.125N (towards q3)]

Electrostatic field, E :
1. Electrostatic field refers to the electric field that surrounds (in 3-D) a stationary

charged body (electric charge) where the electrostatic force is experienced.

EE

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2. Intensity (strength) of an electrostatic field :
The intensity of an electric field at a given point is the ratio of the electrostatic force
exerted on a test charge placed at this point to the value of the charge.

E= F
q

Or E = k Q …..(3.4)
r2

The test charge, q is normally a positive charge accumulated on a ball of infinitely
small size (which is being exerted by other point charge/s).

The direction of the vector of the intensity of the electrostatic field is depending on
the types of the point charge, Q.

Note : The SI unit for electrostatic field is NC-1 and sometimes known as Vm-1.

| Chapter 3 | Electrostatics | page 5/14 |

3. An electrostatic field is represented graphically by means of field lines. Electric
field lines are the lines whose tangents at each point.

Note : Electric field lines do not intersect each other.

Note : The concentration of the field lines indicates the strength of the field
intensity.

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4. Principle of superposition : JOTTING
“The intensity (strength) of an electrostatic field at a given point in space is the SPACE
vector sum of the intensities of the fields generated by the individual charge.”
ET = E1 + E2 + E3 + …+ En

Example :

E1 E1
•A C•
E2
E2

E1 E2

B © Physics Teaching Courseware
• E2

E1

with magnitude of : ( )ET = (∑ E x )2 + ∑ E y 2 …..(3.5)
and direction of : …..(3.6)
tan-1  ∑Ey 
θ = ∑Ex

| Chapter 3 | Electrostatics | page 6/14 |

Steps Taken to Calculate the Resultant Electric Field, ET at a Point O :
(Zahidi, 2006)

1. Draw the figure (if not provided). E1

• E1
C•
A E2 E2

point charge, Q1 E1 E2 point charge,
Q2

B
• E2

E1 JOTTING
2. Label the selected point as O (or A or B ect.) SPACE
3. Draw all individual electric field acting upon the point within direction.

Note : The number of individual electric field is depend on the number of point
charge.
4. Calculate the magnitude for all the individual electric field using equation (3.4) :

E = k Qi
ri 2

5. Draw the ‘free body diagram’ (choose point A as example) :
- The point A acts as the origin.
- Draw the x and y axis.
- Draw all individual electric field acting upon the point A (the arrow
directs away from the origin)
- Determine the angle, θi for each electric field (based on x-axis).

y
E1

θ1 x

A θ2
E2

6. Construct a table consisting x and y components of electric field as below :

E x-component y-component

E1 E1x = + E1cosθ1 = +……………. E1y = + E1sinθ1 = + …………….
E2 E2x = + E2cosθ2 = +……………. E2y = + E2sinθ2 = + …………….
E3 E3x = + E3cosθ3 = +……………. E3y = + E3sinθ3 = + …………….

Ei Eix = + Eicosθi = +……………. Eiy = + Eisinθi = + …………….

∑Ex = ……………. ∑Ey = …………….

Note : + sign for rightwards or upwards direction.
- sign for leftwards or downwards direction.

7. Determine the magnitude of the resultant electric field using equation (3.5), i.e. :

( )ET = (∑ E x )2 + ∑ E y 2

8. Determine the direction of =thteanre-1su∑∑ltaEEnxyt e lectric field using equation (3.6), i.e. :
θ

| Chapter 3 | Electrostatics | page 7/14 |

Note : The angle θ can be stated either in quadrant (e.g. θ = 25o (quadrant III))
or in range of 360o (e.g. θ = 205o (180o + 25o))

Patterns of electrostatic field :

a) Isolated charge : b) Parallel plates of uniform
charges :

EE

JOTTING
SPACE

c) Dipole charges : d) Like charges :
E E

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Moving charge in a Uniform Electric Field :

1. A uniform electrostatic field is one which lines are parallel and which the magnitude
of the force exerted on a test charge is constant at every point of the field.

E

© Physics Teaching Courseware JOTTING
SPACE
2. Consider the motion of a charged particle (electron or proton), q in a uniform
electric field, E. © Physics Teaching Courseware © Physics Teaching Courseware

F

+-

F

Force exerted on the charge along the electric field :

F = qE .....(i)

Note : If the charge is positive, then the direction of F and E are the same.
If the charge is negative, then F and E are directed in opposite direction.

According to the Newton’s 2nd Law : …..(ii)
F = mQa

From equation (i) and (ii), thus the acceleration :
a = qE
mQ

Note : The charge does not undergo a circular motion. If the separation between
both plates is too thick, the charge will undergo a projectile motion since
the force is always directed in one direction.

| Chapter 3 | Electrostatics | page 9/14 |

Example :

Two parallel metal plates separated 2 cm are connected to a 120 V battery. Calculate (a)
the electric field intensity between the plates (b) the force exerts on an electron in between
the plates. [Ans : (a) E = 6kVm-1 (b) F = 9.6x10-16N]

Example :
Two point charges of +5 µC and -2 µC respectively are separated by a distance of 5.0 cm.
Determine the electric field at the midpoint between the charges.
[Ans : ET = 100.8x106 NC-1]

Electric Potential, V :

3. Electric potential :

= The work performed (or required) to bring unit electric charge from infinity

to a point (for example : point A) in an electric field at which the potential is

being specified.

V= W ∫where W = F • dr
q
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Or V = kQ JOTTING…..(3.7)
r SPACE

where r = distance between the point and Q.

A∞

Note : Unit electric charge refers a particle of charged 1C.

Note : Electric potential is a scalar quantity.
The SI unit for electric potential is JC-1 or normally known as volt, V.

Note : Work (or energy) is required for two identical charged bodies since they’re
repel each other.

If the charge is moved from infinity to a fixed point, the work done can be written
as :

∫W = F • dr where F = qE

W = ∫ (qE)•dr …..(i)

| Chapter 3 | Electrostatics | page 10/14 |

From expression : …..(ii)
V= W
q …..(3.8)
…..(3.9)
Substitute equation (i) into (ii) :

V = ∫E• dr

Or E = dV
dr

Electric Potential of a System of Point Charges

© Physics Teaching Courseware r1 r2
JOTTINGr3
SPACE

The total electric potential at a certain point (eg. point O) is given by the following
equation :

V = k  Q1 + Q2 + Q3 + .... …..(3.10)
r1 r2 r3

where r1, r2 and r3 respectively are the distance from the each point charge to
point O.

Note : The magnitude also depends on the type of charge of the point charges.

4. The potential difference, ∆V between two points (for example : A and B) in an
electric field is the difference in the values of the electric potentials at the two
points.

∆V = VB - VA …..(3.11)

| Chapter 3 | Electrostatics | page 11/14 |

In other word, the potential difference refers to the work done in moving unit
charge from one point the other.

V = WAB …..(3.12)
q

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Figure : The test charge, q is moved from point A to B.

5. Potential energy, U : JOTTING
= The work performed (or energy required) to bring an electric charge from SPACE
infinity to a point (for example : point A) in an electric field at which the
potential is being specified.

Based on work-energy theorem :

V = kQq …..(3.13)
r

where r = distance between the point and Q.

© Physics Teaching Courseware A∞

Note : Potential energy is a scalar quantity.
The SI unit for electric potential is J.

The relationship between potential energy, U and electric potential, V :

U = qV …..(3.14)

| Chapter 3 | Electrostatics | page 12/14 |

The relationship between potential energy, U and distance, r in a system of
different and identical charges is shown in the graph below :

Qq Qq

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Figure 1 Figure 2

Figure 1 : U is –ve (the work is performed by the test charge as two
Figure 2 : dipole charges attract each other.

U is +ve (the work is performed upon the test charge as two JOTTING
like charges repel each other. SPACE

Electric Potential Energy of a System of Point Charges

© Physics Teaching Courseware r1 r2
qo
r3

The total electric potential energy upon a test charge, qo is given by the following
equation :

U = k qo  Q1 + Q2 + Q3 + .... …..(3.15)
r1 r2 r3

where r1, r2 and r3 respectively are the distance from the each point charge to the test
charge, qo.

Note : The magnitude also depends on the type of charge of the point charges as
well as the test charge itself.

| Chapter 3 | Electrostatics | page 13/14 |

Equipotential Surface :

1. An equipotential surface is a set of points which have the same potential (also
known as the closed area).

All points on the equipotential surface are having the same electric potential and
the electric field at any point on the surface always perpendicular to the surface.

© Physics Teaching Courseware Equipotential
JOTTINGsurfaces
SPACE
equipotential surfaces of an isolated charge.

In the case of a point charge Q which is generating an electric field, the
potential of the field at a distance r is :

V = kQ
r

Or V= Q
4πε o r

The work performed when a charge is moved within the equipotential surface :

WAB = UB – UA ; VB = VA
= q(VB – VA)
=0

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Figure : Equipotential surfaces of an electric dipole and a uniform electric field.

Example :

Calculate the electric potential at a point (a) 15 cm and (b) 50 cm from a point charge of
3µC. (c) Calculate the work done to move a charge of 0.5µC from a point 50cm from the
point charge 3µC to a point 15cm from the point charge. [Ans: (a) V = 180kV (b) V = 54kV
(c) W = 63mJ]

| Chapter 3 | Electrostatics | page 14/14 |