Note Physics 2 SP025

PHYSICS LECTURE NOTES

PHYSICS 2

SP025

SESSION 2020/2021

UNIT FIZIK
JOHOR MATRICULATION COLLEGE

CHAPTER 1: Electrostatics is a
ELECTROSTATIC branch of physics

2 HOURS that studies
electric charges
1.1 Coulomb’s Law.
1.2 Electric field. at rest.
1.3 Electric potential
1.4 Charge in a uniform electric field

MODE Face to Face Non Face to face SP025
SLT SLT PHYSICS UNIT
Lecture 2 2 KOLEJ MATRIKULASI JOHOR
Tutorial 6 6
1

LEARNING OUTCOMES

At the end of this topic, students should be able to:
1.1 Coulomb’s Law

a) State Coulomb’s Law,

(Lecture : C2,PLO1, MQF LOD1)

b) Sketch the electric force diagram.

(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

c) Apply Coulomb’s Law for a system of point charges

(*simple configuration of charges with a maximum of four charges in 2D)

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

2

LEARNING OUTCOMES

1.2 Electric Field

a) Define and use electric field strength,
⃗
=
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

b) Use for point charge,

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

c) Sketch the electric field strength diagram

(*simple configuration of charges with a maximum of four charges in 2D) (Tutorial : C4, PLO4, CTPS3, MQF LOD6)

d) Determine electric field strength for a system

of charges. (*simple configuration of charges with a maximum of four charges in 2D)

(Tutorial : C4, PLO4, CTPS3, MQ3 F LOD6)

LEARNING OUTCOMES

1.3 Electric Potential

a) Define electric potential,


b) Define and explain equipotential lines and surfaces of an isolated
charge in a uniform electric field

(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)

c) Use for a point charge and a system of charges. Maximum

four charges in 2D.

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

d) Calculate potential difference between two points,

,

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

4

LEARNING OUTCOMES

1.3 Electric Potential

e) Deduce the change in potential energy, between two points in
electric field,

(Tutorial : C5, CLO3,PLO4, CTPS3, MQF LOD6)

f) Calculate potential energy of a system of point charges :

Up to maximum three charges

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

5

LEARNING OUTCOMES

1.4 Charge in a uniform electric field

a) Explain qualitatively with the aid of a diagram the motion of a
charge in a uniform electric field

(Lecture : C2,PLO1, MQF LOD1)

b) Use for uniform E in:

i. stationary charge
ii. charge moving perpendicularly to the field
iii. charge moving parallel to the field
iv. charge in dynamic equilibrium

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

•6

1.1 COULOMB’S LAW

1.1a) State Coulomb’s Law

Coulomb’s Law states that the electrostatic force, F between two charges separated
by a distance, r, is
• directly proportional to the product of the magnitudes of the charges, Q1 and Q2.;
• inversely proportional to the square of the separation distance, r between the two
charges
Where charge Q

Mathematically; Q = ne represents the total
number of
electrons x

magnitude of

electrons.

e = 1.6 x 10-19 C

where : k  1  9.0 109 Nm2C-2
k = Coulomb constant which has the value of 9 × 109 N m2 C–2 4
Q1 = magnitude of charge Q1 7
Q2 = magnitude of charge Q2 where :
r = distance between the two charges. εo = permittivity of free
space 8.85  10–12 F m–1

• Figures (a) and (b) show the variation of electrostatic force with
the distance between two charges.

F F
Gradient,
0 r m  kQ1Q2 1
r2
Figure a 0 Figure b
8

1.1b) Sketch the electric force diagram

• This electrostatic force is directed along the line joining the charges.

F12 F21 + F12 + + F21

Q1 Q2 Q1 Q2

Attractive electrostatic force if Repulsive electrostatic force if two
two charge have opposite sign charge have same sign

• The notation F12 denotes the force exerted on charge 1 by charge 2 and F21 is the
force exerted on charge 2 by charge 1.

• Since electric forces obey Newton’s third law, therefore the forces F12 and F21 are
equal in magnitude but opposite in direction.

• Hence, it can be written as F12 = – F21

Note :
The sign of the charge can be ignored when substituting into the Coulomb’s law equation.

The sign of the charges is important in distinguishing the direction of the electric force when we draw9

the electric force vector.

Example 1 Q2

QUESTION : Sketch the electrostatic force on charges.

Q1

r

Direction of force & F12 = – F21 Direction of force &
the notation between the notation between

2 +ve charges 2 +ve charges

10

Example 2 Q2

QUESTION : Sketch the electrostatic force charge Q2.

Q1

r

Direction of force & 11
the notation between
2 charges for charge q2

only.

Example 3

Three point charges are firmly held on a straight line of 4 cm in length as shown in
the figure below.

Q1 = +10 μC Q2 = +5 μC Q3 = -8 μC

2 cm 2 cm

Find the resultant electric force acting on charge Q2

The electrostatic force is a vector quantity and

has a direction as well as magnitude. When adding

electrostatic forces, must take into account the

directions of all forces, using vector components as

needed 12

Solution
Step 1 : Draw the electric force vectors diagram

Q1 = +10 μC Q2 = +5 μC Q3 = −8 μC

The force acting on Q2 due to Q1 is repulsive because Q1 and Q2 have
the same sign, therefore the direction of F21 is to the right.

The force acting on Q2 due to Q3 is attractive because Q2 and Q3 have
the opposite sign, therefore the direction of F23 is also to the right.

Step 2 : Use Coulomb equation, find the magnitude of each of the
electric forces.

The magnitudes of F21 and F23 are given by :

F 21  k Q 1Q 2 13
r2

Step 3 : Electric force adds as vector (consider the direction)
Therefore, the resultant electric force acting on charge Q2 is ;

14

1.2 ELECTRIC FIELD TEST PHYSICAL
CHARGE MEANING OF AN
1.2a) Define & Use electric field POINT CHARGE ELECTRIC FIELD ;

Electric Field Strength, E is a region in which
an electric force will
Definition : The electric force, F act on a charge that
acting on a test charge that is is placed in the
placed in the electric field region region.
divided by the magnitude charge
of the test charge, q0.

Where, • E is a vector quantity. SI unit for E is N C−1. 15
= electric field
= electric force • Electric field patterns can be represented by electric field
= test charge lines which is drawn pointing in the direction of the E vector
at any point.

Properties of Electric field lines

(1) These electric filed lines never cross each other
and the number of lines determine the strength of the
electric field.
(2) E is large when the field lines are close together
and small when they are far apart.

(3) A positive and a negative charge can produce
electric field.

(4) The number of electric field lines entering or
leaving a charge is proportional to the magnitude of
the charge.

16

(1) Electric field lines pattern for a stationary charge (+ / -)

FIGURE A FIGURE (B)

(5) The direction of E for a positive (6) The direction of E for a negative
point charge is outward from the point charge is toward the charge in
charge in all direction (3 dimension) all direction (3 dimension) (FIGURE B)
(FIGURE A)

17

(2) Electric field lines pattern between TWO charges

Electric field lines Electric field lines Electric field lines
(patterns) for two (patterns) for two (patterns) for a
equal positive point point charge +2q
equal and and a second
opposite point charges. point charge –q.

charges.

18

(3) Uniform electric field lines in parallel plate

Value of E is always constant Electric field lines (patterns) for two
++++++++++++++ opposite charged parallel metal plates.

E • The electric field lines are perpendicular to

-- -- -- -- -- -- -- -- -- -- -- -- the surface of the metal plates.

* Minimum 3 lines • The lines go directly from positive plate to
for electric field lines the negative plate.
/ direction of E for a
+ or – charge. • The field lines are parallel and equally
+ + + + + + + + + + ++ spaced in the central region far from the edges
-- -- -- -- -- -- -- -- -- -- but fringe outward near the edges.

Parallel plate can be • In the central region, the electric field has
horizontal or vertical the same magnitude at all points.

E 19

1.2c) Sketch the electric field strength diagram

QUESTION : Sketch the electric field at point P

a) P E1 Qc) The electric field
2 lines must be
Q1 - drawn from the
point
+
E2 E1
b) E2 P Q1
P
Q2 +
20
-

1.2b) Use E for a point charge

Electric field strength, E for a point charge.
Consider a test charge, q0 located at a distance r from a point charge Q,

According to Coulomb’s Law ;   k Q qo  (1)
F r2

From definition :     (2)
E F
qo
where;
By substituting   k Q k – Coulomb constant (9.0×109 N m2 C–2 )
(1) into (2), we get : E r2 Q – point charge that produce electric field
r – distance a point from the point charge
21

Notice that E is inversely proportional to r2 ;
The relationship between E and r can be shown in the graph below.

E

r

The strength of E will decreases when the distance from the point
charge increases.

Note:
The direction of the electric field strength, E depends on the sign of the point charge only. 22

Relation between Direction of E and F0 due to the charge Q

Electric force,F0 = q0E is acting on a test charge q0 that is placed in an electric field E ;

23

Example 4

Determine:

(a) the electric field strength at a point X at a distance of 20 cm from a point charge
Q = +8 μC.

(a) the electric force that acts on a test charge qo = –1 μC placed at point X.

Q = 8 μC X

+ 20 cm

24

Solution

Step 1 : Draw the electric field diagram. Field point away from a + charge, and
towards a ‒ charge.

Q = 8 μC 20 cm X

+ -

Step 2 : Apply to find the magnitude of the each electric field create

the surrounding point charge(s) Notes: Substitute only magnitude of the

charges, leaving out minus sign (‒) for negative charge

a) b)

E  1 .8  10 6 NC 1 (right) F  1 .8 N (left)

25

Example 5

Two point charges, Q1 = +7 μC and Q2 = -5 μC are
separated by a distance of 0.3 m between each other
as in figure below. Determine the resultant E
produced by these two charges at point P.

P

0.4 m

+ 0.3 m −

Solution Q1 Q2

STEP 1: Draw the electric field, E diagram produced
by Q1 and Q2 at point P.

E1 is produced by Q1 and E2 is produced by Q2.

26

How we draw the electric field
line, E that exist at point P ?

P Use this concept :
E is outward for + charge
 E is inward for − charge

Vector E is draw along the line that
joining point P and the charge.

0.4 m

0.5 m

+Q1 0.3 m − 27
Q2

STEP 2 : Apply to find the magnitude of the each electric

field create the surrounding point charge(s)

E1  kQ1
r1 2

E2  kQ2 Left out the (−) sign of the
r2 2 charge for Q2. Substitute the
magnitude of the charge only

28

STEP 3 : Adds vectorially (consider the direction) to get the resultant.

E1 Given that ;
tan θ = 0.4 / 0.3
E2x
θ = 53.13°
Pθ

E2y E2

Resolve E2 into
comp x & y

Since E is a vector quantity, so we have to resolve E1 and E2 into x and y
component and find the summation of each of the component.

Vector x-comp.(N C-1) y-comp.(N C-1)


E1


E2

29

Therefore, the magnitude of the resultant E is ;

E x2  E y2
 E 

Direction of the resultant E is given by ;

tan   Ey
Ex

30

1.3 ELECTRIC POTENTIAL, V 1.3a) Define electric potential

Electric potential, V is defined as the work done by external force per unit charge
to bring a test charge ( positive charge ) from infinity (∞) to a point in an electric field
produced by a source point charge, Q.

V  W  r
qo

Electric potential, V is a scalar quantity Include minus
SI Unit is J C-1 or Volt (V). sign (‒) of
charge in the
substitution.

Electric potential, V where;
due to a single k – Coulomb constant (9.0×109 N m2 C–2 )
point charge is Q – point charge
given by: r – distance from the point charge to a point at which
the potential is evaluated

31

Equipotential Lines and Surfaces 1.3b) Define & explain equipotential lines & surface

Equipotential line or surface is a line or surface on which all points are at the
same potential.

(a) A point charge (b) Uniform electric field Each
equipotential
lines has its

electric
potential, V

Equipotential lines : circle Equipotential lines : parallel lines 32
Equipotential surface : sphere Equipotential surface : flat parallel plates

The important properties of equipotential surfaces are

(1) The electric field at every point on an equipotential surface
is perpendicular to the surface.

(2) The electric field points in the direction of decreasing
electric potential.

(3) The surfaces are closer together where the electric field
is stronger, and farther apart where the field is weaker

(4) No two equipotential surfaces can intersect each other.
An equipotential surface is normal to the electric field. If two
equipotential surfaces intersect each other then at the point
of intersection in means there will be two directions of
electric field at that point which is impossible.

33

Electric Potential Difference, ∆V between two points :

(Or simply) The Potential difference between two
difference in points B & A in an electric field is
potential defined as the work done in
between 2 bringing a positive point charge
points. from point A to point B in an
electric field per unit charge.

∆VBA = Vfinal – Vinitial
= VB ‒ VA

34

Example 6

A point charge q1 = +5.0 μC is at the origin and a point charge q2 = -2.0 μC is on the
x-axis at (3,0) m as in figure below.

Find the total electric potential due to these charges at point P (0,4) m.

y (m) Include minus
P sign (‒) of
charge in the
(0,4) m substitution.

Total electric
potential at point P
is due to q1 and q2
VP = V1+ V2
{Scalar summation}

q1 q2 x (m)
(3,0) m

35

Solution

V at P due to each charge can be calculated from ;

V k Q
r

The total V is the algebraic sum of these two potentials.

Include minus
sign (‒) of charge
in the substitution.

36

1.3d) Calculate the potential difference

The potential difference between any two points on an equipotential surface is

zero. Hence, no work is required to move a charge at constant speed along

an equipotential surface.

∆VBA = Vfinal – Vinitial VA = VC = VD
VB = VE = VF

Test Point Charge
Charge

37

1.3e) Deduce the change in potential energy

• To find work done to move a charge, rewrite the equation above as:

Point Charge • The work is done either by electric force (field)
or by an external force when a charged
Test particle moves in an electric field region.
Charge
• If W = positive  the work is done by
external force, the potential energy (U)
increases.

• If W = negative  the work is done by

electric force (field), the potential 38
energy decreases.

• If the electric potential at a point in an electric field is V, then a test charge q located
at that point will has electric potential energy, U of :

Change in Electric Potential Energy between 2 points in electric field.
• When a charge, q moves from a point where the potential is VA to a point where its

potential is VB, the change in potential energy, ∆U is given by

∆U =UB ‒ UA {* A : initial ; B : final}
=qVB ‒ qVA
=q(VB‒VA )

∆U = q∆V ….(1)

Knowing that : W = q∆V … (2)

Thus : ∆U = q∆V = W 39

W done by Fe (E)

∆V -ve W done by Fext ∆V +ve
W -ve W +ve
∆U -ve ∆U +ve
U↓ U↑

40

W done by Fext

∆V +ve W done by Fe (E) ∆V -ve
W +ve W -ve
∆U +ve ∆U -ve
U↑ U↓

41

Electric Potential Energy, U 1.3 f) Potential energy of the system of point
charges
of two point charges

where

k – Coulomb constant
q1 – charge 1
q2 – charge 2
r – distance between q1 & q2

Include minus sign (‒)
of charge in the
substitution.

42

Electric Potential Energy, U of a system of point charges

Consider a system of three point charges as shown in Figure below:

The total electric potential energy, U can be expressed as

CAUTION :
Algebraic sum with the
sign of charge included

43

Example 7
Given distance d = 14.0 cm & charge q = 150 nC
What is the electric potential
energy of this system of
charges ?

44

Solution

45

1.4 CHARGE IN A UNIFORM ELECTRIC FIELD

1.4a) i. Stationary Charge

• Lets consider a stationary particle of charge q and mass m is placed in a uniform

electric field E.

• The electric force Fe exerted on the charge is given by Only if m

++++++++++++++ Because the weight of the charge is very
small like
can be negligible, the only force proton

+ Fe a exerted on the charge would be Fe.

Fe - This nett F causes the charge to
a accelerated.
E
From Newton’s 2nd Law;

-- -- -- -- -- -- -- -- -- -- -- -- Direction a is the

Direction of a & F are either parallel same as direction 46
or anti-parallel to the field E
F

1.4a) ii. Charge moving perpendicularly to the field
• But….what if the charge is released perpendicular to the field?

++++++++++++++

q, m u But which
axis?
+

a
Fe E

-- -- -- -- -- -- -- -- -- -- -- -- v

The direction of Fe will cause it to move in a parabolic path

• If charge is released perpendicular to the field, E (2-D solution)

47

1.4a) iii. Charge moving parallel to the field

• If the particle has a positive charge, its acceleration is in the direction of the
electric field (Figure a). If the particle has a negative charge (electron), its
acceleration is in the direction opposite the electric field (Figure b)

• If charge is released parallel to the field, E (1-D solution)

48

1.4a) iv. Charge in dynamic equlibrium -- -- -- -- -- -- -- -- -- -- -- --

++++++++++++++

q, m - u E q, m + u E

V constant, a=0 V constant, a=0

W W

-- -- -- -- -- -- -- -- -- -- -- -- ++++++++++++++

49