CHAPTER 1: ELECTROSTATICS

prepared by Yew Sze Ling@Fiona, KMPP

Chapter 1: Electrostatics
Subtopic C1 & C2 C3 & C4
1.1 Coulomb’s  States Coulomb’s Law,  Sketch the electric force diagram
Law kQq Qq  Apply Coulomb’s Law for a
F 2 
r 4 0 r 2 system of point charges
1.2 Electric Field  Define electric field strength,  Use electric field strength,
 
 
F F
E E
q0 q0
kQ
 Use E  for a point charge
r2
 Sketch the electric field strength
diagram
 Determine the electric field
strength for a system of charges
1.3 Electric W kQ
Potential  Define electric potential, V   Use V  for a point charge
q0 r
 Define and explain equipotential and a system of charges
lines and surfaces of an isolated  Calculate potential difference
charge in a uniform electric field between two points:
W
V  V final  Vinitial 
q0
 Deduce the change in potential
energy between two points in
electric field: U  qV
 Calculate the potential energy for
a system of point charges:
q q qq q q 
U  k  1 2  1 3  2 3 
 r12 r13 r23 
1.4 Charge in a  Explain qualitatively with the aid V
uniform  Use E  for uniform E in:
of diagram the motion of charge d
electric field in uniform electric field  stationary charge
 charge moving
perpendicularly to the field
 charge moving parallel to the
field
 charge in dynamic
equilibrium
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Introduction
 The electrical nature of matter is inherent in atomic structure. An atom consists of a small,
relatively massive nucleus that contains particles called protons and neutrons. A proton has a mass
of 1.673×1027 kg, and a neutron has mass of 1.675×1027 kg. Surrounding the nucleus is a diffuse
cloud of orbiting particles called electrons. An electron has a mass of 9.11×10 31 kg.
 Like mass, electric charge is an intrinsic property of protons and electrons, and only two types of
charge have been discovered, positive and negative. A proton has a positive charge, and an
electron has a negative charge. A neutron has no net electric charge.
 Experiment reveals that the magnitude of the charge on the proton exactly equals the
magnitude of the charge on the electron; the proton carries a charge +e, and the electron carries
a charge −e. The SI unit for measuring the magnitude of an electric charge is the coulomb (C),
and e has been determined experimentally to have the value 1.60×10−19 C.
 Charges of larger magnitude than the charge on an electron or on a proton are built up on an object
by adding or removing electrons. Thus, any charge of magnitude q is an integer multiple of e; that
is, q = Ne, where N is an integer.

Example
How many electrons are there in one coulomb of negative charge?
q = 1 C, e = 1.60×10−19 C
q 1
q = Ne → N   19
 6.25  1018 electrons
e 1.60  10

1.1 Coulomb’s Law

Coulomb’s law states that the magnitude F of the electrostatic force exerted by one point charge
q1 on another point charge q2 is directly proportional to the magnitudes q1 and q2 of the charges
and inversely proportional to the square of the distance r between them:

q1 q2
F
r2
Mathematically,
Notes:
kq q 1  The sign of the charge can be
F  12 2 , k
r 4 0 ignored when substituting into
the Coulomb’s law equation.
q1q2  The sign of the charges is
F important in distinguishing the
4 0 r 2 direction of the electric force.

where F : magnitude of electrostatic force

k : electrostatic (Coulomb) constant = 9.0 × 109 N m2 C-2
q1, q2 : magnitude of charge (without sign)
r : distance between two point charges
ε0 : permittivity of free space = 8.85× 10-12 C2 N-1 m-2
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Equation above gives only the magnitude of the electrostatic force that each point charge exerts on
the other; it does not give the direction. The electrostatic force is directed along the line joining the
charges, and the force is attractive if the charges have unlike signs and repulsive if the charges have
like signs:

Unlike charge: ATTARCTIVE Like charge: REPULSIVE

The two forces obey Newton’s third law; they are always equal in magnitude and opposite in
direction.

Example: 2 charges
Two objects, whose charges are +1.0 and −1.0 C, are separated by 1.0 km. Compared to 1.0 km, the
sizes of the objects are small. Find the magnitude of the attractive force that either charge exerts on
the other.
Ignore negative sign of the charge
kq1q2
F 2 

9  109 1.01.0  9.0  103 N
r 1  10 3 2
 
Example: 3 charges

Determine the magnitude and direction of the net electrostatic force on q1.
Step 1: Sketch force diagram
Unlike charge: Unlike charge:
Attract Attract

Step 2: Apply Coulomb’s law equation

F12 
kq1q2

  
9  109 3  106 4  106
 2.

7 N ; F 
kq1 q3

  
9  10 9 3  10 6 7  10 6
 8.4 N

r122 0.202 13
r132 0.152

Step 3: Add electric forces as vector

  
F  F12  F13   2.7   8.4  5.7 N (to the right)
To the left (−ve) To the right (+ve)
The magnitude of the net electrostatic force acting on q1 is 5.7 N, and it points to the right.
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

1.2 Electric Field

 
The electric field E that exists at a point is the electrostatic force F experienced by a small test
charge q0 placed at that point divided by the charge itself.


F
E
q0
The electric field is a vector, and its direction is the same as the direction of the force on a positive
test charge, or opposite to the direction of the force on a negative test charge. Test charge (q0) is
the charge whose magnitude is very small, in fact negligible, as compared to that of the point charge,
and which does not affect the electric field of the point charge. To find out whether there is an electric
field at a particular point, test charged is placed at that point. If the test charge experiences an electric
force, then there is an electric field at that point.

SI unit of electric field: newton per coulomb (N C-1).

Positive Point Charge

A positive test charge A negative test charge

Negative Point Charge

A positive test charge A negative test charge

*The direction of electric field strength, E depends on sign of isolated point charge.
* The direction of the electric force, F depends on the sign of isolated point charge and test charge.

Example
Given an electric field has a magnitude of 2.0 N C-1 and is directed to the right. Determine the force
on a charge placed at that spot, if the charge has a value of
(a) q0 = +18×10-8 C and
(b) q0 = −24×10-8 C.

a)  
F  q0 E  18  108 2.0  36  108 N
Since q0 is positive, the force points in the same direction as the electric field, which is to the right.
b)  
F  q0 E  24  10 8 2.0  48  10 8 N
Since q0 is negative, the force points in the direction opposite to the electric field, which is to the
left.
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Since

 kQq
F
E and F  2 0 Notes:
q0 r
Q  q0
The electric field for a point charge can be calculated by using  Q is isolated point charge
kQ  q0 is test charge
E
r2
where E : magnitude of electric field strength (intensity)
Q : magnitude of isolated point charge (without sign)
r : distance between a point and isolated point charge

Electric field around charges can be represented by drawing a series of lines. These lines are called
electric field lines (lines of force). The direction of electric field is tangent to the electric field line at
each point. The shape of electric field is sphere.

Isolated point charge

Single positive charge Single negative charge

+ –

The lines point radially outward from the The lines point radially inward from the
charge charge

Two equal point charges

Unlike charges, +Q and −Q Like charges, +Q and +Q

The lines are curved and they are directed Neutral point, X is defined as a point
from the positive charge to the negative (region) where the total electric force is
charge. zero.

It lies along the vertical dash line.

 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Two opposite unequal charges, +2q and –q

Note that twice as many lines leave +2q as
there are lines entering –q

Number of lines is proportional to

magnitude of charge.

Two opposite charged parallel metal plate

The lines go directly from positive plate to
the negative plate.

The field lines are parallel and equally

spaced in the central region far from the
edges but fringe outward near the edges.

Thus, in the central region, the electric field

has the same magnitude at all points.

The fringing of the field near the edges can

be ignored because the separation of the
plates is small compared to their size.

Notes:
 The closer the lines, the stronger the field.
 Electric field lines start on positive charges and end on negative charges.
 The field lines never cross because the electric field doesn’t have two values at the same point.

Example: 2 point charges

Two point charges are separated by a distance of 10.0 cm. One has a charge of −25 µC and the other
+25 µC. Determine the direction and magnitude of the electric field at a point P between the two
charges that is 2.0 cm from the negative charge.

Step 1: Sketch field diagram

−ve charge: Electric

+ve charge: Electric field
field line points inward line points outward
(E1 toward charge Q1) (E2 away from charge Q2)
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Step 2: Apply electric field equation

E1 
k Q1

9  10 25  10   5.63  10
9 6
8
N C -1
r
1
2
0.02 2

E2 
k Q2

9  10 50  10   7.03  10
9 6
7
N C -1
r 2
2
0.08 2

Step 3: Add electric fields as vector

   
  
E  E1  E 2   5.63  10 8   7.03  10 7  6.33  10 8 N C -1 (to the left)
The magnitude of the net electric at point P is 6.33×108 N C-1, and it points to the left.

1.3 Electric Potential

Electric potential, V is defined as the work done in bringing positive test charge from infinity to
that point in the electric field per unit test charge, q0.

W
V
q0

where W∞ : work done in bringing positive test charge from infinity to that point
q0 : value of test charge (including sign)

 Electric potential is a scalar quantity.

 The SI unit of electric potential is volt (V) or joule per coulomb (J C-1).
 The total electric potential at a point in space is equal to the algebraic sum of the constituent
potentials at that point. In other words, when two or more charges are present, the potential
due to all the charges is obtained by adding together the individual potentials :

VTotal at P = VA + VB

 The electric potential at infinity is considered zero.

 The electric potential can be positive, negative or zero depending on the signs and magnitudes
of W∞.
 If the value of work done is negative – work done by the electric force (system).
 If the value of work done is positive –work done by the external force or on the system.
 In the calculation of W and V, the sign of the charge MUST be substituted in the related
equations.
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Equipotential Lines and Surfaces

Equipotential lines and surfaces is defined as a surface where all points on the surface that have the
same electric potential.

A point charge A Uniform Electric Field

 The dashed lines represent the equipotential surface (line).

 The equipotential surfaces (lines) always perpendicular to the electric field lines passing through
them and points in the direction of decreasing potential.
V A  VB  VC
 From the figures, then the work done to bring a test charge from B to A is given by
WBA  q oV AB No work is done in moving a
 q o V A  VB  charge along the same
equipotential surface.
0

Electric Potential Created by a Point Charge

Extra Notes:

dW  Fext dr
dW   FE dr
xr xr
kQq0
 dW 
x 
  F dr
x 
E where FE 
r2
r
W   kQq0  r  2 dr



W   kQq0  r 1  
r

kQq0
W
r
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Since
W kQq0
V and W 
q0 r
Thus, electric potential created by a point charge Q can be written as
kQ
V
r
where V : electric potential at a point in electric field
Q : value of a point charge (including sign)
r : distance between a point and isolated point charge

Potential Difference
Potential difference between two points in an electric field is defined as the work done in bringing a
positive test charge from a point to another point in the electric field per unit test charge.

Mathematically,

W WBA
V  V final  Vinitial  OR VAB  VA  VB 
q0 q0

where WBA : work done in bringing positive test charge from point B to point A
VA : total electric potential at point A (final state)
VB : total electric potential at point B (initial state)
q0 : value of test charge (including sign)

The work done to bring a charge from one point to another in the field does not depend on the path
taken (because the work done by conservative force).

Example: Example:

The work done to bring a

charge from B to A along the
paths 1,2 or 3 is the same.

Where WBA is the work done to

bring a charge q from B to A.
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Example:

Calculate the potential difference between point A and B.

Step 1: Calculate total potential at point A
Must include the sign

VA  V ve  V ve   

kQ ve kQ ve 9  109 8  109


9  109  8  109  
 240 V

r ve r ve 0.2 0.4
Step 2: Calculate total potential at point B

VB  Vve  Vve 
kQve kQve
 

9  10 9 8  10 9

  
9  10 9  8  10 9
 0V
 
rve rve 0.4 0.4
Step 3: Apply potential difference equation
VAB  VA  VB  240  0  240 V

Change in Potential Energy

Extra Notes:
The total work done W in bringing the test charge (+q0) from B to A is given by
xr x A
kQq 0
 dW 
x 
 F
xB
E dr where FE 
r2
rA

W  kQq 0  r  2 dr
rB

W  kQq 0  r 1  
rA
rB

1 1 kQq 0
W  kQq 0    ,U
 rA rB  r
WBA  U A  U B  U AB  U

W W
Since V  or VAB  BA , thus
q0 q0
 U  q0  V OR U AB  q0VAB

where VAB = Vfinal – Vinitial

UAB = Ufinal – Uinitial
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Potential Energy of a System of Point Charges

The total electric potential energy of the system of charges is the total work done to bring all the
charges from infinity to their final positions.

A
C

In the system of charges, suppose there were originally no charges at the points A, B and C as in the
figure above.

Step 1:
The charge Q1 is brought from infinity and placed at point A. Since originally there were no charges,
the charge Q1 does not experience any electric force when it is brought from infinity, that is F = 0.
No work is done to place the charge at point A. Hence VA = 0. Since U1 = Q1VA, hence
U1 = 0
Step 2:
With the charge Q1 fixed at point A, an electric field is produced by the Q1. Work is needed to place
Q2 at point B. Hence, the electric potential to bring the charge Q2 from infinity to the point B at a
distance of r12 from A is
kQ
VB  1
r12
Since U2 = Q2VB, hence
kQ1Q2
U2  A system of TWO charges
r12
Step 3:
With the charge Q1 fixed at point A and Q2 at the point B, the electric potential at the point C is
kQ kQ
VC  1  2
r13 r23
Since U3 = Q3VC, hence
kQ1Q3 kQ2Q3
U3  
r13 r23

The total potential energy, U can be expressed as

U  U1  U 2  U 3
kQ1Q2 kQ1Q3 kQ2 Q3
U  0  
r12 r13 r23
Q Q QQ QQ 
U  k  1 2  1 3  2 3  A system of THREE charges
 r12 r13 r23 
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

1.4 Charge in a Uniform Electric Field

Case 1: Stationary charge
Fe  W → Static Equilibrium
Positive Charge Negative Charge

 Consider a particle of charge q0 and mass m is stationary and placed in a uniform electric field
E, the electric force Fe exerted on the charge is given by
Fe  q0 E
 Since the particle is stationary the net force acting on the particle is zero. So the electric force
on the particle, Fe is equal to the weight, W of the particle.
F  0
Fe  W  0
Fe  W
q0 E  mg
Fe  W
 If Fe  W , the charged particle will be attracted toward the direction with greater force.
Fe  W Fe  W

 F  ma  F  ma
Fe  W  ma W  Fe  ma
q0 E  mg  ma mg  q0 E  ma
Fe  W → W = 0 and Fe  q0 E
 If the mass of the charged particle is negligible, the particle will be attracted toward the plate
with opposite charge.
Positive Charge Negative Charge
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Case 2: Charge moving perpendicularly to the field

 Consider a particle of charge q0 and mass m enters a uniform electric field, E perpendicularly
with an initial velocity u, the upward electric force will cause the charged particle to move
along a parabolic path towards the upper plate.

 From Newton’s second law

 F  ma
 The electric force Fe exerted on the charge is
Fe  q0 E

 Since the path makes by the electron is similar to the motion of a ball projected horizontally
above the ground,
ax  0 , u x  u , u y  0
 Therefore the magnitude of the electron’s acceleration is given by
qE
a  ay  o
m
 The components of electron’s velocity after pass through the electric field are given by
vx  u x  u ; v y  u y  a yt

q E
vy   0  t
 m 
 The position of the electron is
1
s x  u xt ; s y  u yt  a yt 2
2
1  q0 E  2
sy   t
2 m 

Notes: Charged particle is moving in a straight

path before it enters the plate and after it leaves
the plate.

Straight Parabolic Straight

path path path
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Case 3: Charge moving parallel to the field

Positive Charge Negative Charge

 If the electric force on the positive charge is  If the electric force on the negative charge
in the same direction as to its motion, the (electron) is in the same direction as to its
positive charge accelerates along a straight motion, the positive charge accelerates
line. along a straight line.
 If the electric force on the positive charge is  If the electric force on the negative charge
in the opposite direction to its motion, the (electron) is in the opposite direction to its
positive charge decelerates along a straight motion, the positive charge decelerates
line. along a straight line
 Consider a particle of charge q0 and mass m is moving parallel to the uniform electric field E,
the electric force Fe exerted on the charge is given by
Fe  q0 E

 Since only electric force Fe exerted on the particle, thus this force contributes the net force, F
and causes the particle to accelerate.
 According to Newton’s second law,

 F  ma
Fe  ma
q0 E  ma
q0 E
a
m
 Because the electric field is uniform (constant in magnitude and direction) then the
acceleration of the particle is constant.
 CHAPTER 1: ELECTROSTATICS
prepared by Yew Sze Ling@Fiona, KMPP

Case 4: Charge in dynamic equilibrium

 Dynamic equilibrium means the charge moves with constant velocity perpendicularly to the
direction of electric field and the total force acting on charge is zero,  F  0 .
 If the particle travels in a straight line with constant velocity hence the electric and its weight
are equal in magnitude.

F  0
Fe  W  0
Fe  W
q0 E  mg

 Only particles with this constant speed can pass through without being deflected by the fields.
 If Fe  W , the charged particle will be reflected toward the direction with greater force.

Uniform Electric Field (eg. between Two Parallel Plates)

Consider a uniform electric field is produced by a pair of flat metal plates, one at which is earthed
and the other is at a potential of +V as shown in figure.
A B

As the positive test charge travel from A to B :

W
V  and W  Fs  q0 Ed
q0
V  Ed
V
E
d