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Chapter 1
BP3 1 FYSL
Chapter 1: Electrostatic
1.1 Coulomb’s Law
L.O 1.1.1 State Coulomb’s law
Coulomb’s law states that the magnitude of the electrostatic (Coulomb/electric) force
between two point charges is directly proportional to the product of the charges and
inversely proportional to the square of the distance between them.
Mathematically,
The S.I. unit of charge is coulomb (C).
LO 1.1.2 Sketch the electric force diagram and apply Coulomb’s law for a system
of point charges. (2D, maximum four charges)
Like charges – REPEL Unlike charges – ATTRACT
Example:
Sketch the force diagram for q1.
Sketch the force diagram for q1.
2
4 2
0 r
Qqk
r
QqF
Electrostatic constant
(9.0 × 109
N m2
C-2
)
Permittivity of free space
(8.85× 10-12
C2
N-1
m-2
)
where
Q,q : magnitude of charges
r : distance between two point charges
+ –
+
+ – q1
q1
q2
q3
–
Notes: – 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.
2r
QqF
Chapter 1
BP3 2 FYSL
Example
Question Solution
Two point charges, q1 = −20 nC and q2 = 90 nC, are
separated by a distance of 4.0 cm as shown in
figure below.
Find the magnitude and direction of
a. the electric force that q1exerts on q2.
b. the electric force that q2 exerts on q1.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
Three point charges lie along the x-axis as shown in
figure below.
Calculate the magnitude and direction of the total
electric force exerted on q2.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
Figure below shows the three point charges are
placed in the shape of triangular.
Determine the magnitude and direction of the
resultant electric force exerted on q1.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
– +
q1 q2 4.0 cm
q1 = 2C q2 = 4C q3 = 6C
3.0 cm 5.0 cm
+ + –
–
– +
q1 = 1.2 µC q2 = 3.7 µC
q3 = 2.3 µC
15 cm
10 cm
32°
Chapter 1
BP3 3 FYSL
Example
Question Solution
Two point charges, q1 = +4.0µC and q2 = + 6.0µC,
are separated by a distance of 50 cm as shown in
figure below.
Determine the position of a point charge q that is
placed on the line joining q1 and q2 such that the net
force acting on it is zero.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
Exercise
Question
The electron in the hydrogen atom orbits around the proton in an orbit of radius 5.3×10-11
m.
Determine the force between the electron and the proton in a hydrogen atom.
(Given Charge of proton = +1.60×10-19
C, Charge of electron = − 1.60×10-19
C)
Answer: − 8.2×10-8
N
Point charges Q1 = +10 µC, Q2 = − 15 µC and Q3 = + 5.0 µC are placed at the corners of a
triangle as shown in figure below. Determine the resultant force that acts on charge Q3.
Answer: 0.037 N
Four identical point charges (q = +10.0 C) are located on the corners of a rectangle as
shown in figure below.
Calculate the magnitude and direction of the resultant electric force exerted on the charge at
the lower left corner by the other three charges. Answer: 40.9 N, 83° below –ve x-axis)
+ +
q1 q2 50 cm
Q1
Q2 Q3
3 m
4 m +
+
–
+
+
+
+
60 cm
15 cm
Chapter 1
BP3 4 FYSL
1.2 Electric Field
LO 1.2.1 Define and Use Electric Field Strength, oq
FE
LO 1.2.2 Use 2r
kQE for a point charge
Electric field strength (intensity) is defined as the electric (electrostatic) force per unit
positive test charge that acts at that point.
For a point charge,
The S.I. unit is N C1
OR V m 1
.
LO 1.2.3 Sketch the electric field strength diagram and determine electric field
strength E for a system of charges. (2D, maximum 4 charges)
Electric field is defined as a region of space around isolated charge where an electric
force is experienced if a (positive) test charge is placed in the region.
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
charge
The lines point radially inward from the
charge
where
F : magnitude of the electric force
qo : magnitude of test charge
22 4 r
Q
r
kQE
o
where
Q : magnitude of the isolated charge
r : distance between the point and isolated
point charge
+ –
oq
FE
Notes: In the calculation of magnitude E, substitute the magnitude of the charge only.
Chapter 1
BP3 5 FYSL
Two equal point charges
Unlike charges, +Q and −Q Like charges, +Q and +Q
The lines are curved and they are directed
from the positive charge to the negative
charge.
Neutral point, X is defined as a point
(region) where the total electric force is
zero.
It lies along the vertical dash line.
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.
Chapter 1
BP3 6 FYSL
Example
– – – –
− Q − Q − Q − 3Q
Note!!! • 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.
Since both F and E are vectors, how
to determine the direction of F and E
for a 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.
Chapter 1
BP3 7 FYSL
Example
Question Solution
A metal sphere can be regarded as a point object in
space. It carries a charge of +6.0 µC. Find the
electric field that the sphere generates at a distance
of 10 cm around it.
Two point charges, q1= − 1 C and q2= + 4 C, are
placed 2 cm and 3 cm from the point A respectively
as shown in figure below.
Find
a. the magnitude and direction of the electric field
intensity at point A.
b. the total electric force exerted on q0=-4 C if it
is placed at point A.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
Two point charges, q1= − 12nC and q2= 12nC, are
placed 0.10 m apart. Calculate the total electric
field at point a and point b.
– +
q1 q2 2 cm 3 cm
A
6 cm
– +
q1 q2 4 cm
b
a
13 cm 13 cm
Chapter 1
BP3 8 FYSL
Exercise
Question
Two point charges of +2.0 µC and −4.0 µC are separated by a distance of 5.0 cm. Determine
the electric field at the midpoint between two charges.
Point charges of +2.0 µC and −2.0µC are placed at corners A and B respectively of a right-
angle triangle ABC, as shown in the figure below. Determine the resultant electric field
strength at corner C. Answer: 6.85 × 106 N C
-1 , 81.79° above +vex-axis
Find the magnitude and direction of the electric field at the centre of the square in figure
below if q = 1.0×10-8
C and a = 5 cm. Answer: 1.02×105 N C
-1, upwards.
– +
A B
C
12.0 cm
5.0 cm
Chapter 1
BP3 9 FYSL
1.3 Electric Potential
LO 1.3.1& 1.33: Define Electric Potential & Use r
kQV
Electric potential, V of a point in the electric field is defined as the work done in bringing
positive test charge from infinity to that point in the electric field per unit test charge.
OR Electric potential, V of a point in an electric field is defined as the potential energy per
unit positive charge at that point in the electric field.
The S.I. unit is V or J C-1
.
Since r
kQqU o , then the equation of electric potential can be written as
Note!!
The total electric potential at a point in space is equal to the algebraic sum of the
constituent potentials at that point. BA VVV Total
The theoretical zero of electric potential of a charge is at infinity.
The electric potential energy of a positively charged particle increases
when it moves to a point of higher potential.
The electric potential energy of a negatively charged particle increases
when it moves to a point of lower potential.
The electric potential can be positive, negative or zero depending on the signs and
magnitudes of qo and 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 U, W and V, the sign of the charge MUST be substituted in the
related equations.
where
W∞ : work done
qo : magnitude of test charge oq
WV
oq
UV
where
U : potential energy
qo : magnitude of test charge
VqU o
r
kQV
Chapter 1
BP3 10 FYSL
LO 1.3.2: Define and sketch equipotential lines and surfaces of an isolated charge
and a uniform electric field.
Equipotential surface (line)is defined as the locus of points 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.
CBA VVV
From the figures, then the work done to bring a test charge from B to A is given by
0
BAo
ABoBA
VVq
VqW
LO 1.3.4: Calculate potential difference between two points
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,
OR
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).
No work is done in moving a
charge along the same
equipotential surface.
oq
WVVV initialfinal
o
BABAAB
q
WVVV
Chapter 1
BP3 11 FYSL
LO 1.3.5: Deduce the change in potential energy between two points in electric field.
From the definition of electric potential difference,
oq
WV and UW
LO 1.3.6: Calculate 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.
The total potential energy, U can be expressed as
23
32
13
31
12
21
23
32
13
31
12
21
231312
r
r
r
QQkU
r
QkQ
r
QkQ
r
QkQU
UUUU
Example:
Where WBA is the work done to
bring a charge q from B to A.
Example:
The work done to bring a
charge from B to A along the
paths 1,2 or 3 is the same.
VqU o
Chapter 1
BP3 12 FYSL
Example
Question Solution
Figure below shows a point A at distance 10 m
from the positive point charge, q = 5C.
Calculate the electric potential at point A and
describe the meaning of the answer.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
Two point charges, q1=+0.3 C and q2= −0.4 C are
separated by a distance of 6 m as shown in figure
below. Calculate the electric potential at point A if
pointA is at the midpoint of q1 and q2.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
Two point charges, q1=+12 nC and q2= −12 nC are
separated by a distance of 8 cm as shown in figure
below.
Determine the electric potential at point P.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
+
10 m
q A
+
6.0 m
q1
A −
+
8.0 cm
P
−
q2
q1 q2
6.0 cm
Chapter 1
BP3 13 FYSL
Example
Question Solution
Two points, S and T are located around a point
charge of +5.4 nC as shown in figure below.
Calculate
a. the electric potential difference between points
S and T.
b. the work done in bringing a charge of 1.5 nC
from point T to point S.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
Two point charges, Q1= +2.0 C and Q2= 6.0 C,
are placed 4.0 cm and 5.0 cm from a point
respectively as shown in figure below.
a. Calculate the electric potential at P due to the
charges.
b. If a charge Q3= +3.0 C moves from infinity to
P, determine the change in electric potential
energy for this charge.
c. When the charge Q3 at point P, calculate the
electric potential energy for the system of
charges.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
+
6.0 cm
S 8.0 cm T
+
5.0 cm
4.0 cm P
−
Q1
Q2
Chapter 1
BP3 14 FYSL
Exercise
Question Solution
Two point charges, Q1= 40 C and Q2= 30 C
are separated by a distance of 15 cm as shown in
figure below.
Calculate
a. the electric potential at point A and describe
the meaning of the answer,
b. the electric potential at point B.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
Answer: VA = −9.89×106 V , VB = −3.88×10
6 V
Two point charges q1=+2.40 nC and q2= −6.50 nC
are 0.1 m apart. Point A is midway between them;
point B is 8 cm from q1 and 6cm from q2 as shown
in figure below.
Find
a. the electric potential at point A,
b. the electric potential at point B,
c. the work done by the electric field on a charge
of 2.5 nC that travels from point B to point A.
(Given Coulomb’s constant, k = 9.0×109 N m
2 C
-2)
Answer: VA = −738 V , VB = −705 ,
WBA = −8.25×10-8
J
− 5 cm
13 cm
B
− A 10 cm
5 cm
+ −
q1 q2 5 cm
B
A
6 cm 8 cm
Q1 Q2
Chapter 1
BP3 15 FYSL
1.4 Charge In A Uniform Electric Field
LO 1.4.1: Explain quantitatively with the aid of a diagram the motion of charge in a
uniform electric field.
Case 1 : Stationary Charge
Positive stationary charge Negative stationary charge
Force experienced by charge is in the same
direction as electric field, E.
Force experienced by charge is in the
opposite direction as electric field, E.
• Consider a stationary particle of charge qo and mass m is placed in a uniform electric
field E, the electric force Fe exerted on the charge is given by
EqF oe
• Since only electric force exerted on the particle, thus this force contributes the nett force,
F and causes the particle to accelerate.
• According to Newton’s second law, then the magnitude of the acceleration of the
particle is
maEq
maFF
o
e
m
Eqa e
• Because the electric field is uniform (constant in magnitude and direction) then the
acceleration of the particle is constant.
• If the particle has a positive charge, its acceleration is in the direction of the electric
field. If the particle has a negative charge (electron), its acceleration is in the direction
opposite the electric field.
− − − − −
+ + + + +
− − − − −
+ + + + +
+
−
F
F
Chapter 1
BP3 16 FYSL
Case 2 : Charge moving perpendicularly to the field
Positive charge Negative charge
• The positive charge will be deflected and
moves along a parabolic path towards
the negative plate.
• The positive charge moves under the
influence of the electric force which is at
the same direction as electric field
lines.
• The negative charge will be deflected and
moves along a parabolic path towards
the positive plate.
• The negative charge moves under the
influence of the electric force which is
opposite direction to the electric field
lines.
• Consider an electron (e) with mass, me enters a uniform electric field, E perpendicularly
with an initial velocity u, the upward electric force will cause the electron to move
along a parabolic path towards the upper plate.
• From Newton’s second law,
maF
• The electric force Fe exerted on the charge,
EqF oe
• Therefore the magnitude of the electron’s acceleration is given by
upward) :(direction m
eEa y since 0xa
• The path makes by the electron is similar to the motion of a ball projected horizontally
above the ground.
x - component y – component
u
v
s
a
− − − − −
+ + + + +
− − − − −
+ + + + +
+
− +
−
v
v
Chapter 1
BP3 17 FYSL
Case 3 : Charge moving parallel to the field
Fe and v → in the same direction Fe and v → in the opposite direction
• The electric force on the positive charge
is in the same direction as to its motion.
• The positive charge accelerates along a
straight line.
• The electric force on the positive charge
is in the opposite direction to its motion.
• The positive charge decelerates along a
straight line.
Case 4 : Charge in dynamic equilibrium
Between electric force and weight
Particle weight is at the opposite
direction to the electric force.
WFe
Dynamic equilibrium means the charge moves with constant velocity.
Only particles with this constant speed can pass through without being deflected by the
fields.
− − − − −
+ + + + +
+
v
v
F
F
+
− − − − −
+ + + + +
F
W
−
− − − − −
+ + + + +
− −
Chapter 1
BP3 18 FYSL
LO 1.4.2: Use d
VE
for uniform electric field.
The graph is a straight line with negative constant gradient, thus
d
VE
d
V
r
VE
0
0
Example
Question Solution
Two parallel plates are separated 5.0 mm apart. The
electric field strength between the plates is 1.0
104 N C
1. A small charge of +4.0 nC is moved
from one conducting plate to another. Calculate
a. the potential difference between the plates
b. the work done on the charge.
Chapter 1
BP3 19 FYSL
Question Solution
If the plates are horizontal and separated by 1.0 cm,
the plates are connected to 100 V battery, the
magnitude of the electric field is 1.0×104 N C
-1. If
an electron is released from the rest at the upper
plate, determine
a. the acceleration of the electron.
b. the speed and the kinetic energy required to
travel to the lower plate.
c. the time required to travel to the lower plate.
Question Solution
The figure shows a section of the deflection system
of a cathode ray oscilloscope. An electron
travelling at a speed of 1.5×107 m s
-1 enters the
space between two parallel metal plates 60 mm
long. The electric field between the plates is
4.0×103 V m
-1.
a. Copy the figure, sketch the path of the electron
in between plates, and after emerging from the
space between the plates.
b. Find the acceleration of the electron between
the plates.
c. Determine the velocity when it emerges from
the space between the plates.
20 mm
60 mm
80 V
0 V
1.5×107 m s
-1
Chapter 1
BP3 20 FYSL
Exercise
Question Solution
Two parallel electrodes X and Y are separated by a
distance of 1.80 cm. The electric field between the
electrodes is 2.4 x 104 NC
-1. Electrons are emitted
with negligible velocity from the electrode X.
Determine
a. the force in the electric field
b. the acceleration of the electron in the electric
field.
c. the velocity of the electron when it reaches the
electrode Y.
d. the time taken by the electron to travel from
the electrode X to Y.
(Given e=1.60 1019
C and me=9.11 1031
kg)
Answer: 3.84 x 10-15
N, 4.22 x 1015
ms-2
,
1.23 x 107 ms
-1, 2.93 x 10
-9 s
Question Solution
Figure shows an electron entering charged parallel
plates with a speed of 5.45 106 m s
1. The electric
field produces by the parallel plates has deflected
the electron downward by a distance of 0.618 cm at
the point where the electron exits. Determine
a. the magnitude of the electric field,
b. the speed of the electron when it exits the
parallel plates.
(Given e=1.60 1019
C and me=9.11 1031
kg)
Answer: 4126 N C-1
, 6.22×106 m s
-1
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