UNIT 16 : ELECTROSTATICS
(4 hours)
16.1
16.2
16.3
16.4
Coulomb’s Law
Electric Field
Charge in a uniform electric field
Electric Potential
Electrostatics - the study of electric
charges at rest, the forces between
them and the electric fields
associated with them.
1
SUBTOPIC :
16.1 Coulomb’s Law (1 hour)
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
Qq
kQq
 2
a) State Coulomb’s law, F 
2
4o r
r
b) Sketch the force diagram and apply Coulomb’s law
for a system of point charges.
2
16.1 Coulomb’s Law
• Coulomb’s law states that the force,F between two point
charges q1 and q2 is directly proportional to the product
of their magnitudes and inversely proportional to the
square of the distance separating them, r.
F  q1q2
• In equation form :
F
F
kq1q2
r2
electrostatic force
• Unit : N(Newton)
• Vector quantity
1
r2
attractive force
repulsive force
Coulomb’s law equation
3
k
F
1
4 0
16.1 Coulomb’s Law
kq1q2
r2
 0 : permittivi ty of free space
(called proportion ality constant
or Coulomb' s constant)
(vacuum or air)
( 0  8.85x1012 C2 N1 m2 )
k  9.0 x 10 9 N m 2 C 2
q1
+
F12
q2
F21
-
F12
r
q2
q1
+
+
F21
r
attractive force
repulsive force
F12 : the force on charge q1 due to charge q2.
F21 : the force on charge q2 due to charge q1.
F12  F21 
kq1q2
r2
4
Example of force diagram
5
F
F
16.1 Coulomb’s Law
kq1q2
r
F
2
Gradient,
M  kq1q2
r
0
0
1
r2
Graphs above show the variation of electrostatic force
with the distance between two charges.
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.
6
16.1 Coulomb’s Law
Example 16.1
Two point charges, q1=-20 nC and q2=90 nC, are separated
by a distance of 4.0 cm as shown in figure below.
q1 -
+ q2
4.0 cm
Find the magnitude and direction of
a) the electric force on q1 due to q2.
b) the electric force on q2 due to q1.
c) the electric force on each charge.
(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)
7
Solution 16.1
16.1 Coulomb’s Law
q1= 2.0 x 10-8 C, q2= 9.0 x 10-8 C, r = 4.0 x 10-2 m


F21
F12
q1 + q2

4.0 cm
kq1q2
a) F12  2
r

kq2 q1
b) F21  2
r
8
16.1 Coulomb’s Law
Example 16.2
Three point charges lie along the x-axis as shown in figure
below.
q2  4 C
q3  6 C
q1  2 C
+
+
3.0 cm
5.0 cm
Calculate the magnitude and direction of the net electric
force exerted on q2.
(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)
9
16.1 Coulomb’s Law
Solution 16.2
r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m

q2  4 C F23
q3  6 C
q1  2 C

+
+
F21
3.0 cm
5.0 cm

kq2 q1
F21  2
r21

F21 

kq2 q3
F23  2
r23

F23 
10
16.1 Coulomb’s Law
Solution 16.2
r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m
q1  2 C
+

q2  4 C F23

+
F21
3.0 cm
Magnitude ;
q3  6 C
-
5.0 cm



F2  F21  F23



F2  F21  F23
11
16.1 Coulomb’s Law
Example 16.3
Three point charges lie along the x-axis as shown in figure
below.
q2  4 C
q3  6 C
q1  2 C
+
3.0 cm
5.0 cm
Calculate the magnitude and direction of the net electric
force exerted on q2.
(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)
12
16.1 Coulomb’s Law
Solution 16.3
r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m

q3  6 C
q1  2 C F23 q2  4 C
+ F21 -

kq2 q1
F21  2
r21
3.0 cm
5.0 cm
Magnitude ;



F2  F21  F23



F2  F21  F23

kq2 q3
F23  2
r23
13
16.1 Coulomb’s Law
Example 16.4
Three point charges lie along the x-axis as shown in figure
below.
q2  4 C
q3  6 C
q1  2 C
+
+
3.0 cm
5.0 cm
Calculate the magnitude and direction of the net electric
force exerted on q1.
(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)
14
16.1 Coulomb’s Law
Solution 16.4
r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m
 q  4C
q1  2 C F 2
13
 +
+
F12
3.0 cm

kq1q2
F12  2
r12
q3  6 C
-
5.0 cm
Magnitude ;
 

F1  F12  F13



F1  F12  F13

kq1q3
F13  2
r13
15
16.1 Coulomb’s Law
Example 16.5
Figure below shows the three point charges are placed in the
shape of triangular.
q3 -
r13
q1
-
r12
+
q2
Determine the magnitude and direction of the resultant
electric force exerted on q1.
Given q1 = -1.2 C, q2 =+3.7 C, q3 =-2.3 C, r12 =15 cm,
r13=10 cm, and k = 9.0 x 109 N m2 C-2.
16
16.1 Coulomb’s Law
Solution 16.5
q1= 1.2x10-6 C, q2= 3.7x10-6 C, q3= 2.3x10-6 C,
r12= 15x10-2 m, r13= 10x10-2 m

kq1q2
F12  2
r12
q3 r13

kq1q3
F13  2
r13
q1
-
F13
F12
r12
+
q2
16.1 Coulomb’s Law
Magnitude ,

2
2
F1   Fx    Fy 

F1 
q3 r13
q1
-
r12
+
q2
Direction,
F

tan  
F
y
x
18
Example 16.6
16.1 Coulomb’s Law
Figure below shows the three point charges are placed in the
shape of triangular.
q3

r13
+
r
12
q1
q2
Determine the magnitude and direction of the resultant
electric force exerted on q1.
Given q1=-1.2 C, q2 =+3.7 C, q3 =-2.3 C, r12 =15 cm,
r13=10 cm,  =32 and k = 9.0 x 109 N m2 C-2.
19
Solution 16.6
16.1 Coulomb’s Law
q1= 1.2x10-6 C, q2= 3.7x10-6 C, q3= 2.3x10-6 C,
r12= 15x10-2 m, r13= 10x10-2 m

kq q
F12  12 2
q3


r12
r13
F12
q1 - 58  r12

F13

kq1q3
F13  2
r13
+
q2
16.1 Coulomb’s Law
Solution 16.6
q1= 1.2x10-6 C, q2= 3.7x10-6 C, q3= 2.3x10-6 C,
r12= 15x10-2 m, r13= 10x10-2 m



F

F

F
cos
58
 x 12 13
F
F
F
x
q3
r13

y

 0  F13 sin 58
y

 F    F 
Direction ;
-
r12
+
q2

F1  3.74 N
2
x
  34.2 

q1
Magnitude ;

F1 
-
2
y

F

tan  
F

F1  3.74 N
y
x
21
16.1 Coulomb’s Law
Exercise
1. Two point charges of -1.0 x 10 -6 C and +2.0 x 10 -6 C
are separated by a distance of 0.30 m. What is the
electrostatics force on each particle ?
(0.20 N , directed to one another)
2. Calculate the net electrostatics force on charge
i ) q1 ii) q2 iii) q3 q  5 C
q2  2 C q3  - 3 C
1
+
(18.75 N , to the left , +
191.25 N to the right ,
4.0 cm
2.0 cm
172.51 N to the left)
3. Calculate the net electrostatics force on charge q2 .
q1  5C
q2  2C
+
+
4m
3m
- q3  3C
(4.11 x10-3 N , 18.4 o below the x-axis)
22
16.1 Coulomb’s Law
4. What is the electrostatics force on charge q3 ?
q1  2.5C
(0,+0.30 m) +
(0.43 N in the +x direction)
+
q3  3C
(0.40 m,0)
(0,-0.30 m) +
q2  2.5C
5. Four identical point charges (q = +10.0 C) are located
on the corners of a rectangle as shown in figure below.
The dimension of the rectangle
are l = 60.0 cm and w = 15.0
cm. Calculate the magnitude
and direction of the resultant
electric force exerted on the
charge at the lower left corner
by the other three charges.
+ q
q +
w
q+
l
(40.9 N at 263)
+ q23
SUBTOPIC :
16.2 Electric field (1 hour)
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Define electric field.
F
.
qo
c) Sketch the electric field lines of isolated point charge,
b) Define and use electric field strength, E 
two charges and uniformly charged parallel plates.
d) Sketch the electric field strength diagram and
determine electric field strength for a system of
charges.
24
16.2 Electric field
Definition of electric field :
A region of space around isolated charge
where an electric force is experienced if a
positive test charge 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.
25
16.2 Electric field
• Figures below show the electric field patterns around the
charge.
Field direction
+q
a. Single positive charge
the lines point radially
outward from the
charge
-q
b. Single negative charge
the lines point radially
inward toward the
charge
26
16.2 Electric field
c. Two equal point charges of opposite sign, +q and -q
Field direction
+q
-q
the lines are curved and
they are directed from the
positive charge to the
negative charge.
d. Two equal positive charges, +q and + q
+q
Field
direction
X
(point X is neutral point )
+q

Is defined as a point
(region) where the
resultant electric force
is zero.

It lies along the vertical
27
dash line.
e. Two opposite unequal charges, +2q and -q
16.2 Electric field
Field direction
-q
+2q
• note that twice as many lines
leave +2q as there are lines
entering –q.
• number of lines is
proportional to magnitude of
charge.
f. Two opposite charged parallel metal plates

The electric field lines are perpendicular to the surface
of the metal plates.

The lines go directly from positive plate to the negative
plate.

The field lines are parallel and equally spaced in the
central region.Thus, in the central region, the electric
field has the same magnitude at all points (uniform)
except near the edges.
28
16.2 Electric field
The properties of electric field lines:
•The field lines indicate the direction of the electric field
(the field points in the direction tangent to the field
line at any point).
•The lines are drawn so that the magnitude of electric
field is proportional to the number of lines crossing unit
area perpendicular to the lines. The closer the lines, the
stronger the field.
•Electric field lines start on positive charges and end
on negative charges, and the number starting or ending
is proportional to the magnitude of the charge.
•The field lines never cross because the electric field
does not have two values at the same point.
29
Electric field strength
16.2 Electric field
• The electric field strength at a point is defined as
the electric force (electrostatic) per unit positive
test charge that acts at that point in the same direction
as the force.
where
F
E
q0
E : magnitude of the electric field strength
F : magnitude of the electric force
q0 : magnitude of test charge
• It is a vector quantity.
• The units of electric field strength is N C-1 or V m-1.
• The direction of the electric field strength, E depends
on the sign of isolated charge.
30
16.2 Electric field

F
From
E
q0 (ve)
q
and
r
A positive isolated point charge.
thus

F

E
q
q0 (ve)
r
r : distance between the
point and isolated point charge
A negative isolated point charge.
F
E
q0
kqq0
F 2
r
 kqq0
 2
E  r
 q0








kq
E 2
r
In the calculation of
magnitude E, substitute
the MAGNITUDE of the
charge only.
31
16.2 Electric field
Example 16.7
2.0 cm
+
q1  5.0C
A
Calculate the electric field strength at point A, 2.0 cm
from a point charge q1.
(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)
Solution
32
16.2 Electric field
Example 16.8
2.0 cm
q1  5.0C
A
Calculate the electric field strength at point A, 2.0 cm
from a point charge q1.
Solution
33
16.2 Electric field
Example 16.9
A
q1 +
2.0 cm
- q2
3.0 cm
Two point charges, q1=1.0 C and q2=-4.0 C, are placed 2.0
cm and 3.0 cm from the point A respectively as shown in
figure above.
Find
a) the magnitude and direction of the electric field
intensity at point A.
b) the resultant electric force exerted on q0=4.0 C if it is
placed at point A.
(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)
34
16.2 Electric field
Solution 16.9
q1 +
A
2.0 cm
a)

 E A2
q
E A1 - 2
3.0 cm
kq1
E A1  2 
r1
E A2
kq2
 2 
r2
The electric field strength at point A due to the charges is



given by E  E  E
A
A1
A2
35
16.2 Electric field
Solution 16.9
q1 +
- q2
A
2.0 cm
3.0 cm
b)
FA
EA 
q0
FA  q0 E A
36
16.2 Electric field
Example 16.10
A
q1 2.0 cm
- q2
3.0 cm
Two point charges, q1=-1.0 C and q2=-4.0 C, are placed 2.0
cm and 3.0 cm from the point A respectively as shown in
figure above.
Find
a) the magnitude and direction of the electric field
intensity at point A.
b) the resultant electric force exerted on q0=4.0 C if it is
placed at point A.
(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)
37
16.2 Electric field
Solution 16.10
q1 -

E A1
A
2.0 cm
a)

E A 2 - q2
3.0 cm
kq1
E A1  2 
r1
E A2
kq2
 2 
r2
The electric field strength at point A due to the charges is



given by E  E  E
A
A1
A2
38
16.2 Electric field
Solution 16.10
q1 -
- q2
A
2.0 cm
3.0 cm
b)
FA
EA 
q0
FA  q0 E A
39
16.2 Electric field
Example 16.11
4.0 cm
q1 -
- q2
3.0 cm
A
Two point charges, q1= -1.0 C and q2= -4.0 C, are separated
by 4.0 cm from each other as shown in figure above.
Find the magnitude and direction of the electric field
intensity at point A 3.0 cm from charge q1.
(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)
40
16.2 Electric field
Solution 16.11
q1 3.0 cm
A
4.0 cm

E A1

- q2

E A2
  tan
kq1
E A1  2 
r1
E A2
kq2
 2 
r2
4
 53o
5
16.2 Electric field
Solution 16.11
q1 3.0 cm
A
4.0 cm

E A1
  53o 
E A2
- q2
42
16.2 Electric field
Solution 16.11
q1 
Ey
A
- q2

EA
  58.7 oE
x
The electric field strength at point A due to the charges is
given by
Direction ;
43
16.2 Electric field
Example 16.12
q1  8.00C +
q3  4.00C -
- q2  5.00C
A
Three charges are placed on three corners of a square,
as shown above. Each side of the square is 30.0 cm.
Calculate the electric field strength at point A. What
would be the force on a 6.00 µC charge placed at the
point A?
44
16.2 Electric field
Solution16.12
q1  8.00C +
30.0 cm
- q2  5.00C
30.0 cm
30.0 cm
q3  4.00C -
EA2
42.4 cm
EA3
A
EA1
kq1 (9.0  109 )(8.00  106 )
E A1  2 
r1
( 42.4  102 )2
E A2
E A3
kq2 (9.0  109 )(5.00  106 )
 2 
r2
(30.0  102 )2
kq3 (9.0  109 )( 4.00  106 )
 2 
r3
(30.0  102 )2
45
16.2 Electric field
Solution 16.12
q1  8.00C +
30.0 cm
- q2  5.00C
E A1  4.00 105 N C-1
30.0 cm
30.0 cm
q3  4.00C -
E A2  5.00  105 N C-1
EA2
42.4 cm
EA3
E A3  4.00  105 N C-1
A
45o
EA1
46
16.2 Electric field
Solution 16.12
E
E
AY
=61.7o
Magnitude ;
E
AX
Direction ;
47
16.2 Electric field
Exercise
1. Determine
a) the electric field strength at a point X at a distance 20 cm from a
point charge Q = + 6µC. (1.4 x 10 6 N/C)
b) the electric force that acts on a point charge q= -0.20 µC placed at
point X. (0.28 N towards Q)
2.
X
q1 +
20 cm
- q2
20 cm
Two point charges, q1 = +2.0 C and q2 = -3.0 C, are separeted by a
distance of 40 cm, as shown in figure above. Determine
a) The resultant electric field strength at point X.
(1.13 x 103 kN C-1 towards q2)
b) The electric force that acts on a point charge q = 0.50 µC placed at X.
(0.57 N)
48
16.2 Electric field
Exercise
3.
Find the magnitude of the electric field at point P due to the four
point charges as shown in the figure below if q=1 nC and d=1 cm.
(Given 0=8.85 x 10-12 C2 N-1 m-2)
(HRW. pg. 540.11)
Ans. : zero.
4.
Find the magnitude and direction
of the electric field at the centre of
the square in figure below if
q=1.0x10-8 C and a= 5cm.
(Given 0=8.85 x 10-12 C2 N-1 m-2)
(HRW. pg. 540.13)
Ans. : 1.02x105 N C-1, upwards.
49
SUBTOPIC :
16.3 Charge in a uniform electric field
(1 hour)
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Explain quantitatively with the aid of a diagram the
motion of a charge in a uniform electric field.
Cases : 1. stationary charge
2. charge moving perpendicularly to the field
3. charge moving parallel to the field
4. charge in dynamic equilibrium
50
16.3 Charge in a uniform electric field
Case 1 : Stationary charge
Case 4 : Charge in dynamic equilibrium
(moves in straight line, consider the weigh, W of the particle)

E

Fe
v =0, a=0
W
Figure a : Case 1
vo

E

Fe
v  0, a  0
W
vo , constant,

a 0
Figure b : Case 4
• Figure a and Figure b show a particle with positive charge q is held stationary
and moves at constant speed respectively, in a uniform electric field, E .
• The forces acted on the particle are electrostatic force (upwards)
and weight (downwards).
• For the particle in static equilibrium (Figure a) and dynamic
equilibrium : moves horizontally in straight line (Figure b),
electrostatic force = weight
FE= W
51
qE = mg
16.3 Charge in a uniform electric field
Case 2 : Charge moving perpendicularly to the field
x
v0
q0
-

E
y
FE
θ
vy
vx
v
• When the electron enters the electric field , E the only
force that acts on the electron is electrostatic force , FE=qE.
52
16.3 Charge in a uniform electric field
• This causes the electron of mass m to accelerate downwards
with an acceleration a.
q0 E
ay 
m
• Since the horizontal component of the velocity of the
electron remains unchanged as vo , the path of the electron
in the uniform electric field is a parabola.
• The time taken for the electron to tranverse the electric field
is given by
x
t
vo
From R  S x  vo t
53
16.3 Charge in a uniform electric field
• The vertical component of the velocity vy , when the electron
emerges from the electric field is given by
v y  uy  a y t
 qo E   x 
vy  0  
 v 
m

 o 
q Ex
vy  o
mvo
• After emerging from the electric field, the electron travels
with constant velocity v, where
v  vx2  v y2
54
16.3 Charge in a uniform electric field
• The direction of the velocity v is at an angle
 vy 
  tan   to the horizontal.
 vx 
1
• The position of the electron at time t is
s x  v0 t
or
1 2
s y  uy t  a y t
2
1 2
sy   ayt
2
v y 2  u y 2  2a y s y where u y  0
sy 
v y 2
2a y
55
16.3 Charge in a uniform electric field
Example 16.13
+
20 mm
uo = 1.5 x 107m/s
60 mm
An electron travelling at speed of 1.5 x 107 m/s enters the
space between two parallel metal plates 60 mm long. The
electric field between the plates is 4.0 x 103 V/m.
a) Sketch the path of the electron in between plates, and
after emerging from the space between the plates.
b) Find the magnitude and direction of the acceleration of
the electron in between the plates.
c) Calculate the vertical and horizontal components of the
electron velocity when it emerges from the space
between the plates.
56
d) Find the angle of deflection of the electron beam.
16.3 Charge in a uniform electric field
Solution 16.13
+
20 mm
uo = 1.5 x 107m/s
60 mm
a) DIY
b)
57
16.3 Charge in a uniform electric field
Solution 16.13
c) Time taken for electron to travel through the space between
l 60 103
9
the plates is t 


4
.
0

10
s
7
v x 1.5 10
Vertical component of velocity,
v y  u y  at



v y  0  7.03 1014 4.0  09  2.81106 m s-1
Horizontal component of velocity,
v x= 1.5 x 107 m s-1
2.81106
0


10
.
6
d) Angle of deflection,   tan
v x 1.5 107
v
1 y
58
16.3 Charge in a uniform electric field
Case 3 : Charge moving parallel to the field
Consider 2 cases as shown in figure 1 and figure 2

E

a

Fe
 v0
Figure 1

E

a v
0
Fe
Figure 2
59
16.3 Charge in a uniform electric field
Figure 1
• A particle with positive charge q moves with initial velocity vo
towards negative plate (upwards).
• For a positive charge, its acceleration a is in
the direction of the electric field is given by
qE
a
m
• Since the direction of electric force is upward
(same direction and parallel to the direction of motion of
particle), the particle is then accelerated in straight line
towards negative plate with speed v, where v > vo .
60
16.3 Charge in a uniform electric field
• The velocity v of the particle after time t is given by
v  vo  at ,
 qE 
v  vo  
t
 m
• The displacement s after time t is given by
1 2
s  vot  at ,
2
 1 qE  2
s  vot  
t
2 m 
• The displacement s in terms of velocity v is given by
v 2  vo  2as,
2
v 2  vo
v 2  vo
s

2a
 qE 
2

 m
2
2
61
16.3 Charge in a uniform electric field
Figure 2
• A particle with negative charge q moves with initial velocity vo
towards positive plate (downwards).
• For a negative charge, its acceleration is in
the direction opposite the electric field is given by
qE
a
m
• Since the direction of electric force is downwards,
(same direction and parallel to the direction of motion of
particle), the particle is then accelerated in straight line
towards positive plate with speed v, where v > vo
62
16.3 Charge in a uniform electric field
• The velocity v of the particle after time t is given by
v  vo  at ,
 qE 
v  vo  
t
 m
• The displacement s after time t is given by
1 2
s  vot  at ,
2
 1 qE  2
s  vot  
t
2 m 
• The displacement s in terms of velocity v is given by
v 2  vo  2as,
2
vo  v 2 vo  v 2
s

2a
 qE 
2

 m
2
2
63
Example 16.14
16.3 Charge in a uniform electric field
A pair of flat parallel metal plates A and B, are separated by
1.0 cm and the electric field strength that exists between the
plates is 10 kV m-1. An electron emerges from plate A at a
speed 5.0 x 106 m s-1 and moves towards B in a linear path
which is perpendicular to B. Determine
a) the acceleration of the electron
b) the speed of the electron when it reaches B
c) the time taken by the electron to travel to B.
64
16.3 Charge in a uniform electric field
Solution 16.14
s= 1.0 x 10 -2 m , vo = 5.0 x 106 m s-1,
q = 1.6 x 10 -19 C, me = 9.11 x 10 -31 kg
qE 1.6  10 1.0  10 
a

 1.8  10
19
a)
4
15
m
-2
m
s
9.11 1031
a is in opposite direction to E
b)
c)
65
16.3 Charge in a uniform electric field
Example 16.15
An electron is released from rest and allowed to accelerate in
a straight line in a uniform electric field of strength 3.0 kV m-1.
Determine
a) its acceleration
b) its speed after 3.0 s.
q = 1.6 x 10 -19 C,
me = 9.11 x 10 -31 kg
Solution
qE
a) a 

m


1.6  1019 3.0  103
9.11 1031
  5.3 10
14
m s-2
a is in opposite direction to E
b)
66
SUBTOPIC :
16.4 Electric Potential (1 hour)
LEARNING OUTCOMES :
At the end of this lesson, the students should be able to :
a) Define electric potential.
b) Define and sketch equipotential lines and surfaces of
i) an isolated charge ,
ii) a uniform electric field
Q
for a point charge and a system of charges.
r
d) Calculate potential difference between two points. V  V  V
AB
A
B
c) Use V  k
e) Use E 
V
for uniform E.
d
VAB 
WBA
q
f) Deduce the change in potential energy,U between two points in
electric field. U  qV
g) Calculate potential energy of a system of point charges.
 q1q2 q1q3 q2q3 
U  k



r
r
r
13
23 
 12
67
16.4 Electric Potential
• The electric potential V at a point in an electric field is the
work done to bring a unit positive charge from infinity to
that point. (energy required to bring 1 C of positive charge
from infinity to that point)
• The electric potential at infinity is considered zero. V  0
• Scalar quantity.
• Its unit is volt (V) or J/C.
Q
• Its formula is V  k
r
68
16.4 Electric Potential
Case B
Case A
Q
Q
• A
+
• A
r
r
The electric potential at
point A at distance r from
a positive point charge Q
is
The electric potential at
point A at distance r from
a negative point charge Q
is
Q 

k
Q 

k
VA
VA  k
r
Q
r
VA
r
Q
VA  k
r
69
16.4 Electric Potential
Case A
Q
Fext
• A
+
FE
+qo
r
r 
The electric potential V at a point in an electric field is the work done to bring
a unit positive charge from infinity to that point.
dW  Fext dr
kQq0
but FE  2
r
dW  FE dr

r r
r 
dW  
r r
r 
 FE dr
r
W  kQqo  r 2dr

r
W  kQqo  r 1 

Qqo
W k
r
Wr
VA 
q0
kQqo
VA 
qo r
Q
VA  k
r
70
16.4 Electric Potential
Case B
Q
FE
• A
-
+qo
r
r 
The electric potential V at a point in an electric field is the work done to bring
a unit positive charge from infinity to that point.
kQq0
but FE  2
r
dW  FE dr

r r
r 
dW  
r r
r 
FE dr
r
W  kQqo  r 2dr

r
W  kQqo  r 1 

Qqo
W  k
r
Wr
VA 
q0
kQqo
VA 
qo r
Q
VA  k
r
71
16.4 Electric Potential
Example 16.16
A
q+
10 m
Figure above 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 x 109 N m2 C-2)
kq (9.0 109 )(5)
VA 

r
(10)
72
16.4 Electric Potential
Example 16.17
Q1 =2µC
+
Q2 =2µC
+
3 cm
5 cm
A•
Calculate the electric potential at point A .
73
16.4 Electric Potential
Example 16.18
Q1 =2µC
+
Q2 =2µC
-
3 cm
5 cm
A•
Calculate the electric potential at point A .
74
16.4 Electric Potential
Example 16.19
+0.01 µC
+0.02 µC
+
+
10 cm
o
+
+0.01 µC
10 cm
-0.02 µC
Calculate the electric potential at point O .
VA = 2546 V
75
Potential Difference (a)
•
16.4 Electric Potential
The potential difference between points A and B, VAB is
given by
 WBA
VAB  VA  VB 
q
WBA : work done in bringing a charge from point B to point A
VA : electric potential at point A
VB : electric potential at point B
q : a charge
VAB
 WBA

q
In calculation if
i) W positive : work done by electric field
or
WBA  qVAB
(work done by electric force)
i)
W negative : work done on electric field
(work done by external force)
76
Potential Difference (b)
•
16.4 Electric Potential
The potential difference between points A and B, VAB is
given by
WBA
VAB  VA  VB 
q
WBA : work done in bringing a charge from point B to point A
VA : electric potential at point A
VB : electric potential at point B
q : a charge
VAB
WBA

q
In calculation if
i) W positive : work done on electric field
or
WBA  qVAB
(work done by external force)
i)
W negative : work done by electric field
(work done by electric force)
77
Example 16.20
16.4 Electric Potential
A point charge of q = +50 µC moves from point A to point B in
an electric field. The work done by the field is 10 µJ.
Determine the electric p.d crossed by the charge.
78
Example 16.21
16.4 Electric Potential
A point charge of q =-2.0 µC travels from point X to point Y in
an electric field, crossing a potential rise of 200 V.
a) Determine the work done in transferring the charge.
b) Is the work done on or by the electric field ?
c) An electric potential of -20 V exists at point Y. Determine
the potential at point X.
Solution
a)
79
16.4 Electric Potential
Example 16.22
Two point charges q1=+2.40 nC and q2=-6.50 nC are 0.100 m
apart. Point A is midway between them, point B is 0.080 m
from q1 and 0.060 m from q2 as shown in figure below.
B
0.080 m
q1
0.060 m
A
+
0.050 m
0.050 m
q2
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
80
that travels from point B to point A.
16.4 Electric Potential
Solution 16.22
B
q1=+2.40 nC , q2=-6.50 nC
0.080 m
q1
0.060 m
A
+
0.050 m
0.050 m
q2
a) VA  VA1  VA2
c)
b) VB  VB1  VB 2
81
Example 16.23
16.4 Electric Potential
a) What is the electric potential at the point
i) 10 cm and
ii) 50 cm from a point charge of 2 µC ?
b) Find the work done to move a charge of 0.05 µC from
a point 50 cm from the point charge 2 µC to a point 10 cm
from the point charge. State whether the work done by
or on the electric field.
a)
b)
82
Exercise
16.4 Electric Potential
1. What is the electric potential 11.0 cm from a 4.25 µC point charge ?
(3.47 x 10 5 V)
2. What is the electric potential 2.0 x 10 -10 m from a proton (charge +e) ?
(7.2 V)
3. The electric potential and the electric field strength at a point in an
electric field produced by a point charge Q are + 500 V and 150 V/m
respectively. Determine
a) the distance of the point from the point charge (3.33 m)
b) the charge Q. (1.85 x 10 -7 C)
4. A negative point charge of 0.75 mC travelling from point X to point Y
in an electric field, experiences a pd drop of 200 V.
a) Determine the potential at point Y if a potential of +20 V exists at
point X. (-180 V)
b) Determine the work done in this charge transfer.
Is the work done on or by the field ? (-0.15 J, work done on the field)
5. A rectangle ABCD has length AB = 4.0 m and breadth AD = 3.0 m.
Point charges Q1 = + 0.08µC and Q2 = - 0.03 µC are placed at A and C
respectively. Determine
a) the electric potential VB-VD across B and D (-72.8 V)
83
b) the work required in moving a point charge q = =0.50 µC from D to B.
(+36.4 J)
Relation Between V and E
16.4 Electric Potential
• The relationship between electric field intensity E at a
point in a uniform electric field and electric potential V is
given by
V
E
d
d
E (uniform)
V
• Unit ; Vm-1
• Vector quantity : directed towards negative plate
84
Example 16.24
16.4 Electric Potential
Two parallel plates are charged to a voltage of 50 V.
If the separation between the plates is 0.050 m,
calculate the electric field strength between them.
Exercise
1. How strong is the electric field between twp parallel plates 5.0 mm
apart if the potential difference between them is 110 V ? (22 kV m-1)
2. An electric field of 800 V/m is desired between two parallel plates
6.0 cm apart. How large a voltage should be applied ? (48 V)
3. The electric field between two parallel plates connected to a
85
45-V battery is 600 V/m. How far apart are the plates ? (75 mm)
16.4 Electric Potential
Equipotential Lines and Surfaces
• The electric potential can be represented graphically by
drawing equipotential lines or in three dimensions,
equipotential surfaces.
• An equipotential surface is a surface on which all points
are at the same potential.
• An equipotential line is a line on which all points
are at the same potential.
• No work is done when a charge moves from one point on
an equipotential surface to another point on the same
surface (because the potential difference is zero)
• An equipotential surface must be perpendicularly to the
86
electric field at any point.
16.4 Electric Potential
i) An isolated charge
A

E
VA  VB  VC
B
WBA  qVAB  q(VA  VB )
C
WBA  0
The dashed lines represent the
equipotential surface (line).
87
16.4 Electric Potential
ii) A uniform electric field
+
+
+
+
+
-
A
-
C
-
B
-

E
VA  VB  VC
WBA  qVAB  q(VA  VB )
WBA  0
-
The dashed lines represent the
equipotential surface (line).
88
16.4 Electric Potential
Electric Potential Energy, U
• The electric potential energy U of a charge q at a point
in an electric field is given by U  qV
Q
A
q•
+
r
U A  qVA
Q
A
q•
r
U A  q( VA )
• Scalar quantity.
• Its unit is joule (J).
• The sign of charges must be substituted in the calculation.
Exercise
What is the potential energy of an electron that is 0.53 x 10 -10 m
from a proton ? (-4.35 x 10-18 J)
89
16.4 Electric Potential
The Change in Electric Potential Energy, ∆U
U  U AB
 U A  UB
 qVA  qVB
The difference between two electric potential energy
The difference between electric potential energy at
point A and electric potential energy at point B
90
Example 16.25
16.4 Electric Potential
a) What is the electric potential energy of a charge 0.05µC at the point
i) 10 cm and
ii) 50 cm from a point charge of 2 µC ?
b) Find the change in electric potential energy between i) and ii).
a)
b)
91
16.4 Electric Potential
Electrical Potential Energy of A System of Point Charges
• Consider a system with three point charges, q1 , q2 and q3.
• The potential energy of the system is given by
U  U12  U13  U 23
The electric potential
energy of the system of
charges is the work done
to move all the
charges from infinity to the
points where the charges
are placed.
 q1q2 q1q3 q2q3 
U  k



r
r
r
13
23 
 12
q2 +
r23
q3 +
r12
r12
+
q1
• The sign for the charge (+ or -) must be substituted in the calculation.
92
16.4 Electric Potential
Example 16.26
+
q2 = 10µC
r23 = 5.0 cm
+
q3 = 10µC
r12= 4.0 cm
r13 = 5.0 cm
-
q1 = - 10µC
Calculate the potential energy for the system of charges
shown above.
93
16.4 Electric Potential
Exercise
1. A point charge Q = +9.10 µC is held fixed at the origin. A second point
charge with a charge of q = -0.420 µC is placed on the x -axis, 0.960 m
from the origin. Calculate the electric potential energy of the pair of
charges. (-3.58 x 10-2 J)
2.
+200 pC
•
A
20 cm
•
D
•
C
60 cm
-100 pC
•
B
20 cm
In figure above, the charge at A is +200 pC, while the charge at B is
-100 pC.
a) Calculate the electric potentials at points C and D. (-2.25 V, +7.88 V)
b) If a charge of +500 µC is placed at points C and D,
calculate the electric potential energy at points C and D.
(- 1.13 x 10 -3 J, 3.94 x 10-3 J)
c) The change in potential energy between points C and D. (-5.07 mJ)
94
Exercise
16.4 Electric Potential
3. A point charge q1 = +2.00 nC is placed at the origin and a second point
charge q2 = -3.00 nC is placed on the x -axis, at x = +20.0 cm. A third
point charge q3 = 5.00 nC is to be placed on the x -axis between q1 and q2.
What is the electric potential energy of the system of the three charges if
q3 is placed at x = +10.0 cm ? (-7.20 x 10 -7 J)
4. Three equal point charges q = 8.40 x 10-7 C are placed at the corner of an
equilateral triangle whose side is 1.00 m. What is the potential energy of the
system? ( 19.1 mJ)
95