Topic

--- Electrostatics
The study of electric
charges at rest, the
forces between them
and the electric fields
associated with them.

16.1 Coulumbs Law

16.2 Electric Field
16.3 Electric Potential

(1 Hour)
(1 Hour)
(1 Hour)
16.4 Charge In A Uniform Electric Field (1 Hour)

Electrostatics

Topic

--- Electrostatics

Coulombs Law
Electric field

Equipotential
surface

Electric field
strength

Charge in uniform
electric field
Electric potential
Potential energy

Topic

--- Electrostatics

16.1 COULOMBS LAW

(a) State Coulombs law,
Qq
kQq
F
2
2
4 0 r
r

(b) Sketch the force diagram and apply Coulombs law

for a system of point charges

Topic

--- Electrostatics

Two kind of charges:

(+)ve & ()ve charges
Charges of
opposite
sign attract
one another
attractive
force

Charges of
the same
sign repel
one another
repulsive
force
Figure 16.1

16.1

COULUMBS LAW

Topic

--- Electrostatics

Principle of
conservation of
charges
the total charge
in an isolated
system is
constant
(conserved)
electric
charge can neither
be created nor
destroyed

Charge is
quantized
Electric charge
exists as discrete
packets

Q ne

where: Q = electric charge

n = positive integer number = 1, 2,
e = fundamental amount of charge = 1.6 x 10-19 C
16.1

COULUMBS LAW

Charge, Q is a
scalar quantity
S.I unit:
Coulumb (C)
1 C is defined

as the total
charge
transferred by a
current of 1
Ampere in 1
second

How many electrons in 1 C

and what are the total
mass of these electrons?

Topic

--- Electrostatics

states that the magnitude of

the electrostatic (Coulomb or
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

Q1Q2
F 2
r

kQ1Q2
F
r2

F = magnitude of electrostatic force

Q1, Q2 = magnitude of charges
r = distance between two point
charges
k = electrostatic (Coulumb)
constant = 9.0 x 109 N m2 C-2
16.1

COULUMBS LAW

Since

1
4 0

hence the Coulombs law can

be written as

1 Q1Q2
2
F
4 0 r
= permittivity of free space (vacuum or
air) = 8.85 x 10 12 C2 N-1 m-2
Permittivity is a property of a material that is
indicative of how well it supports an electric
field

Topic

--- Electrostatics

What happens to the force

between two charges, if
(a) the distance between
them is doubled?
(b) the distance between
them is cut in half?
(c) the magnitude of one
charge is doubled?
(d) the magnitude of both
charges is doubled?

Topic

--- Electrostatics

Graphically

3 CASES

repulsive force

attractive force

0
F

Figure 16.3
Gradient,
m = kQ1Q2

0
16.1

1
r2

Figure 16.2

COULUMBS LAW

F12 : the force on charge q1 due

to charge q2
F21 : the force on charge q2 due
to charge q1
Simulation 16.1

Topic

--- Electrostatics

Figure 16.4

16.1

COULUMBS LAW

In electromagnetism, permittivity is the

measure of how much resistance is
encountered when forming an electric field in
a medium. In other words, permittivity is a
measure of how an electric field affects, and is
affected by, a dielectric medium. Permittivity is
determined by the ability of a material to
polarize in response to the field, and thereby
reduce the total electric field inside the
material. Thus, permittivity relates to a
material's ability to transmit (or "permit") an
electric field.
It is directly related to electric susceptibility,
which is a measure of how easily a
dielectric polarizes in response to an electric
field.

Topic

--- Electrostatics

Points to bring out about Coulombs law:

The form is exactly the
same as Newtons law of
universal gravitation; in
particular, it is an
inverse-square law
This force can be
attractive or repulsive
The magnitude of the
force can be calculated
by this equation, and the
direction should be
obvious from the signs of
the interacting charges

Actually, if you include

the signs of the charges
in the equation, then
whenever you get a
negative answer for the
force, there is an
attraction, whereas a
positive answer indicates
repulsion

Topic

--- Electrostatics

Coulombs Law VS Newtons Law of Gravitation

16.1

Coulombs Law

Newtons Law of Gravitation

Attractive or repulsive force

Only attractive force

Force due to charge

interaction

Force due to mass

interaction

The force is a short-range

force

The force is a long-range

forces

kQ Q
Fe 12 2
r

Gm1m2
Fg
r2

COULUMBS LAW

Topic

--- Electrostatics

Two point charges, Q1 =

85 C and Q2 = 50 C
are separated by a
distance of 3.5 cm as
shown in Figure 16.5.
Determine the magnitude
and direction of
(a) the electric force that
Q2 exerts on Q1
(b) the electric force that
Q1 exerts on Q2
(Given Coulombs
constant, k = 9.0 109
N m2 C2)
16.1

COULUMBS LAW

Q1

Q2

3.5 cm
Figure 16.5

Topic

Solution:

--- Electrostatics
Q1 85 10 6 C; Q2 50 10 6 C; r 3.5 10 2 m

STEP 1: Draw the electric force vectors

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

Q1

F21

Q2

F : force by charge 2
21

on charge 1

3.5 cm
STEP 2: Use Coulombs equation to calculate electric force

9.0 109 85 10 6 50 10 6
kQ1Q2
F21
F21
2
2 2
r
3.5 10
F21 3.12 10 4 N
Direction: to the right (towards Q2)
16.1

COULUMBS LAW

Since electric
forces obey
Newtons Law,
therefore the
forces F21 and
F12 are equal in
magnitude but
opposite in
direction
F12 = F21
F12 = 3.12 x
104 N

Topic

--- Electrostatics

Three point charges lie along the x-axis as shown in Figure

16.6.
Q1 12 C
Q2 20 C
Q3 36 C

+
12 cm

+
20 cm

Figure 16.6

(a) Calculate the magnitude and direction of the total

electrostatic force exerted on Q1
(b) Suppose the charge Q2 can be moved left or right along
the line connecting the charges Q1 and Q3. Determine the
distance from Q3 where Q2 experiences a nett
electrostatic force of zero
(Given permittivity of free space, 0 = 8.85 1012 C2 N1 m2)
16.1

COULUMBS LAW

--- Electrostatics

Topic

Solution:

Q1 12 10 6 C; Q2 20 10 6 C; r12 12 10 2 m
Q3 36 10 6 C; r13 32 10 2 m

STEP 1: Draw the electric force vectors

(a) Q1 will experience of forces by Q2 and Q3

F31

Q1

F21

r12

Q2

Q3

r13
The force acting on Q1 due to Q2 is attractive because Q1 and Q2 have the
opposite sign, therefore the direction of F21 is to the right

The force acting on Q1 due to Q3 is repulsive because Q1 and Q3 have the

same sign, therefore the direction of F31 is to the left

16.1

COULUMBS LAW

Topic

--- Electrostatics

STEP 2: Use Coulombs equation to calculate electric force

By applying the Coulombs law equation, thus

Q1Q2
F21
2
4 0 r12

12 10 20 10

4 8.85 10 12 10
6

F21

12

2 2

F21 150 N

Direction : to the right (towards Q2)

12 10 36 10

4 8.85 10 32 10
6

F31

12

2 2

F31 37.9 N
Direction : to the left

STEP 3: Electric force adds as vectors

Q1
F21 150 N
F31 37.9 N

16.1

COULUMBS LAW

The total electrostatic force on Q1

F1 F21 F31
F1 150 37.9 Direction: to the right
(towards Q2)
F1 112 N

Topic

Solution:
(b)

--- Electrostatics
Q1 12 10 6 C; Q2 20 10 6 C; r12 12 10 2 m
Q3 36 10 6 C; r13 32 10 2 m

Q1
Q2 F
Q3
F12
32

r13 x

r13

The nett force acting on Q2 is zero thus

F12 F32
Q2Q3
Q1Q2

2
2
4 0 r12
4 0 r23
12 10 6
36 10 6

2
2
2
x
32 10 x
x 0.203 m OR 20.3 cm

16.1

COULUMBS LAW

Topic

--- Electrostatics

For each diagram below, draw the

direction of electric force acting on Q1.

16.1

COULUMBS LAW

Topic

--- Electrostatics

Figure 16.7 shows three

point charges that lie in
the x-y plane in a
vacuum.
Calculate the magnitude
and direction of the nett
electrostatic force on Q2.
(Given electrostatic
constant, k = 9.00 109 N
m2 C2)

Q1 6.0 C

20 cm

17

12 cm

+
Q2 4.0 C

Q3 5.0 C

Figure 16.7
16.1

COULUMBS LAW

Topic

--- Electrostatics

Given 0 = 8.85 1012 C2 N1 m2

1.

16.1

Two point charges are placed

2.
on the x-axis as follows : Charge
Q1 = +4.00 nC is located at x =
0.200 m, charge Q2 = +5.00 nC
is at x = 0.300 m. Determine
the magnitude and direction of
the total electric force exerted
by these two charges on a
negative point charge Q3 =
6.00 nC that is placed at the
origin. (University physics,
11th edition, Young &
Freedman, Q21.20, p.829)
ANS: 2.4 N to the right

COULUMBS LAW

A point charge Q = 0.35 nC is

fixed at the origin. Where a
proton must be placed in
order for the electric force
acting on it to be exactly
opposite to its weight? (Given
charge of proton, Qp= 1.60
1019 C and mass of the
proton, mp = 1.67 1027 kg )
(Physics, 3rd edition, J.S.
Walker, Q18, p.657)
ANS: 5.55 km below Q

Topic

--- Electrostatics

3. Four identical point charges (Q = +10.0 C) are located on the

corners of a rectangle as shown in Figure 16.8.

+Q

Q+

w 15.0cm

Figure 16.8

Q +

l 60.0cm

+Q

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. (Physics for scientists and
engineers, 6th edition,Serway&Jewett, Q57, p.735)
ANS: 40.9 N at 263
16.1

COULUMBS LAW

Topic

--- Electrostatics

16.2 ELECTRIC FIELD

(a) Define and use electric field strength,

F
E
q0

(b) Use E

kQ
r2

for point charge

(c) Sketch the electric field strength diagram and

determine electric field strength E for a system of
charges

Topic

--- Electrostatics

is defined as a region in Electric field around

charges can be
which an electric force
represented by drawing a
will act on a charge
series of lines
that, is place in that
These lines are called
region/ a region of
electric field lines (lines of
space around isolated
force)
charge where an electric
force is experienced if a
positive test charge
placed in the region
16.2

ELECTRIC FIELD

Topic

--- Electrostatics

Simulation 16.2

Topic

--- Electrostatics

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 of starting or ending is proportional to
the magnitude of the charge
The field lines never cross because the electric field dont have two
value at the same point

Topic

--- Electrostatics

Mathematically,
defined as the

electric

F
E
F q0 E
(electrostatic) force
q0

per unit positive

E : electric field strength
charge that acts at
F : electric force
q0 : test charge
that point in the
same direction as the It is a vector quantity.
force
Unit: N C1 OR V m 1

16.2

ELECTRIC FIELD

--- Electrostatics

Topic

Consider a test charge, q0

located at a distance r from a
point charge, Q

q0

F
E
q0

(2)

(1) in (2):

r
Figure 16.11
A test charge is a charge small
enough to leave the main charge
configuration undisturbed

16.2

According to Coulumbs Law

kQ1Q2
F
(1)
2
r
From definition
electric field:

ELECTRIC FIELD

kQ
Q
E 2 OR E
2
4

r
r
0
Q: magnitude of a point charge
r: distance between a point & point
charge

Topic

--- Electrostatics

Determine
(a) the electric field
strength at a point X
at a distance of 20 cm
from a point charge Q
= 8 C

Q = 8C

X
20 cm

16.2

ELECTRIC FIELD

(b) the electric force that

acts on a point charge
q = 1 C placed at
point X.

q = 1 C

Q = 8C

+
20 cm

Topic

--- Electrostatics

The direction
of the electric
field strength,
E depends on
the sign of
the point
charge only
of the electric
force, F
depends on
both signs of
the point
charge and
the test
charge
16.2

ELECTRIC FIELD

Figure 16.12
Simulation 16.3

Topic

--- Electrostatics

Two point charges, Q1= 3.0 C and Q2= 5.0 C, are placed 12
cm and 30 cm from the point P respectively as shown in Figure
16.13.

Q1

Q2

12 cm

30 cm

Figure 16.13

Determine
(a) the magnitude and direction of the electric field intensity at P,
(b) the nett electric force exerted on q0= +1 C if it is placed at P,
(c) the distance of a point from Q1 where the electric field intensity
is zero.
(Given electrostatic constant, k = 9.00 109 N m2 C2)
16.2

ELECTRIC FIELD

Topic

--- Electrostatics

Two point charges, Q1= 2.0 nC and Q2= +3.2 nC, are placed
3.0 cm apart as shown in Figure 16.14.

Q1
Figure 16.14

3.0 cm

4.0 cm
Determine the magnitude and direction of the resultant
electric field intensity at point M.
(Given permittivity of free space, 0 = 8.85 1012 C2 N1 m2)

Q2

16.2

ELECTRIC FIELD

Topic

--- Electrostatics

Given 0 = 8.85 1012 C2 N1 m2

1. Sketch an electric field lines pattern for following cases:
(a) Two equal negative charges, Q and Q.
Q
Q

(b) Two unequal negative charges, 2Q and Q

16.2

ELECTRIC FIELD

2Q

Topic

2.

--- Electrostatics

Determine the
magnitude of the electric
field at point P due to the
four point charges as
shown in Figure 16.15 if
q = 1 nC and d = 1 cm.
(Fundamental of
physics, 6th edition,
Halliday, Resnick &
Walker, Q11, p.540)

ANS: zero

16.2

ELECTRIC FIELD

Figure 16.15

Topic

3.

--- Electrostatics

Calculate the magnitude

and direction of the
electric field at the
centre of the square in
Figure 16.16 if q =1.0
108 C and a = 5 cm.
(Fundamental of
physics,6th edition,
Halliday, Resnick
&Walker, Q13, p.540)

ANS: 1.02 105 N C1 ;

upwards
16.2

ELECTRIC FIELD

Figure 16.16

Topic

--- Electrostatics

16.3 ELECTRIC POTENTIAL

(a) Define electric potential

W
V
q0

(b) Define and sketch equipotential lines and surfaces of

(i) an isolated charge
(ii) a uniform electric field

kQ
(c) Use V
r

for a point charge and a system of charges

(d) Calculate potential difference between

two points

VAB VA VB
WBA
VAB
q0

Topic

--- Electrostatics

16.3 ELECTRIC POTENTIAL

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

U q0 V
(f) Calculate potential energy of a system of point
charges

Q1Q2 Q1Q3 Q2Q3

U k

r13
r23
r12

Topic

--- Electrostatics

W
V
q0

16.3

ELECTRIC POTENTIAL

Topic

--- Electrostatics

is defined as the locus of points that have the same electric

potential

Figure 16.17
16.3

ELECTRIC POTENTIAL

Topic

--- Electrostatics
The electric field, E at every
point on an equipotential
surface is perpendicular to
the surface

The electric field points in

the direction of decreasing
electric potential

A (+)ve
point charge

The surface are closer

together where the electric
field is stronger, and farther
apart where the field is weaker A uniform
alectric

No two equipotential
surfaces can intersect
each other
16.3

ELECTRIC POTENTIAL

field

Topic

--- Electrostatics

APPLICATION: 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)

16.3

ELECTRIC POTENTIAL

--- Electrostatics

Topic

Q
A

+
r

The electric potential at

point A at distance r from a
positive point charge Q is

Q
VA k
r

16.3

ELECTRIC POTENTIAL

The electric potential at

point A at distance r from a
negative point charge Q is

(Q)
VA k
r
Q
V A k
r

Topic

--- Electrostatics

A (+)ve point charge

A (-)ve point charge

For a point charge

Q
kQ
OR V 4 r
V
0
r
16.3

ELECTRIC POTENTIAL

--- Electrostatics

Topic

Fext

+qo

dW Fext dr
kQq0
but FE 2
r

dW FE dr

r r

dW

r r

FE dr
r

W kQqo r dr
2

W kQqo r

Qq
W k o
r
1

16.3

FE

ELECTRIC POTENTIAL

The electric
potential at
point A at
distance r from
a positive point
charge Q

VA

Wr
q0

VA

kQqo
qo r

Q
VA k
r

Topic

--- Electrostatics

Figure 16.18
V = (+)ve W = (+)ve
work is required (need external force
to move the charge
V = ()ve W = ()ve
No work required (work done by the
electric force itself// No external
force is needed to move the charge

16.3

ELECTRIC POTENTIAL

Figure 16.18 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.

Topic

--- Electrostatics

The total electric potential at a point in space is equal to the

algebraic sum of the constituent potentials at that point
In the calculation of U , W and V, the sign of the charge must
be substituted in the related equations

Two point charges, Q1= 40 C and Q2= 30 C are

separated by a distance of 15 cm as shown
in Figure 16.19. Calculate
Figure
16.19

Q1

13 cm
Q2

5 cm

10 cm

(a) the electric potential at point A and describe the meaning of the
answer,
(b) the electric potential at point B.
(Given 0 = 8.85 1012 C2 N1 m2)
16.3

ELECTRIC POTENTIAL

Topic

--- Electrostatics

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

W
V
q0
V V final Vinitial

16.3

ELECTRIC POTENTIAL

The electric field is a

conservative field The work
done to bring a charge from
one point to another point in
an electric field is
independent of the path
If the value of work done is
positive work done on the
system (need external force)
If the value of work done is
negative work done by the
system (no external force is
needed)

Topic

--- Electrostatics

Consider a positive test

charge is moved by the
external force, F from point
A to point B as shown in
Figure 16.20
B

q0

For this (+) charge, +Q

VB > VA
ELECTRIC POTENTIAL

W A B
V
q0
A

Figure 16.20

16.3

The potential difference

between points A and B, VAB
is given by

where

V VB V A
WBA: work done in bringing a (+)ve
test charge from point A to point
B
VA: electric potential at point A
VB: electric potential at point B

Topic

16.3

--- Electrostatics

ELECTRIC POTENTIAL

--- Electrostatics

Topic

Moving the test charge (+)ve

from A to B is harder and will
require work (Need external
force)
WAB = (+)ve
Work done by the test
charge/ on an electric field
V = VB VA = (+)ve VB > VA
(From lower to higher electric
potential, potential energy of
the charge will increase)
16.3

ELECTRIC POTENTIAL

Moving the test charge (+)ve

from B to A is easier because it
will naturally move from B to A in
the direction of electric field. No
external force needed
WBA = ()ve
Work done on the test charge/
by an electric field
V = VA VB = ()ve VA < VB
(From higher to lower electric
potential, potential energy of the
charge will decrease)

--- Electrostatics

Topic

Moving the test charge (+)ve

from A to B is easier because it
will naturally move from A to B in
the direction of electric field. No
external force needed
WBA = ()ve
Work done on the test charge/
by an electric field
V = VB VA = ()ve VB < VA
(From higher to lower electric
potential, potential energy of the
charge will decrease)
16.3

ELECTRIC POTENTIAL

Moving the test charge (+)ve

from B to A is harder and will
require work (Need external
force)
WAB = (+)ve
Work done by the test
charge/ on an electric field
V = VA VB = (+)ve VA > VB
(From lower to higher electric
potential, potential energy of
the charge will increase)

--- Electrostatics

Topic

Moving the test charge (+)ve

from A to B is difficult and will
require work (Need external
force)
WAB = (+)ve
Work done by the test
charge/ on an electric field
V = VB VA = (+)ve VA < VB
(From lower to higher electric
potential, potential energy of
the charge will increase)
16.3

ELECTRIC POTENTIAL

Moving the test charge (+)ve

from B to A is easier because it
will naturally move from B to A in
the direction of electric field. No
external force needed
WBA = ()ve
Work done on the test charge/
by an electric field
V = VA VB = ()ve VB > VA
(From higher to lower electric
potential, potential energy of the
charge will decrease)

Topic

--- Electrostatics

The potential
difference between
any two points on an
equipotential surface
is zero
Hence no work is
required to move a
charge along an
equipotential surface
16.3

ELECTRIC POTENTIAL

Topic

--- Electrostatics

Figure 16.21 show two points A and B are at a distance of

2.0 cm and 3.0 cm respectively from a point charge Q =
100 C.

100
C

Figure 16.21

Determine:
(a) the electric potential at A and B
(b) the work require in moving a point charge q = + 2 C,
from A to B

16.3

ELECTRIC POTENTIAL

Topic

--- Electrostatics

Two points, S and T are located around a point charge of +5.4

nC as shown in Figure 16.22. Calculate
S

8.0 cm

6.0 cm
Figure 16.22

(a)
(b)

16.3

5.4 nC
the electric potential difference between points S and T,
the work done in bringing a charge of 1.5 nC from point
T to point S. (Electrostatic constant, k = 9.00 109 N m2
C2)

ELECTRIC POTENTIAL

Topic

--- Electrostatics

Change in a potential
energy, U
From the definition of
electric potential
difference, V

W
V
q0

and

W U

U
V
q0
16.3

ELECTRIC POTENTIAL

Therefore the change in a

potential energy is given
by
U q0 V
U 2 U1
final

V2 V1
initial

Topic

--- Electrostatics

Consider a system of three point charges as shown in Figure

16.23.
Q2

r12

Figure 16.23

Q1

r23
r13
Q3

The total electric potential energy, U can be expressed as

U U12 U13 U 23
kQ1Q2 kQ1Q3 kQ2 Q3
U

r12
r13
r23

Q1Q2 Q1Q3 Q2 Q3

U k

r
r
r
13
23
12
16.3

ELECTRIC POTENTIAL

Topic

--- Electrostatics

Take distance d = 14.0

cm and charge q = 150
nC. What is the electric
potential energy of this
system of charges?

16.3

ELECTRIC POTENTIAL

Solution:

Q1Q2 Q1Q3 Q2Q3

U k

r
r
r
13
23
12

Topic

--- Electrostatics

0 = 8.85 1012 C2 N1 m2; me= 9.11 1031 kg; e = 1.60

1019 C
1. At a certain distance from a point charge, the magnitude
of the electric field is 500 V m1 and the electric potential
is 3.00 kV. Calculate
(a) the distance to the charge.
(b) the value of the charge.
(Physics for scientists and engineers,6th
edition,Serway & Jewett, Q17, p.788)
ANS: 6.00 m; 2.00 C

16.3

ELECTRIC POTENTIAL

Topic

2.

--- Electrostatics

Four point charges are located at the corners of a square

that is 8.0 cm on a side. The charges, going in rotation
around the square, are Q, 2Q, 3Q and 2Q, where Q = 4.8
C as shown in Figure 16.24.

Q +

+ 2Q

Figure 16.24

3Q
Determine the electric potential at the centre of the
square.
ANS: 1.53 106 V

2Q +

16.3

ELECTRIC POTENTIAL

Topic

--- Electrostatics

3.

Initially two electrons are fixed in place with a separation

of 2.00 m. How much work must we do to bring a third
electron in from infinity to complete an equilateral
triangle?
(Fundamental of physics,7th edition, Halliday, Resnick &
Walker, Q79, p.653) ANS: 2.30 1022 J
4. Two point charges, Q1= +q and Q2= +2q are separated by
1.0 m as shown below.

Q1 +

1.0 m

+ Q2

Determine the position of a point where

(a) the nett electric field intensity is zero,
(b) the electric potential due to the two charges is zero.
(Fundamental of physics,7th edition, Halliday, Resnick
& Walker, Q81, p.653) ANS: 0.41 m, U think
16.3

ELECTRIC POTENTIAL

Topic

--- Electrostatics

Topic

--- Electrostatics

W
V
q0
U U12 U13 U 23

Q1Q2 Q1Q3 Q2 Q3

U k

r
r
r
12
13
23

U
V
q0
K
V
q0

W
V
q0

W ( )ve

+
W ( )ve

W ( )ve

W ( )ve

Topic

--- Electrostatics

16.4 CHARGE IN A UNIFORM

ELECTRIC FIELD
(a) Explain quantitatively with the aid of a
diagram the motion of a charge in a uniform
electric field
(b) Use

V
E
d

for uniform E

Topic

--- Electrostatics

A uniform electric field is

represented by a set of
electric field lines which are
straight, parallel to each
other and equally spaced
It can be produced by two
flat parallel metal plates
which is charged, one with
positive and one is negative
and is separated by a
distance
Direction of E: (+)ve plat to
() plat (Figure 16.24)
16.4

CHARGE IN A UNIFORM ELECTRIC FIELD

Figure 16.24

Topic

--- Electrostatics

CASE1: A a particle with

positive charge q is held
stationary

CASE2: A particle with positive

charge moves at constant speed
respectively, in a uniform electric
field, E

Figure 16.23
The forces acted on the particle are electrostatic force (upwards)
and weight (downwards).

Fe qE and W mg
16.4

CHARGE IN A UNIFORM ELECTRIC FIELD

Fe W , qE W

Topic

--- Electrostatics

When a test charge is placed

between two oppositely charged
plates there will be a push force
from the positive plate and a pull
force from the negative plate.
The force gets weaker as the charge
gets further from the plate.
The left charge gets a very strong
push (75% of the total forces) from
the nearby positive plate and a
weak pull (25%) from the faraway
negative plate.
The middle charge which is halfway
between the two plates gets half of
the total force from the pushing
positive plate and 50% of the force
from the pulling negative plate.
The right charge gets 25% push
from the positive and 75% pull from
the negative.
16.4

CHARGE IN A UNIFORM ELECTRIC FIELD

Topic

--- Electrostatics

CASE3: Consider an electron (e) with mass, me enters a

uniform electric field, E perpendicularly with an initial
velocity u in Figure 16.24
The upward electric force will cause the electron to move
along a parabolic path towards the upper plate as shown
below
sxx
v

Figure 16.24

q0

16.4

sy

CHARGE IN A UNIFORM ELECTRIC FIELD

Simulation 16.4

Topic

--- Electrostatics

Therefore the magnitude of the

electrons acceleration is given by

eE
a ay
me

direction: upwards
since a x 0

The components of electrons velocity

after pass through the electric field
are given by v u constant
x-component: x
v y u y a y t ,u y 0
y-component:

vy

eE
t
me

The position of the electron is

s x ut

and
1
s y u yt a yt
2

16.4

The path makes by the electron is

similar to the motion of a ball
projected horizontally above the
ground

1 eE
s y
2 me

CHARGE IN A UNIFORM ELECTRIC FIELD

2
t

Topic

--- Electrostatics

Figure 16.25 shows an electron

entering a charged parallel plate
(Physics, 3rd edition, J. S. Walker, Q78, p.661)
with a speed of 5.45 106 m s1.
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.
Figure 16.25
19
(Given e = 1.60 10 C and me =
9.11 1031 kg)
16.4

CHARGE IN A UNIFORM ELECTRIC FIELD

Topic

16.4

--- Electrostatics

CASE4: Charge moving parallel to the field

Since only electric force exerted on the particle, thus this
force contributes the nett force, F and causes the particle to
accelerate
q0 E
q0 E
For a (+)ve charge: a
For a ()ve charge: a
m
m

CHARGE IN A UNIFORM ELECTRIC FIELD

Topic

--- Electrostatics

CASE1: Consider a stationary particle of charge q and mass m is placed in a

uniform electric field, E in Figure 16.26
Figure 16.26

a+

Fe

Fe

a -

The electric force Fe exerted on the charge is given by Fe qE

Since only electric force exerted on the particle, thus this force
contributes the nett force, F and causes the particle to accelerate
From Newtons 2nd law,
For a positive charge, its acceleration
is in the direction of the electric field,
F Fe ma
q0 E
for a negative charge (electron), its
a

q0 E ma
acceleration is in the direction
m
opposite the electric field

16.4

CHARGE IN A UNIFORM ELECTRIC FIELD

--- Electrostatics

Topic

s x xu x t

vx u x

ux ,

uy 0

q0
16.4

v vx 2 v y 2

v y u y a y t ,u y 0
eE

t
me
s y u yt

CHARGE IN A UNIFORM ELECTRIC FIELD

1
a y t ,u y 0
2

1 eE 2
t
s y
2 me

Topic

--- Electrostatics
W K V

W
V ,
q0

K
q0

Fd
W Fd V
q0

1 2
Gain in kinetic energy , K mv
2
arrive at () plat with velocity, v

q0 E
F q0 E , F ma a
m

16.4

v
Fe
a

V
E
d

CHARGE IN A UNIFORM ELECTRIC FIELD

ch arg e at rest, u 0

Topic

--- Electrostatics

Figure 16.27 show a

uniform electric field
between two plats
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 below

16.4

CHARGE IN A UNIFORM ELECTRIC FIELD

Figure 16.27

Topic

--- Electrostatics
The V against r graph for pair
of flat metal plates can be
shown
The graph is a straight line
with negative constant
gradient, thus

16.4

V ( 0 V )
E

r ( d 0 )
For uniform E
V
E
such as in
d
capacitor

CHARGE IN A UNIFORM ELECTRIC FIELD

Topic

--- Electrostatics

Two parallel plates are

separated 5.0 mm apart. The
electric field strength
between the plates is 1.0
104 N C1.
A small charge of +4.0 nC is
moved from one conducting
plate to another. Calculate
(a) the work done on the
charge, and
(b) the potential difference
between the plates.

16.4

CHARGE IN A UNIFORM ELECTRIC FIELD

Topic

--- Electrostatics

0 = 8.85 1012 C2 N1 m2; me = 9.11 1031 kg; e = 1.60 1019 C

An electron beam enters at right angle into a uniform

electric field between two horizontal plates
separated of 5.0 cm apart. The plates are connected
across a potential difference of 1000 V. The length of
the plates is 10.0 cm. The beam is deflected
vertically at the edge of the field by a distance of 2.0
cm. Calculate the speed of the electrons entering the
field.
ANS: 2.97 107 m s1

16.4

CHARGE IN A UNIFORM ELECTRIC FIELD