CHAPTER 1:

ELECTROSTATIC
 WHAT IS ELECTROSTATIC?

Electrostatic is the study of electrical charges in

static condition. It involves the influence of a charge
on other charge, the forces between the charges,
the area of influence (electric field ) and energy
produced and lost in certain operation.
 SUBTOPIC

1.1 Coulomb’s Law

1.2 Electric Field

1.3 Electric Potential

1.4 Charge in a Uniform Electric Field

 Learning Outcome :

1.1 Coulomb’s Law

At the end of this chapter, students should be able to:

a) State Coulomb’s Law,

Qq kQq
F  
4 o r 2
r2

b) Sketch the force diagram

c) apply Coulomb’s law for a system of point

charges.
 COULOMB’S LAW
Coulomb’s Law states that
the electrostatic force, F Q1Q2
F k
between two charges r2
separated by a distance, r,
is where :
(i) inversely proportional to k = Coulomb constant
the square of the (9  109 N m2 C-2)
separation, r between the Q1 = magnitude of
two charges, and ; charge Q1 (unit C)
Q2 = magnitude of
(ii) directly proportional to charge Q2 (unit C)
the product of the r = separation distance
magnitudes of the between the two
charges, Q1 and Q2. charges.
 COULOMB’S LAW

Constant k is also given by Q1Q2

F k
1 r2
k  9.0 109 NC-2 m 2
4 
where :
k = Coulomb constant
Where : (9  109 N m2 C-2)
εo = permittivity of free space Q1 = magnitude of
which has the value of charge Q1 (unit C)
8.85  10-12 F m-1 Q2 = magnitude of
charge Q2 (unit C)
r = separation distance
between the two
charges.
 ELECTROSTATIC FORCE
 Electrostatic force is F21
F12
directed along the line +
joining the charges. Q1 Q2
Force Sign of F12 F21
Charges + +

Fattraction Opposite Q1 Q2

Frepulsion Same
Where:
 Since electric forces obey F12 :the force exerted
Newton’s third law, on charge 1 by
therefore the forces F12 charge 2
and F21 are equal in F21 :the force exerted
magnitude but opposite on charge 2 by
in direction. Hence, it can charge 1
be written as F = - F
 CHARGES, Q

CHARGES, Q

Scalar S.I unit :

Signs
quantity Coulomb (C)

Ignore Noted

When substitute into Distinguish direction

Coulomb’s law equation of electric force, F
 EXAMPLE 1.1 SOLUTION
Three point charges are (a) F23 F21
firmly held on a straight line Q1 = +10 μC Q2 = +5 μC Q3 = -8 μC
of 4 cm in length as shown
in the figure below. Q1 and Q2 have the same
sign, therefore the force
Q1 = +10 μC Q2 = +5 μC Q3 = -8 μC
acting on Q2 due to Q1 is
repulsive.
2 cm 2 cm
F21 is to the right.
Find the resultant electric
force acting on ; Q2 and Q3 have the
(a) charge Q2 opposite sign, therefore
(b) charge Q1 the force acting on Q2
due to Q3 is attractive.
F23 is also to the right.
 EXAMPLE 1.1 SOLUTION
Three point charges are (a) The magnitudes of F21
firmly held on a straight line and F23 are given by ;
of 4 cm in length as shown
Q1Q2
in the figure below. F21  k
r2
6 6
(10  10 )(5  10 )
Q1 = +10 μC Q2 = +5 μC Q3 = -8 μC F21  (9 10 9 )
0.02 2
2 cm 2 cm  1125 N

Find the resultant electric Q2Q3

force acting on ; F23  k 2
r
(a) charge Q2 (5  10 6
)(8  10 6
)
(b) charge Q1 F23  (9 10 9 )
0.02 2
 900 N
 EXAMPLE 1.1 SOLUTION
Three point charges are (a) Therefore, the resultant
firmly held on a straight line electric force acting on
of 4 cm in length as shown charge Q2 is ;
in the figure below.
F = F21 + F23
Q1 = +10 μC Q2 = +5 μC Q3 = -8 μC = (+1125) + (+900)
2 cm 2 cm = 2025 N (to the right)

Find the resultant electric

force acting on ;
(a) charge Q2
(b) charge Q1
 EXAMPLE 1.1 SOLUTION
Three point charges are (b)
firmly held on a straight line F12 F13
of 4 cm in length as shown Q1 = +10 μC Q2 = +5 μC Q3 = -8 μC
in the figure below.
Q1 and Q2 have the
Q1 = +10 μC Q2 = +5 μC Q3 = -8 μC same sign, therefore the
2 cm 2 cm force acting on Q1 due to
Q2 is repulsive.
Find the resultant electric F12 is to the left.
force acting on ;
(a) charge Q2 Q1 and Q3 have the
(b) charge Q1 opposite sign, therefore
the force acting on Q1
due to Q3 is attractive.
F13 is to the right.
 EXAMPLE 1.1 SOLUTION
Three point charges are (b) The magnitudes of F12
firmly held on a straight line and F13 are given by ;
of 4 cm in length as shown
in the figure below. Q1Q2
F12  k
r2
Q1 = +10 μC Q2 = +5 μC Q3 = -8 μC (10 10 6 )(5 10 6 )
F12  (9 10 )9

2 cm 2 cm
0.02 2
 1125 N
Find the resultant electric
force acting on ; Q1Q3
F13  k
(a) charge Q2 r2 6 6
(10  10 )(8  10 )
(b) charge Q1 F13  (9 10 )
9

0.04 2
 450 N
 EXAMPLE 1.1 SOLUTION
Three point charges are (b) Therefore, the resultant
firmly held on a straight line electric force acting on
of 4 cm in length as shown charge Q1 is ;
in the figure below.
F = F12 + F13
Q1 = +10 μC Q2 = +5 μC Q3 = -8 μC = (-1125) + (+450)
2 cm 2 cm = -675 N (to the left)

Find the resultant electric

force acting on ;
(a) charge Q2
(b) charge Q1
 EXAMPLE 1.2 SOLUTION
Figure shows three point
charges that lie in the x, y
plane in a vacuum. Find the
electrostatic force on q1

The magnitude of the

forces are

Q1Q2
F12  k
r12 2
(4 10 6 )(6 10 6 )
 (9 10 9 )
(0.15) 2
F 12  9.6 N
 EXAMPLE 1.2 SOLUTION
Figure shows three point
charges that lie in the x, y
plane in a vacuum. Find the
electrostatic force on q1

The magnitude of the

forces are
Q1Q 3
F 13  k
r 13 2
(4 10 6 )(5 10 6 )
 (9 10 ) 9

(0.10) 2
F 13  18 N
 EXAMPLE 1.2 SOLUTION
Figure shows three point Electrostatic force is a
charges that lie in the x, y vector quantity!!
plane in a vacuum. Find the
Force x - comp y - comp
electrostatic force on q1
F12 +9.6cos73 +9.6sin73
= +2.8 N = +9.2 N
F13 + 18 N 0N

F Fx = + 21 N Fy = +9.2 N

The magnitude of force:

F  Fx 2  Fy 2

 212  9.2 2
 23N
 EXAMPLE 1.2 SOLUTION
Figure shows three point Electrostatic force is a
charges that lie in the x, y vector quantity!!
plane in a vacuum. Find the
Force x - comp y - comp
electrostatic force on q1
F12 +9.6cos73 +9.6sin73
= +2.8 N = +9.2 N
F13 + 18 N 0N

F Fx = + 21 N Fy = +9.2 N

The magnitude of force:

Fy
tan  
Fx
9.2
 
21
  24 above  x
 FOLLOW UP EXERCISE

The figure below shows three point charges that lie

along the x-axis in a vacuum. Determine the magnitude
and direction of the net electrostatic force on q1.

Answer: F = 5.7 N ( to the right )

 Learning Outcome :

1.2 Electric Field

At the end of this chapter, students should be able to:
F
a) Define and use electric field strength, E
q0
kQ
b) Use E  for point charge
2
r
c) Sketch the electric field strength diagram

d) Determine electric field strength E for a system of

charges.
 ELECTRIC FIELD

 is defined as a region of
space around isolated
charge where an electric
force is experienced if a
positive test charge
placed in the region.
 represented by electric
field lines (lines of
force).

 an isolated charge
means that we are only
considering one charge.
 ELECTRIC FIELD LINES

The direction is tangent to

the field line at any point.

number of lines determine

the strength of the electric
field.

Point Direction of
Charges Electric Field
Positive outward
Negative inward

The field lines never cross

ELECTRIC FIELD PATTERNS AROUND THE CHARGE

Conditions Electric Field Lines Notes

 Curved lines.
+Q and Q +Q -Q
 Directed from the
+ positive charge
-
to the negative
charge.

 point X is neutral
point (zero
+Q +Q
+Q and +Q + X +
electric force).
 It lies along the
vertical dash line.

 note that twice as

many lines leave
+2Q and Q Q +2Q +2Q as there are
- +
lines entering –Q
Definition The electric field F
strength at a point is E
defined as the qo
electric (electrostatic)
Where :
force per unit positive 
charge that acts at E : electric field strength
that point in the same F : electric force
direction as the force. q0 : test charge

Quantity Vector quantity

Unit N C1 OR V m 1
 ELECTRIC FIELD STRENGTH AND ELECTRIC FORCE
Since kQ Q
kQq0
F 2 E 2 OR E
r 4 0 r 2
r
Where:
then the equation of Q : magnitude of a point
electric field strength can charge
be written as… r : distance between a point
and point charge

Items Direction
depends on…

Electric field Sign of the point F
strength, E charge + E
Q q0 ( ve)
Electric Sign of the point
Force, F charge and the r
test charge
ELECTRIC FIELD STRENGTH AND ELECTRIC FORCE
Point Charge Test Charge Pattern

F

Positive + E
q0 (  ve)
Q
r
Positive
 
F E
Negative + q0 (  ve)
Q
r

F
Positive - E q0 ( ve )
Q
r
Negative
 
Negative - E F
q0 ( ve )
Q
r
 ELECTRIC FIELD STRENGTH AND DISTANCE

Notice that E is inversely

proportional to r2 kQ
E 2
r
1
E  2
r

The relationship between E E

and r can be shown in the
graph below.

r
 EXAMPLE 1.3 SOLUTION

Determine ;
(a) the electric field (a) From
Q
strength at a point Ek 2
r 6
X at a distance of (8  10 )
20 cm from a point  (9 10 )
9

0.2 2
charge Q= +8 μC.
E = 1.8  106 N C-1
(b) the electric force that (b) Knowing that ;
acts on a point charge F
q = -1 μC placed at E
q
point X.
F = Eq
= (1.8  10 6)(1  10 -6)
= 1.8 N towards Q
 EXAMPLE 1.4 SOLUTION
Point charges q1 and q2 of Point a;
12 nC and -12 nC, q1
Ea1  k
respectively are placed ra21
9
0.10 m apart. Compute the (12  10 C)
 9 109
total electric field at point ‘a’ (0.06) 2
and ‘b’.  3 104 NC 1
Eb
b q2
1
Ea 2 k 2
ra 2
Eb 9
(12  10 C)
13cm 2 13cm  9 109
(0.04) 2
Ea1
+q1 -q  6.8 108 NC 1
a Ea2 2
6cm 4cm 29
 EXAMPLE 1.4 SOLUTION
Point charges q1 and q2 of Point a;
12 nC and -12 nC, 1
Ea1  3 10 NC 4
respectively are placed
0.10 m apart. Compute the Ea 2  6.8 108 NC 1
total electric field at point ‘a’
 Eay  0
and ‘b’.
E ax  Ea1x  Ea 2 x
Eb  (3.0  10 4  6.8  108 )
b 1
 9.8 104 NC 1
Eb (direction +x, right)
13cm 2 13cm

Ea1
+q1 -q
a Ea2 2
6cm 4cm
 EXAMPLE 1.4 SOLUTION
Point charges q1 and q2 of Point b;
12 nC and -12 nC, q1
Eb1  k 2
respectively are placed rb1
9
0.10 m apart. Compute the (12  10 C)
 9 10 9
total electric field at point ‘a’ (0.13) 2
and ‘b’.  6.39 10 3 NC 1

Eb
b q2
1
Eb 2  k 2
rb 2
Eb (12 109 C )
13cm 2 13cm  9 10 9
(0.13) 2
Ea1  6.39 103 NC 1
+q1 -q
a Ea2 2
6cm 4cm 31
 EXAMPLE 1.4 SOLUTION
Point charges q1 and q2 of E x - comp y - comp
12 nC and -12 nC, Eb1 6.39 x 103 cos 6.39 x 103
67.38o sin 67.38o
respectively are placed
0.10 m apart. Compute the
=2.458 x 103 =5.898 x
N C-1 103 N C-1
total electric field at point ‘a’
and ‘b’.
Eb2 6.39 x 103 cos - 6.39 x 103
67.38o sin 67.38o
Eb
b 1
=2.458 x 103 5.898 x
= -

N C-1 103 N C-1

Eb
13cm 2 13cm E Ex = + 4.916 x Ey = O N C-1
103 N C-1
Ea1
+q1 -q
a Ea2 2
6cm 4cm
 FOLLOW UP EXERCISE

Two charges q1 = 12 x 10-9 C and q2 = -8 x 10-9 C with

distances 0.6 m and 0.4 m from point P respectively.
Find net electric field at point P.

Answer: E = 541 N C-1 with  = 33.69

 Learning Outcome :

1.3 Electric Potential, V

At the end of this chapter, students should be able to:

W
a) Define electric potential, V 
qo
b) Define and sketch equipotential lines and surfaces of
an isolated charge in a uniform electric field
kQ
c) Use V  for a point charge and a system of charge.
r
d) Calculate potential difference between two points.
ΔV = Vfinal – Vinitial
W
V 
qo
 Learning Outcome :

1.3 Electric Potential, V

At the end of this chapter, students should be able to:

e) Deduce the change in potential energy, U between

two points in electric field.
U  q V
f) Calculate potential energy of a system of point charges.
 q1q2 q1q3 q2 q3 
U  k    
 r12 r13 r23 
 ELECTRIC POTENTIAL, V

Electric potential, V is defined

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

Electric potential, V is: W

V 
 scalar quantity qo
 unit is J C-1 or Volt (V).
In terms of k, Q and r : Q
V k
r
 POTENTIAL DIFFERENCE, ∆V

WAB
Potential difference between two VBA 
q
points B & A is the work done
per unit charge to bring a where :
positive point charge from point ∆VBA = VB – VA
A to point B in an electric field. = Vfinal – Vinitial
 EXAMPLE 1.5 SOLUTION
(a)
Points A and B are at
Q
distances of 2.0 cm and 3.0 VA  k
cm respectively from a point rA
charge Q = -100 μC.   100 10 6 
 (9 10 )
9

Determine ;  0.02 
(a) the electric potentials  4.5 10 7 V
at A and B,
(b) the work required in Q
moving a point VB  k
rB
charge q = +2.0 μC,   100  10 6

from A to B.  (9 10 )
9

 0.03 
 3.0 10 7 V
 EXAMPLE 1.5 SOLUTION
(b) the work required is
Points A and B are at given by :
distances of 2.0 cm and 3.0
cm respectively from a point
WAB  qVBA
charge Q = -100 μC.
Determine ;  q(VB  VA)

(a) the electric potentials  (2  10 6 )[(3.0)  (4.5)]  10 7

at A and B,  30 J
(b) the work required in  work done is negative –
moving a point work done by the
charge q = +2.0 μC, external force or on the
from A to B. system.
 work done is positive–
work done by the electric
force (system).
 EXAMPLE 1.6 SOLUTION
A point charge q1 = + 5.0 μC V at P due to each charge
is at the origin and a point can be calculated from ;
charge q2 = - 2.0 μC is on
Q
the x-axis at (3,0) m as in V k
figure below. r
y (m) The total V is the scalar
P
(0,4) m sum of these two
potentials.

Q2 x (m)
Q1 (3,0) m

Find the total electric

potential due to these
charges at point P.
 EXAMPLE 1.6 SOLUTION
A point charge q1 = + 5.0 μC Q1
is at the origin and a point V1  k
r1
charge q2 = - 2.0 μC is on   5.0  10 6 
the x-axis at (3,0) m as in  (9  10 )
9


 4 .0 
figure below.
 1.125  10 4 V
y (m)
P
(0,4) m
Q2
V2  k
r2
  2.0  10 6

 (9 10 )
9

Q2 x (m)  5.0 
Q1 (3,0) m  0.36 10 4 V
Find the total electric
potential due to these
charges at point P.
 EXAMPLE 1.6 SOLUTION
A point charge q1 = + 5.0 μC Therefore,
is at the origin and a point Vp = V1 + V2
charge q2 = - 2.0 μC is on = (+ 1.125  104) +
the x-axis at (3,0) m as in (- 0.36  104)
figure below. = 7650 V
y (m)
P
(0,4) m

Q2 x (m)
Q1 (3,0) m

Find the total electric

potential due to these
charges at point P.
 FOLLOW UP EXERCISE

A particle with mass 0.002 kg and charge,q = 3 x 10-9 C,

starts from a rest point and moves in a straight line to
point B under the electric field. Find the potential at A
and B.

Answer: VA = 1800 V, VB = -1800 V

 ELECTRIC POTENTIAL ENERGY, U

The electric potential Wab  ΔU

energy of the system of
charges is the work done to  Ub  Ua
bring all the charges from
infinity to the points where
  U ab
the charges are placed.
U  W

U  qV

In the calculation of U , W and V, the sign of the charge must

be substituted in the related equations.
U depends on the position of the charged particle in the
electric field.
 ELECTRIC POTENTIAL ENERGY, U
A to B (lower to higher V) Difficult to moves

More work is require

B A
+ + Increase potential energy

B to A (higher to lower V) Easy to moves

Test charge naturally

B A moves
+ +
No work required
Decrease potential energy
 ELECTRIC POTENTIAL ENERGY, U
A to B (lower to higher V) Easy to moves

Test charge naturally

B A moves
- +
No work required
Decrease potential energy

B to A (higher to lower V)
Difficult to moves
B A More work is require
- +
Increase potential energy
 CHANGE IN ELECTRIC POTENTIAL ENERGY, ∆U

Va Vb

+ U  Ub - Ua
 ( Ua - Ub)
  Wab
 qVab
Charged particle moves
from a point Va to a point U  q(Va - Vb)
Vb, the change in
potential energy, ∆U is
 ELECTRIC POTENTIAL ENERGY OF TWO POINT
CHARGES

k q1 q 2
U
r

1
U is proportional to . where :
r
k – Coulomb constant
q1 – charge 1
q2 – charge 2
r – distance between q1 & q2

Include the sign of charges

 ELECTRIC POTENTIAL ENERGY OF A SYSTEM OF
POINT CHARGES

The potential energy

associated with the
charge q1 at point a is
the sum of the potential
energy of interaction for
each pair of charges.

 q1 q 2 q1 q 3 q 2 q3 
U  k  
 r12 r13 r 23 
( Algebraic sum with the sign of charge included )
 EXAMPLE 1.7 SOLUTION

 q1 q 2 q1 q3 q 2 q3 
U  k  
 r12 r13 r 23 
  4q - 2q  4q  5q  5q - 2q 
 k  
 d d d 
  8q 2 20q 2  10q 2 
 k   
 d d d 
Take distance d = 14.0   2q 2 
 k 
cm & charge q = 150  d 
nC.
  2(150 x10 ) 
9 2

What is the electric  8.99 x10 

9

 0.14 
potential energy of this
system of charges ?
U  2.9 x10 3 J
 EQUIPOTENTIAL LINES AND SURFACES

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

The potential difference VA

between any two points VB
on an equipotential
surface is zero. Hence, no
work is required to move The electric field at every
a charge at constant point on an equipotential
speed on an equipotential surface is perpendicular
surface. to the surface.
 ELECTRIC FIELD LINES & CROSS SECTIONS OF
EQUIPOTENTIAL SURFACES FOR

(a) A point charge

(b) A uniform field

(c) An electric dipole

 FOLLOW UP EXERCISE

Three point charges of +q, +2q and – 3q are arranged as

shown in the figure. Find the electric potential energy of
the system of three charges.

k q2
Answer:  7.59
a
 Learning Outcome :

1.4 Charge in a Uniform Electric Field

At the end of this chapter, students should be able to:

a) Explain quantitatively with the aid of a diagram the

motion of a charge in a uniform electric field.

b) Use E 
V for uniform E
d
i) stationary charge
ii)charge moving perpendiculary to the field
iii)charge moving parallel to the field
iv)charge in dynamics equilibrium
 CHARGE IN A UNIFORM ELECTRIC FIELD

Uniform electric field

---------------------

Electric field lines E

Lines equally + + + + + + + + + + + + ++ +
Straight line
spaced

Produced by two flat

Lines parallel to each other
parallel metal plates,
one is positive and one
is negative charges and
The direction of a uniform E is separated by a
is from the positive plate to distance as shown in the
the negative plate. figure above.
 ELECTRIC FORCE

If a charged particle (e.g : an    


electron) is held in between Fe
the two plates and then  
released, it will accelerates to E a-
one of the plates (in this
 
case, to the positive plate).

An electric force will be Fe  me a

exerted on the electron and
the magnitude of the electric
Where
force is given by :
a = acceleration of
the electron
 ELECTRIC FORCE

Fe can also be written as ; Fe  qE

By equating these two me a  qE

equations, we get ;  q 
a
m 
E
 e 
Where
me = mass of electron
= 9.11  10-31 kg

To find other kinematics v  u  at

quantities such as v, s and t, v 2  u 2  2as
use the equations of linear 1 2
(straight line ) motion. s  ut  at
2
 ELECTRIC FORCE

If the electron is being shot perpendicular (90°) to the

direction of E, it will be deflected to the positive plate in a
projectile motion and it will follows a parabolic path as
shown.
 ELECTRIC FORCE

Knowing that ;  q 
a 
m E
 e 

By substituting q = e (for eE
a
electron) in the equation me
above, we get ;

We can determine the velocity v x  u x  constant

in the x and y component. v y  u y  at

eE
In y-component, uy = 0, vy  at  t
therefore ; me
 ELECTRIC FORCE

The final position [coordinate s x  vi t

(x,y)] of the electron at a 1 2
certain time can be s y  u y t  at
2
determined from ;

Knowing that uy = 0 in y- 1
sy  at 2
component, therefore ; 2
1 eE 2
sy  t
2 me
RELATIONSHIP BETWEEN UNIFORM E & POTENTIAL
DIFFERENCE, ∆V

The relationship
between uniform
electric field produced
by two parallel plates
separated by a
distance r and potential dV
E  (Potential Gradient )
difference is given by : dr

From this relationship, Where :

the magnitude of E can r = separation distance
also be written as : between the two plates
V V = Potential difference
E 
r between 2 plates
 EXAMPLE 1.8 SOLUTION

A uniform electric field exists in (a) From

the space between two
identical parallel charged metal F = mea and F = eE
plates. The plates are 1.0 cm me a
apart. An electron is released E 
e
from rest at the negatively
charged plate. It arrives at the
From kinematics equation;
positively charged plate 2.0 ns
1
later. Determine ; s  ut  at 2
2
(a) the electric field strength
(b) the speed of the electron Given that u = 0;
1 2
when it arrives at the s  0  at
positively charged plate. 2
(given ; me = 9.11  10-31 kg) a 2
2s
t
 EXAMPLE 1.8 SOLUTION

A uniform electric field exists in (a)

the space between two
identical parallel charged metal Therefore;
plates. The plates are 1.0 cm me 2s
apart. An electron is released E
et 2
from rest at the negatively
charged plate. It arrives at the (9.11 10 3 1)(2  0.01)

positively charged plate 2.0 ns (1.6 10 1 9 )(2 10 9 ) 2
later. Determine ;  2.85 10 4 N C 1
(a) the electric field strength
(b) the speed of the electron
when it arrives at the
positively charged plate.
(given ; me = 9.11  10-31 kg)
 EXAMPLE 1.8 SOLUTION

A uniform electric field exists in (b) From v = u + at

the space between two
identical parallel charged metal
v  0  at
plates. The plates are 1.0 cm
apart. An electron is released  2s 
v  0   2 t
from rest at the negatively t 
charged plate. It arrives at the 2s
v
positively charged plate 2.0 ns t
later. Determine ; 2(0.01)
v
(a) the electric field strength 2 10 9
(b) the speed of the electron
when it arrives at the  110 7 m s 1
positively charged plate.
(given ; me = 9.11  10-31 kg)
 FOLLOW UP EXERCISE

FIGURE shows a section of the deflection system of a cathode

ray oscilloscope. An electron traveling at a speed 1.5 x 107 m s1
enters the space between two parallel plate 60 mm long. The
electric field between the plates is 4.0 x 103 V m1.
a) Sketch the path of the electron in between plates and
after emerging from the space between the plates.
b) Determine the magnitude and direction of the
acceleration of the electron in between the plates.
c) Calculate the velocity of electron when it emerges from
the space between the plates.
d) Calculate the angle of deflection of the electron beam.
[Ans: 7.03 × 1014 m s-2, 1.53 × 107 m s-1, 10.6 ]
 THE END…
Next Chapter…
CAPACITOR AND
DIELECTRICS

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