Download electrostatics - Alfa Tutorials

ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Electrostatics:
It is the branch of physics which deals with the study of electric charges
at rest. But charges being microparticles are physically never at rest except at
0ºK which is never possible. Thus for studying electrostatics we divide a body
into number of small fragments. Due to continuous motion, electrons are
entering a given part and also moving out of it. If the number of electrons
entering a given portion is equal to the number of electrons moving out of it,
then total charge in that portion is constant. In this case we assume that
charge is at rest. The study of its properties in this situation is called
electrostatics.
Electrodynamics :
It deals with the study of electric charges in motion, the magnetic
concepts are to be studied in electrodynamics .
Detection Of Charge
Charge can detected by simple apparatus called electroscope. Electroscope consists
of a metallic rod which is suspended vertically in a box. Two gold leaves are attached
at the lower end as shown in figure. When a charged conducting rod touches the
vertically suspended rod, some charge exchange takes place between both the rods.
Thus vertically rod as well gold leaves suspended at the lower end are charged and
diverges away from each other due to repulsion. The degree of divergence is the
measure of the magnitude of the charge.
Electroscope Experiment 1
Properties of Charge:
Conservation of Charge :
The total charge in an isolated system never changes. [By isolated
system we mean that no matter is allowed to cross the boundary of system].
No exceptions have ever been found to the hypothesis of charge conservation
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Like Charges Repel 1
Chapter: Electrostatics
The most fundamental property of electric charge is its existence in two
varieties i.e. positive and negative. What we call negative charge could just as
well be called positive and vice versa. The choice was based on Benjamin
Franklin experiment. The electricity acquired by ebonite when rubbed with
cat’s fur is arbitrarily termed negative electricity while that acquired by glass
rod when rubbed with flannel is termed as positive electricity (charge). The
sign of charge on any other body can then be decided by using fundamental
law i.e. like charges repel and unlike charges attract each other.
1
Unlike Charges Attract 1
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
and it is true at both macro as well as micro level. This concept is also true in
relativistic conditions. Charge conservation can also be stated as, “Charge can
never be created nor be destroyed, it can only be transferred from one body
to another.”
For eg. if glass rod is rubbed with silk, glass becomes positively charged
and silk becomes negatively charged. But the magnitude of charge on glass is
always equal to the magnitude of charge on silk.
[Will charge conservation holds if two persons are making the
measurements in two different reference frames?]
Quantization Of charge
According to principle of quantization ‘ the charge on the body cannot
have an arbitrary value, but its total charge is always an integral multiple
of the certain minimum charge  e .’ Thus if Q is charge on the body its
value can be
Q =  ne
Quantization implies that ‘n’ can’t be a rational number. For physical and
chemical properties of individual atoms quantization is used. But is ignored
in case of electrically charged bodies and current carrying conductors that
involve large scale transfer of electric charge. This is done b ecause the
magnitude of charge on electron is negligibly small as compared to
magnitude of charge on a macroscopic body.
Quarks: Quarks are subatomic particles of which proton and neutron[ all
hadrons] are made off. Neutron consists of 2 down quarks and 1 up quark
whereas proton consists of 2 up and 1 down quark.
2
2
1
e e e e
3
3
3
1
3
1
3
n [ d d u] =  e  e 
2
e0
3
Quarks don’t break the quantization principle although they have
fractional charges because they don’t have independent exist ence of their
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
p [ u u d] = 
2
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
own, they always exists in combination and the charge on the combination
is quantized.
[The very unlikely event of proton decay can lead us to stage
where this quantization may not hold true . We call it unlikely
because all HEP theories find the decay improbable. But theory
can always be replaced by a new theory. ]
Additive Nature of Charge :
The electric charge is additive in nature. It implies that total charge on
any system is the scalar addition of all the charges in it.
Relativistic Invariance of Charge:
By relativistic invariance of charge we mean that the charge is invariant
physical quantity and magnitude of charge on a body doesn’t vary with the
change in speed of the body i.e. if charge on electron is 1.6x10-19C it will
remain same whatever be the speed of the electron. This is different from
other physical quantities like mass, length or time. Einstein found that mass of
the body increases with the increase in velocity of the body. If m 0 is mass of
the body at rest, therefore its mass when it is moving with velocity v is
M=
m0
1
v2
c2
As specific charge on body is ratio of charge to mass i.e. q/m therefore we can
say that specific charge decreases with the increase in velocity of the mass
because ‘q’ remains constant with velocity and ‘m’ increases with the increase
in velocity.
[Will the magnitude of charge be same in different inertial frame? Also, will
the charge be same if the reference frame is non-inertial?]
This phenomenon can be observed in daily life. For eg. people using
woolen rugs or carpets sometimes experience a shock when after walking
briskly along rug they touch a metallic object. The metallic bodies of cars and
trucks also gets charged due to friction between them and air. This charge is
considerable and can even produce a shock or spark. That is why bodies of
petrol tankers have chain dragging along the ground. These chains will earth
the charge produced on the body of tanker. This phenomenon is also noticed
in the case of gramophone records. They gets covered with dust very easily
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Frictional Electricity :
3
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
because frictional electricity produce charge on record which attract dust
particles.
[Why this electrostatics experiment can be easily demonstrated by
a Jalandhar teacher as compared to his counterpart in Mumbai?
Why fans in houses accumulate more dust in Summers than in
Winters?]
Charging By Induction
It was found that whenever two bodies are rubbed electrons are transferred from
one body to the other, thus both the bodies gets charged. Another method of
charging the body without placing it in direct contact with any other charged body is
called charging by induction.
As shown in figure two neutral metal spheres are in contact, both supported
on insulating stands. When a negatively charged rod is brought near one of the
spheres but without touching it, as in [b], the free electrons from the metal are
repelled and drift slightly away from the rod, towards the right. This leaves a positive
charge on the left sphere. This negative charge on right sphere and positive charge
on left sphere are called induced charges. These induced charges remain on the
surfaces of the sphere as long as the rod is held nearby. [When the rod is removed ,
the electron cloud in right moves to the left and original neutral condition is
restored.] But suppose the electrons are separated slightly as shown in [c] while the
plastic rod is nearby. If the rod is now removed the opposite charges on two spheres
will attract each other. When the two spheres are separated by large distance each
of the two spheres will have uniform
Change in mass on electrification
[Try to find the change in mass whi ch takes place in the body for
every 1C of charge given to it?]
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Whenever two bodies are rubbed which results in electron transfer from one body to
the other. The body, which gains electrons, experiences an increase in mass and the
body, which looses electrons, experience a decrease in mass. The increase in mass of
one body is always equal to the decrease in mass of another body.
4
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Point Charges :
A charged particle whose size is very small in comparison with other
distances involved in the problem is called a point charge. This is a relative
concept and a large body like earth can also be considered as point object.
Coulomb's Law :
According to this law, the force of interaction between two point
charges is directly proportional to the product of the two charges and
inversely proportional to the square of the distance between them. If two
point charges be q1 and q2 and r is the distance between them, then,
F 
q1q2
r2
F  k
or
q1q2
r2
...(1)
where k is the constant of proportionality and depends on the medium
in which the charges are placed and the system of units selected.
If the charges are placed in vacuum, in SI, k 
1
4 0
, where 0 is the
absolute permittivity of vacuum. Therefore,
Fvac 
1
q1q2
...( 2)
4 0 r 2
From eq.(1),
0 
q1q2
4 Fr 2
Thus the unit of 0 will be,
ε0 
Coulomb  Coulomb
Newton  (meter)
2
 C 2 N 1m 2
and its value is found to be 8.854  10—12 C2 N—1 m—2
Therefore,
1
4 0
 9  10 9 Nm 2 C  2
Units of Charge :
In S.I., the unit of charge is Coulombs.
One Coulomb is that much charge which when placed in vacuum at a
distance of 1m from an equal and similar charge would repel it with a force of
9 x 109 Newton.
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
In vacuum and in cgs system k = 1.
5
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
In cgs system, the unit of charge is stat coulomb or e.s.u. (electrostatic
unit)
1 Coulomb = 3 x 109 stat coulomb
Another unit is e.m.u. and 1 e.m.u. of charge = 1/10 Coulomb.
[Coulombic force is said to be central force, is this a general
statement or there are some cases in Physics where this force is
non-central?
Try to find some limitations of Coulombic force? ]
We know even if the electrons are at rest th ey still spin about
their own axis, does that effect the force between the charges?
Explain]
Dielectric Constant or Relative Permittivity :
The force between two charges depend upon the medium between the two
charges. The force between two charges q1 and q2 located at a distance r in
some medium is,
Fm 
q1 q 2
4 m r 2
1
...(3)
where m is the absolute permittivity of that medium.
The equation (2) gives the force between charges in vacuum,
q1 q 2
4 0 r 2


 m   r ...(4)
1 q1 q 2
0
2
4 m r
1
Fv
Fm
This ratio m/0 is called relative permittivity or dielectric constant of the
medium. Also, m = 0r (From 4)
Equn. (3) gives
q1 q 2
4 0  m r 2
Fm 
1
Fvac
r
[We say the the force in medium is always K times less than the
force in vacuum, but this is true only if the whole space between
the two charges is filled with some medium, but will be the effect
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Fm 
6
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
if the distance between the two charges is partially filled with
dielectric?]
Vector Form of Coulomb’s Law


Consider two point charges q 1 and q 2 located P and Q with r1 and r2 be

the position vectors of two points. Let F21 be the force exerted by q 1 on q 2
,

F21  k
q1q2
r̂12
r122



where r̂21 is the unit vector along AB and r12  r2  r1 .

q1 q 2 
r12
r123
F21  k

qq
 
F21  k  1 2 3 r2  r1 
r2  r1
...(1)
Similarly, force exerted by q 2 on q 1 is,

q1 q 2 
r21
r213
F12  k



qq
F12  k  1 2
r1  r2

3
r1  r2 
...(2)

As (r2  r1 )   (r1  r2 )
From (1) and (2)


F21   F12
This is an experimental law and according to this principle if a
charge q is acted upon by the electric forces from charge q 1 , q 2 ,
......, q n , the total force on q is the vector sum of all the forces.
The mutual interaction of a pair of charges is unaffected by the
presence of other charges.
Consider a configuration of charges q 1 , q 2 , ............ , q n situated



with position vectors r1 , r2 , . . . . . . . . , rn . The force acting on q 1 due



to q 2 , q 3 , . . . . , q n is given by F12 , F13 , . . . . . . . , F1n respectively
such that,
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Principle Of Superposition
7
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES

 
qq
F12  k  1 2 3 r1  r2 
r1  r2

qq
F1n  k  1 2
r1  rn

3
r1  rn 


qq
 k  1 3
r1  r3
3
The net force on q 1 due to all the
charges will be,

F1  F12  F13        F1n
qq
 k  1 2
r1  r2
3
r1  r2 

Fi  k
1
2
r1  r3 
N

j 1
j i
qi q j
 
ri  r j
qq
   k  1 n
r1  rn
3
3
r1  rn 
r  r 
i
j
[ Principle of superposition is important law in physics, but is it
applicable for all distances or it breaks down in some domain?
Have you ever thaught of reason why the electron or any
microscopic lepton doesn’t split due to mutual repulsion ? ]
Differences between Charge and mass
5. Force between charges can be
attractive as well as repulsive.
Mass
1. mass on the body is always
positive
2. Mass is not a quantized physical
quantity
3. Mass of the body is not conserved
as conversion between mass and
energy is possible
4. Mass is not relativistically
invariant as it changes with speed
5. Gravitational force between
masses is always attractive.
Charge Distributions :
The concept of charge distributions is essential because we usually deal
with a vast number of elemental charges distributed in finite region of space.
The macroscopic effects of such a vast number of elementary charges can be
calculated by assuming them to be smeared out in the form of continuous
charge distribution. It is usual to define charge distributions in terms of
charge density function.
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Electric Charge
1. Charge exists in both positive and
negative values
2. Charge is a quantized physical
quantity
3. Charge is conserved on a system
only transfer of charge takes place
within isolated system
4. Charge is relativistically invariant
8
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
(1) Volume Charge Density :
It is the ratio of sum of all the charges q present inside the volume V
to the volume V
q
V
q   V
 
To define charge density at a point, we let V in the above equation to
shrink about the point at which  is to be evaluated and take V  0.
q
dq

V
dV
q    dV
  Lt
.
V 0
(2) Surface Charge Density :
This is the ratio of total charge on the surface element S to the area
S.
 
q
S
or
q    dS
(3) Linear Charge Density :
Whenever charge is distributed along a linear body then it is the ratio of
charge to its length.
q 
  dl
A field is a physical quantity that can be associated with position. For eg.
temperature of the air in the room has specific value at each point in the
room. If we let T represent the temperature, then there exists a function that
gives us the value of temperature at any point. This is an example of scalar
field.
Now electrostatics can also be easily dealt with if we introduce the
concept of field. According to field theory, the two charges should need not
be in contact with each other for electric interactions between them. The
charges are capable of directly influencing the other charges through
intervening medium. Thus electrostatic field can be defined as the region
around the charge in which electrostatic force due to this charge could be
experienced.
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
The Electric Field:
9
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Electric Field Intensity :
It is the force experience by unit positive charge when placed at that
point (point where intensity is to be determined). The direction of electric
field intensity is the direction in which unit positive charge will move if free to
do so. Mathematically, intensity E is the quotient obtained by dividing force
acting on the test charge to the magnitude of test charge,


F
E 
q
Units of Electric Field Intensity:
In SI the electric field intensity is measured in Newton/Coulomb and in
cgs system it is measured in dyne/stat coulomb.
One difficulty with above definition is that force exerted by test may
change the charge distribution of the body generating the electric field,
especially if body is a conductor in which charges are free to move.

Therefore, q’ is to be taken as small as possible. Hence mathematically E
can be written as,


E  Lt
q 0
F
q'
Electric Lines of Force:
These are the imaginary lines drawn on the paper tangent to which at
any point gives us the direction of electric field. The lines of force will be
straight if they are due to isolated charge and are generally curved for
system of two or more charges.
Properties of Electric Lines of Force:
2.
The number of lines of force per unit area is the measure of electric field
intensity. Larger the number of lines per unit area larger will be the electric
field.
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
1.
They always starts from positive charge and ends on negative
charge. For positive charge the electric lines of force are radially
outwards and for isolated negative charge they are radially inwards
10
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
3.
The electric lines of force never cross each other, otherwise at the point
of intersection of two lines we have two directions of intensity which is never
possible. As Shown in the figure if they intersect at point P, we can draw two
tangents A and B to the point P signifying two directions of electric lines of
force which is never possible.
Electric Lines of force for two positive charges and one positive and one negative
charge is as shown in the figure
Electric Field Due to a Monopole :
Consider a point charge Q situated at any point A in space, we have to
find electric field due to this charge at any point B such that AB = r. Place a
test charge q0 at B.
Then force of interaction,

F 
1
4 0
Qq0
rˆ
r2



F
1
Q
E  Lt.

rˆ
q 0
q0
4 0 r 2
0

Similarly, if we assume that charge Q is having position vector r1 and

electric field is to be determined at B with position vector r1 . The test charge
q0 is placed at B and force exerted by Q on q0.

qQ
F  k  0
r2  r1
3
r
2

 r1 

Q
E  k  
r2  r1
3
r

 r1 
and the electric field
2
Electric dipole is a system of two equal and opposite charges separated
by a certain fixed distance.
Electric Dipolemoment is defined as the product of either charge and

the distance between the two charges. It is denoted by p and is a vector
quantity with direction always pointing from negative to positive charge.
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Electric Dipole and Electric Dipolemoment :
11
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES

p  Q (2a)  2aQ
Electric field on the Axial Line of Dipole:
Axial line is the line passing through (joining) the two charges and
extended on both sides.
Let Q and Q be two charges with ‘2a’ as the distance between them.
We have to find the electric field at P which is at a distance of r from the
centre. The magnitude of electric field due to positive charge,

kQ
(r  a) 2
E1 
Magnitude of electric field due to negative charge,

E2 
kQ
(r  a) 2
Net Electric field =
Electric field On Axial Line 1


E 2  E1
 1
1 
kQ 

2
(r  a) 2 
 (r  a)
 4ar 
kQ  2
2 2 
 (r  a ) 




E

k
2 pr
(r 2  a 2 ) 2



E  k
2 pr
 a2 
r 4 1  2 
 r 
Neglecting a2/r2 , we get,


2p
E 
4 0 r 3
1
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
and as E2 is greater than E1 , the direction will be same as that of E2 .
Special Case :
If r >> a, then,
12
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Electric Field on Equitorial Line of Electric Dipole:
Equitorial line is the line passing through center and perpendicular
to the axial line. We have to find the electric field at a distance r from the
centre of dipole of length 2a.
Electric field due to positive charge +Q is,
Electric Field On Equatorial Line 1
kQ
(r 2  a 2 )
E1 
along AP
Electric field due to negative charge Q is,
E2 
kQ
(r  a 2 )
2
along PB.
Dividing into component form. The sine components being oppositely
directed cancel out and cosine components gets added up. Thus,
E = E1 cos + E2 cos
E = 2E1 cos

2kQ
a

2
r a
r 2  a2
2Q a
 k
(r 2  a 2 ) 3 / 2
p
 k
(r 2  a 2 ) 3 / 2
2
Special Case:
For r>>a,
E  k
p
 a2 
r 1  2 
r 

3/ 2
2
E 
1
p
4 0 r 3
Electric Field At Any Point Due to Dipole :
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
a2/r2 can be neglected, so
13
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Consider a dipole of dipole moment p, and electric field intensity is to be
calculated at a point k such that its position vector makes an angle θ with the
dipole moment. We make the components of dipole moment such that pcosθ
is along the position vector of point k and psinθ is perpendicular to it. Thus,
point k can be assumed to lying on the axial line of dipole with dipole moment
pcosθ and on equatorial line of dipole with dipole moment psinθ.
Electric field due to dipole of moment pcosθ is
1 2 p cos
E1 =
40
r3
The direction of field E1 is same as pcosθ, along KL
Similarly, the electric field intensity due to dipole of moment psinθ is
1 p sin 
E 2=
4 0 r 3
The direction of E2 is antiparallel to psinθ and is along KM. Thus as E1 and E2
are perpendicular, thus the resultant electric is given by
E= √𝐸12 + 𝐸22
E=
 2 p cos    p sin  

 

4 0  r 3   r 3 
p
E=
3 cos 2   1
4 0 r 3
2
1
2
The direction of resultant electric field is along the diagonal KN. It makes an
angle α with the line joining center of dipole with point k,such that
E
p cos 
Tan α = 2 
E1 2 p sin 
Tan α =
1
cot 
2
Consider an electric dipole placed in electric field making an angle q with
the electric field lines. The electric field is assumed to be uniform so field lined
are straight parallel and equidistant . The two charges will experience an
equal and opposite force therefore as the line of action is different we say the
system forms a couple and torque acts on it. The magnitude of torque is ,
 = either force   distance between line of action
= qE x 2a sin = 2 a  E sin
= pEsin



In vector form this can be written as   p  E
The direction of torque can be determined using right hand screw
rule. The torque is maximum when dipole is perpendicular to the electric
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Torque Acting on a Dipole:
14
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
field and zero if dipolemoment is either parallel or antiparallel to the electric
field.
Potential Energy of a Dipole :
Consider an electric dipole placed in uniform electric field, the work
done to rotate it through a small angle d is given by,
dW =  d
As
 = pE sin
dW = pE sin d
Total work done in rotating from initial position q 1 to the final position
q2 is given by,
W 
2
 pE sin  d
1
 pE  cos   2

1
=  pE ( cos2 - cos1 )
If whole of the work done is converted into potential energy then
potential energy,
U =  pE (cos2 - cos1)
Let the dipole be initially perpendicular to the electric field, then 1 =
90º and let 2 = 
 
U   pE cos   p . E
Special Cases :
1.
Potential energy is minimum if  = 0º ; In this case U =  pE and stability
is maximum.
2.
If  = 180º, U = pE, potential energy is maximum and stability is
minimum.
Electric Field Intensity on the Axis of Charged Circular Ring :
Consider a charged circular ring of negligible thickness and radius a . We
have to find electric field at a distance x from the centre of charged ring on
the axial line. Let the loop carry total charge of q. Consider a
small charge dq at the top of ring. Electric field due to this
charge at P will be,
dq
a  x2
2
Similarly, electric field due to charge dq at the bottom,
[by bottom we mean diametrically opposite point.
dE 2  k
dq
a  x2
2
As the magnitude of two electric field is same only direction is different,
we divide them into component form. Sine components being oppositely
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
dE1  k
15
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
directed cancel out and cosine components added up. Hence, resultant
electric field at P,
E 
 dE cos 
 k

dq
x
 k (a 2  x 2 ) (a 2  x 2 ) 1 / 2
qx
(a  x 2 ) 3 / 2
2
The direction of electric field is parallel to axial line and pointing away
from the centre of the ring.
Special Cases:
1.
When the point is at the centre of the loop. In this case the distance x
becomes zero, thus,
E = 0
2.
When the ring is very short as compared to the distance x i.e. x >> a or
x/a >> 1 or a/x << 1, then a2 can be neglected in comparison to x2
E  k
q
x2
Chapter: Electrostatics
Note: If a dipole is placed in non-uniform electric field, both force as well as
torque acts on it.
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
16
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Potential:
It is the term in electrostatics analogous to the temperature in heat and
thermodynamics. As temperature determines the direction of flow of heat,
potential determines the direction of charge flow. As heat always flow from
higher to lower temperature, charge also flows from higher to lower potential
(negative charge flows from lower to higher potential). The potential are
classified into three categories: positive, negative and zero potential.
Potential of a body is determined by assuming earth to be at zero
potential. If we connect any body to earth and positive charge flows from the
body to earth then the body is at higher potential than earth or body is at
positive potential. Similarly, if charge flows from the earth to body on
connecting body to earth then body is at lower potential than earth or body is
at negative potential.
Mathematically potential at a point is defined as the amount of work
done in bringing a unit positive charge from infinity to that point or potential
is the work done per unit charge to bring the charge from infinity to that
point,
V 
dW
dq
Units :
The units of electric potential are Joules / Coulomb (Common name is
volt) or ergs/statcoulomb (common name statvolt)
Relation between Volt and Stat Volt :
1 Volt 
1 Joule
107 ergs
1


statVolt
1 Coulomb
300
3  109 stat  Coulomb
Potential due to Monopole:
Consider a charge particle Q placed at a point A. We have to find the
potential at any point P which is at a distance r from Q. Now as potential is
the work done in bringing unit positive charge from infinity to that point.
Imagine a small displacement dx in the path of the charge. Work done to
move this small distance dx will be,
 
Force between charge Q and unit charge at distance x is,
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
dW  F . dx  F dx cos 1800   F dx
17
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
F  k
Q (1)
dW   k
x2
Q
x2
dx
Total work done in moving from infinity to r is,
W 

r

k
Q
Q
dx  k
2
r
x
Potential At a Point Due to An Electric Dipole
Consider an electric dipole consisting of charge q and –q kept at a
distance 2a from each other. The electric potential is to be calculated
at point P. Let AP = r 1 and BP = r 2 and angle POB =  where O is the
mid point of dipole. Draw BN  PO and AM  PO. On = OB cos  = a
cos  and OM = AO cos  = a cos .
Thus, r 1 = r + a cos  and r 2 = r – acos . And the potential at P is given
by
V=
q
1 q

4 0 r2
4 0 r1
V=
q q
  
4 0  r2 r1 
V=
q
q





4 0  r  a cos r  a cos 
V =
 r  a cos  r  a cos  


4 0 
r 2  a 2 cos 2 

V=
p cos 


 2

2
2
4 0  r  a cos  
1
1
1
1
But, if acos  << r, then neglecting a cos in comparison to r, we get
potential as
V=
1
4 0
p cos
r2
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
1
18
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Special Cases: [1] If point P lies on the axial line of dipole, then  = 0
V=
1
4 0
p
r2
[2] if point P lies on the equatorial line of dipole, then  = 90
1
V=
4 0
p cos 90
=0
r2
Potential Due to a Number of Charges:
Potential at a point due to number of charges is the sum of potential at
that point due to individual charges. For e.g. if we have number of charges
Q1, Q2, ......, Qn and we have to find the potential at P which is at a distance
of r1, r2, ......, rn from charges Q1, Q2, ......, Qn respectively. The potential at P
is,
V = V1 + V2 + V3 + - - - - - - - -+ Vn
Q Q
Q
Q 
 k  1  2  3   n 
r2
r3
rn 
 r1
The potential is a scalar quantity, so simple addition is required without
considering directions.
Potential Energy For A System Of Two Charges :
It is defined as the amount of work done in bringing the charges from
infinity to their respective positions in the absence of any other charge. For
eg. consider two charges q1 and q2 situated a distance ‘r’ apart. To find the
potential energy of the system of charges, first find the work done in bringing
q1 from infinity to A, this is equal to zero, as in absence of any other charge
force acting on it is zero. Similarly we find the work done in bringing q2 from
infinity to B in the presence of q1. We imagine a small path ‘dx’ in the path at
a distance ‘x’. Work done to move this small distance is,


dW = F . dx = F dx cos180º = - Fdx
where F  k
q1q2
x2
is the force acting on q2 at P.
To find the total work done, integrate from  to r,
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Potential Energy:
It is the work done in bringing number of charges from infinity to their
respective positions in the absence of any other charges.
19
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
W 
r
k

q1q 2
x2
r
qq
  1
dx   k q1q 2    k 1 2
r
 x 
This work done is equal to potential energy and hence
U  k
q1q2
r
Potential Energy for System of Three Charges:
Consider three charges to be situated as shown in the figure. Work done
in bringing q1 in absence of any other charge is zero. Similarly as in the above
article work done in bringing q2 in the presence of q1 is,
W1  k
q1q 2
r12
Work done in bringing q3 in presence of q1 and q2 is,
W2  k
q1q2
qq
k 1 2
r13
r23
Thus, total work done is,
W  W1  W 2
q q
qq
q q 
 k  1 2  1 3  2 3
r
r
r23 
13
 12
This total work done is the potential energy of the system. Hence for a
system of n-charges potential energy in generalised form is,
U  k

ij
i j
qi q j
rij
Potential Difference:
It is the difference of potential between any two points and is defined as
the amount of work done in moving a unit positive charge from one point to
another.
Consider two points A and B with potential VA and VB. The work done in
moving a charge particle from A to B is given by,
B 
A E . dl
  (VB  V A )
Electric field is Conservative in Nature:
The force or field is said to be conservative in nature if work done in
moving a particle in this field around any closed path is zero or work done in
moving from one point to another is independent of the path followed. For
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
W AB 
20
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
example, if unit positive charge is to be moved from A to B via path ACB, then
work done will be,
W AB 
B 
A E . dl
Work done in moving the particle back from B to A via some other path
ACB will be
W BA 
A 
B E . dl
Thus, total work done in moving a particle around any closed loop
BCBCA is
W AB  W BA 
B 
A 
B 
B 
A
B
A
A
 E . dl   E . dl 
 E . dl   E . dl  0
Hence as total work done is zero therefore, electric field must be
conservative in nature.
Electric Potential Difference and Electric Field Intensity:
Consider any path AB in non uniform electric field and P be any point on
this curve. We know that potential difference between A and B,
A  
V B  V A    E . dl
B
If the point A is moved to infinity, the potential at any point is given by,

r  
V (r )    E . dl

Let Q be another point situaed very close to P at a very small distance
‘dl’ so that the field between P and Q is practically the same. In that case
potential difference dV between these points is
 
dV   E . dl   E .dl cos    ET dl
where ET is the tangential component of electric field in direction of dl or
ET =  dV/dl and the term is known as potential gradient. Hence if
potential is constant in a certain region of space then electric field is zero.
:An equipotential surface is that o which potential everywhere on the
surface is same.
From the defination of electric potential surface
dV =
dW
q0
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Equipotential Surface
21
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Thus, if the potential difference between two points on the surface is zero
i.e. dV =0, it implies that no work is done in moving the charge between
the two points dW =0
Work done to move a charge between two points is also given by
 
dW = F.dx  q0 Edx cos =0
As q 0 , E and dx are not zero this implies that cos =0 or angle between
electric field intensity and displacement vector is zero. Thus, electric field
intensity is always perpendicular to the equipotential surface. Thus, no two
equipotential surfaces can intersect, because if they do intersect there will
be two direction of electric field intensity corresponding to two surfaces at
the same point which is never possible.
Note: [a] If the electric field intensity is uniform, then the equipotential
surfaces are planes with their surfaces perpendicular to the electric lines
of force.
[b] If we have isolated point charge positive or negative then the
equipotential surfaces are concentric spheres with their center coinciding
with the position of charge.
Solid Angle:
Just as we talk about the angles in two dimensions we talk about solid
angle in three dimensions. The solid angle is the measure of total opening of
cone around its vertex. We measure it by drawing a sphere centered at the
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
[c] equipotential surfaces are crowded in the region of strong electric
fields and are far apart in the weak fields.
22
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
vertex of cone ‘a’ of the sphere intercepted by cone and dividing it by the
square of distance
a

r
a'

2
r2
The complete solid angle corresponds to the case for which ‘a’ = surface
area of sphere i.e.
 
4r 2
r2
 4
Electric Flux:

Consider a closed surface S placed in uniform electric field E , and divide

this surface into infinitesimally small parts of dS each. (The surface area is a
vector and its direction is same as the direction perpendicular to the plane of
area)


Electric flux is thus defined as the sum of dot products of E and dS for
all elementary areas constituting the surface. It is denoted by  and
 
 
 E . dS
or
 


 E . dS
or
 
 E dS cos 
S
where  is the angle between electric field and the area vector.
Positive Electric Flux:
If the angle between E and dS is acute then flux is said to be positive or
if number of field lines leaving the surface are more than the number of
electric lines of force entering the given surface.
Negative Electric Flux :
If the number of electric lines of force entering the surface are more
than the field lines leaving the surface or if the angle between E and dS is
obtuse, the the flux is said to be negative flux.
Electric field intensity can be calculated from Coulomb’s law for point
charges only, but if we have some complex configuration of charges the field
intensity can be computed using Gauss Law. It states that
“ for any distribution of charges, the total electric flux linked with a
closed surface is
1
0
times the total charge within the surface ".
Mathematically,
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
GAUSS LAW:
23
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES


 E . dS


q
0
or

 E . dS

1
0
  dV
where the first equality applies if the surface enclosed discrete charges
and the second applies if the surface encloses continuous charge distribution.
Proof of Gauss’s Law:
To prove gauss law, consider a single point charge q enclosed in a closed
surface of arbitrary shape. For positive charge, the electric field is pointing

radially outwards. Imagine any infinitesimally small surface dS such that,


E . dS  E dS cos   k
But
dS cos 
r2
q
r2
dS cos 
 d
where d is the solid angle subtended at O by the surface area dS.


 E . dS
q
4 0

C

 d
C

 E . dS

C

where

q
( 4 )
4 0
q
0
represents closed integral
C
If there are number of charges q1, q2, …, qn then we can write gauss law




q1  q 2    q n  . In cgs system, Gauss Law can be stated
as  E . dS 
0
C
1
as  E . dS  4q .
C
Derivation of Coulomb’s Law from Gauss Law :
Consider a point charge q and we have to find the electric field at a
distance ‘r’ from charge. The gaussian surface is the spherical surface of
radius ‘r’ with centre on charge q. From symmetry electric field must have


 
 E . dS
E
 dS
C

q
0

 E dS cos 0

E  k

q
0
E (4r 2 ) 
q
0
q
r2
If a test charge q is located at point where electric field is determined,
then,
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
same value at all the points on the surface. Thus angle between E and dS is
zero,
24
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
F  q' E  k
q q'
r2
which is nothing but Coulomb’s law.
Selecting a Gaussian Surface:
In application of gauss’s laws to field calculations, some judgment is
required in choosing the surface. Two useful guiding principles are that the
point or points at which the field is to be determined must lie on the surface
and the surface must have enough symmetry so that it is possible to evaluate
the integral. Thus if a problem has spherical or cylindrical symmetry, the
gaussian surface will usually be spherical or cylindrical respectively.
Applications of Gauss Theorem :
1.
Electric Field Due to Infinitely Long Wire of Uniform Charge
Density () :
In order to find electric field due to wire at P, select cylindrical surface of
radius r and height  to be gaussian surface. The electric field lines are parallel
to upper and lower surface of cylinder and hence makes no contribution to
the electric flux. It is the curved surface which contributes to the electric flux.


 E . dS

C
 E dS
 E (2rl )
C
where r is the radius and l is the length of the cylinder.
E ( 2rl ) 
E 
q
0


2 0 r
E ( 2rl ) 
E 
i .e
l
0
1
r
2.
Electric Field Due to Infinite Sheet of Charge with Uniform
Charge Density (  ) :
 
q
0
2  E . dS 
2 ( ES ) 
E 
S
0

2 0
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Electric field is to be determined at P. To apply gauss theorem, let us
consider a cylinder to be a gaussian surface with ends on each side of sheet as
shown. Let S be the surface area of the two end surfaces. In this case field
lines are parallel to curved surface, hence their contribution to flux is zero.
Only end surfaces will contribute.
25
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Chapter: Electrostatics
Thus, we see that magnitude of electric field is independent of the
distance from the sheet.
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
26
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
3. Electric Field At Any Point Due to Two Charged Conducting
Plates :
Consider two parallel plane conductors P and Q given opposite charges
with 1 and 2 be the surface charge density for positive and negative
conductor respectively.
Electric field at A due to P and Q will be,
E1 
1 4


4 0 2 0
2 0
along AB
E2 

2 0
along AC
Thus, net electric field,
1   2
2 0
E 
If 1 = 2, then, E = 0
Point lying inside two conductors:
If point lies outside the two conductors then direction of electric field
due to both the conductors is same and net electric field is,
E
1

 2
2 0 2 0
If 1 = 2 , then E = /0.
4.
Electric field At Point Inside A uniformly Charged Sphere:
Let us consider a sphere in which charge is uniformly distributed. Let r
be the charge density. To find the electric field at any point inside the sphere
at a distance of r from the centre. The gaussian surface is thus a sphere of
radius r,
 
 E . dS 
ES 
C
q
0
Charge inside gaussian sphere = Charge per unit volume  volume
q 
4 3
r 
3
E (4r 2 ) 
4 r 3

3 0

E 
r
3 0
For point lying on the surface of the sphere,
E 
q
q
r
R
1


4
3 0
3 0
4 0 R 2
R 3
3
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
i.e
27
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
5.
Electric Field Due To Uniformly Charged Spherical Shell :
Consider a shell of radius R with charge uniformly distributed over its
surface.
Electric Field At Any Outside Point :
Consider any point lying at a distance r from the centre of the shell of
radius R such that ( r > R). The gaussian surface in this case is a spherical shell
of radius r. At all points on this sphere electric field is same.


 E . dS
 E ( S )  E (4r 2 )
C
According to Gauss Law,


 E . dS
C
q
0
E (4r 2 ) 

q
0

E 
kq
r2
Electric Field At A Point On the Surface :
In case the point P lies on the shell i.e. if r = R, the gaussian surface
is assumed to be sphere of radius R. Hence E = k q/r2
Electric Field At Point Inside The Shell :
If r < R, the electric field is zero because charge enclosed inside
gaussian surface is zero.
Relation Between Surface Density Of Charge & Radius of
Curvature :
Consider two spherical conductors with radius r1 and r2 and having
charges q1 and q2. If the two spheres are connected by wire the potential will
become same, i.e.
V1 = V2
k
q1
q
 k 2
r1
r2
If the electric field at the two surfaces be E1 and E2, then
E1

E2
k
q1
r12
q r2
r
 1 22  2
q1
r1
q2  r1
k 2
r2
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
q1
r
 1
q2
r2
28
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Similarly, the ratio of surface charge densities on the two spheres will
be,
q1
1
4r12
r

 2
q
2
r1
2
4r22
or surface density of charge is inversely proportional to the radius of
curvature or we can say that charge always tends to concentrate towards the
region with low value of radius of curvature or pointed ends.
Conductor in An Electric Field:
In any conductor or piece of metal electrons are always acts as charge
carriers and the number of electrons are always equal to the number of
positive ions in it. When conductor is placed in an electric field, the positive
charge carriers moves in the direction of the field and negative charge carriers
in direction opposite to the field. Therefore, there is an accumulation of
positive charges on one side and accumulation of negative charges on the
other. These charges are called induced charges and they generate an electric
field whose direction is opposite to the applied field. This electric field is
called induced electric field. The accumulation of charges will keep on
increasing unless applied and induced field totally balance each other. At this
point, net field inside conductor becomes zero, hence force (F = qE) acting on
the conductor is also reduced to zero. This whole process completes in a
fraction of a second. Hence, we assume, E = 0 inside a conductor.


Resultant Field  E  E 1


If E  E 1 then resultant field  0
If we go on adding charge to a given body its potential goes on
increasing i.e. charge on the body and potential are directly proportional to
each other.
Charge  Potential
or
Q = CV
where C is the constant of proportionality called the capacitance of the
body. We can also define capacitance as the ratio of charge on the body to its
potential. The capacitance of a capacitor depends upon:
(i) size and shape of the conductor and nature of medium surrounding the
conductor
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Capacitance:
29
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
(ii) it also depends on the position of charges present in the neighborhood.
It however is not dependent on the material of which conductor is made off.
Units of Capacitance:
The SI unit of capacitance is Farad and capacitance is said to be one
farad if
a charge of one coulomb is sufficient to raise the potential through one volt.
1 Farad 
1 Coulomb
1 Volt
The cgs unit of capacitance is statfarad and capacitance is said to be one
statfarad if a charge of one e.s.u. is sufficient to raise the body potential by
one statvolt.
1 Farad = 9 x 1011 stat farad
Capacity of Isolated Spherical Conductor::
Let us consider a charge spherical body of radius r insulated from other
charged bodies. If total charge on the body is q , then potential on the surface
of sphere
= k
q
r
Capacity,
C 
Q

V
q
 4 0 r
q
k
r
In cgs system,
C 
q
q

 r
q
V
r
or capacitance of a body is numerically equal to its radius.
A capacitor consists of two conductors separated by a certain distance
with insulating medium called dielectric in between. Its main function is to
increase the ability of body to take up charge. The basic principle is that the
capacity of an insulated charged conductor is increased appreciably by
bringing it near an earth connected uncharged conductor.
Consider a plate A having charge +Q and potential V when another
uncharged plate B is brought near this charged plate. Negative charge is
induced on the inner side of this plate and positive charge is induced on the
outer side of the plate B.
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Capacitor and Its Principle:
30
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
The negative charge tries to decrease the potential and positive charge
tends to increase it. On the whole there is net decrease in the potential of A
because negative charge is near to the plate as compared with the positive
charge.
On the other hand, if we connect the outer side of plate B to earth, the free
positive charge on the surface disappears thereby causing a further reduction
in potential. Hence as V goes on decreasing, to bring it back to the original
potential we have to add lot more charge to it and thus capacity of the system
increases further.
As C = Q/V, if V decreases capacitance increases.
Parallel Plate Capacitor:
Parallel plate capacitor consists of parallel plates of conducting material
seperated by certain fixed distance. The space in between the two plates
consists of some insulating material called dielectric.
Consider two such plates of area A with distance `d' between
them. Then the electric field between the two oppositely charged
plates will be,
E 

0
Chapter: Electrostatics
Imagine a point P in between the two plates, potential
difference between the two closely situated points around A is,
dV =  E dr
Potential difference between two plates,
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
31
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
V
0
V 
0
dV    E dr
d

0
V 
d
0 dr
V 

d
0

qd
0 A
and Capacitanc e
 A
q
 0
qd
d
0 A
C 
If instead of air there is some medium between the plates of capacitor,
then
C
 0 r A
d
where r is the relative permittivity of the medium.
Potential Energy of Capacitor :
Whenever a charge is added to the plates of capacitor, it increases its
potential. To add more charge to it, we have to do work against coulombic
repulsive force. This work done gets stored in the form of potential energy. If
at any instant charge on the plates of the capacitor is q, the work done to add
additional charge dq is given by,
dW  V dq 
q
dq
C
Total work done in increasing the charge from 0 to Q, we get,
W  dW 
Q
0
q
1
dq 
C
C
Q
0
q dq 
Q2
2C
If V is the final potential of capacitor, then,
Q  CV or W 
1
CV 2
2
Consider a parallel plate capacitor with distance between the plates of
capacitor ‘d’. To increase this distance from d to d + d, the work has to be
performed which is equivalent to the change in the potential energy,


Q2 Q2
Q2  1
1 
Q 2 d
dW 




 
0 A 
2C ' 2C
2  0 A
2 0 A
d 
 d  d
Also work done is given by,
dW = F d
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Force between Two Parallel Plates:
32
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
Equating the two values,
F d 
E 
Q 2 d
2 0 A

Q
A 0

Q2
2 0 A
...( i )
Q   0 AE
...( ii )
F 
Substitute (ii) in (i),
F 
( 0 AE ) 2
1
  0 AE 2
2 0 A
2
Grouping of Capacitors:
Series of Grouping:
Capacitors are said to be connected in series if second plate of first
capacitor is connected with first plate of second capacitor and so on. The
charge on the plates of all the capacitors is same, but potential difference will
be different across different capacitors such that,
V = V1 + V2 + . . . . . . . . .+Vn
Q
Q
Q
Q


 ........
CS
C1 C 2
Cn
1
1
1
1


 ........
CS
C1 C 2
Cn
For two capacitors, C1 and C2,
CS 
C1C 2
C1  C 2
Capacitors are said to connected in parallel if positive plate of
all the capacitors is connected to one point and negative plate to the other
point. The potential difference across all the capacitors is same but the charge
on the plates of capacitors is different, i.e.,
Q = Q 1 + Q2 + . . . . . . . . . + Q n
Cp V = C1 V + C2 V + . . . . . . . . . + Cn V
Cp = C 1 + C 2 + . . . . . . . . . . + C n
i.e. the net capacitance is the sum of individual capacitance of all the
capacitors.
Capacity of a Spherical Condenser :
Consider any two spherical shells of radius r1 and r2. The inner sphere is
given a charge q and outer sphere is earthed. If inner sphere is given a
positive charge q, there will be negative charge on the inner side of outer
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Parallel Grouping:
33
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
shell. If P be any point lying between two shells, then dV is the potential
difference between two points situated a distance dr apart around P, then
dV =  Edr where E = kq/r2
Potential difference V between A and B is,
V   kq
r1
r
r 2 dr
2
1 1
kq ( r2  r1 )
V  kq    
r1 r2
 r1 r2 
r1 r2
1
C 
k ( r2  r1 )
When inner sphere is earthed,
If instead of earthing outer sphere, we give a charge q to outer sphere
and earth the inner sphere. If the charge induced on the inner sphere be q1,
hence a charge +q1 will be present on inner surface of outer sphere, while +q2
is distributed over outer surface.
q = q1 + q2
Thus two condensers are formed (i) between spheres A and B having
capacity C1 given by 40 ab/(b-a) (ii) between outer sphere and earth having
capacity 40b.
Net Capacity,
4 0 b 2
 ab

C '  4 0 
 b 
ba
b  a

Capacitance of Parallel Plate Capacitor with Dielectric Between
Its Plates:
Polar and Non Polar Molecules :
Polar molecules are those which are formed by the combination of two
atoms having different electronegativities or we can say that molecules in
which the centre of gravity of positive and negative charge do not coincide. As
on the whole molecule is to be neutral therefore the magnitude of positive
charge is equal to the magnitude of negative charge. Thus this system
resembles a dipole and possesses a dipolemoment called its natural
dipolemoment. For eg. HCl
Non Polar Molecules are formed by joining atoms having same
electronegativity. In these molecules centre of gravity of positive and negative
charge coincide. Thus they do not possess dipolemoment of their own. But
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
Before finding the capacitance with dielectric between its plates we
must know the behaviour of dielectric in the presence of electric field.
34
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES
when non polar molecule is placed in an electric field, positive and negative
charges experiences force in two opposite direction. Thus, molecules now
also resembles a dipole and have dipolemoment which is called their induced
dipolemoment. At some stage the electric force pulling the charges apart and
the electrostatic attractive force balance each other and molecules is said to
be polarised.
Now, if a non polar dielectric slab is placed in an electric field, the atoms

get polarised in the direction of E0 . If q is the charge induced in any atom
with d be distance between the two charges then total induced
dipolemoment
will
be


p  ( Nqd ) V or P  Nqd
where

p
is
the

dipolemoment and P is the dipolemoment per unit volume called electric
polarisation. As field acts on a dielectric a layer of positive charge is formed
on the one side and a layer of negative charge on the other, this positive and

negative layer generates an induced electric field E i . Thus net field inside the
dielectric is,



E'  E  E i
Also,

Ei

electric dipolemome nt density
P


0
0
Also the ratio of applied electric field to reduced electric field
is called the dielectric constant of the medium. The polarisation is also
found to be proportional to E or







E'  E  E i  E   0  E
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
Chapter: Electrostatics
P  0  E
35
ELECTROSTATICS 2016
ALFA PHYSICS CLASSES





E'  E   E'
E  E ' (1   )

E

 1 
or
  1 
E'
where K is called the dielectric constant of the material.
Capacitor with Dielectric Slab:
Consider a parallel plate capacitor with plate area A and distance
between capacitor plates d. Its capacitance is given by,
C
0 A
d
Also the potential difference between the plates is given by,
V = Ed where E is the electric field between capacitor plates. Now if we
introduce a dielectric slab of thickness t between the plates of capacitor, the
potential difference V between plates
V = E (d  t) + (E  Ei) t
C' 
q
q

V'
E ( d  t )  ( E  E i )t
E ( d  t )  ( E  E i )t
C
q /V


C'
q /V '
Ed
C'
Ed

C
E ( d  t )  ( E  E i )t
C' 
C' 
 0 AE
E ( d  t )  ( E  E i )t
0 A
0 A

t
 E  Ei 
(d  t ) 
(d  t )  
t
k
 E 
If instead of dielectric conducting plate is present between the plates of
capacitor, then E  Ei = 0 or
0 A
d t
for conductors.
Chapter: Electrostatics
C' 
42, SUS Nagar, Jalandhar, 98152- 15362 http://www.alfatutorials.in/
36