Download f. Physics notes 2 (DOC).

. Gauss's
Law
We already know about electric field lines and electric flux. Electricflux through a closed surface S is
which is the number of field lines passing through surface S.
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Statement of Gauss's Law
“Electric flux through any surface enclosing charge is equal to q/ε0 , where q is the net charge enclosed by
the surface”
mathematically,
where qenc is the net charge enclosed by the surface and E is the total electric field at each point on the
surface under consideration.
It is the net charge enclosed in the surface that matters in Gauss's law but the total flux of electric field E
depends also on the surface chosen not merely on the charge enclosed.
So if you have information about distribution of electric charge inside the surface you can find
electric flux through that surface using Gauss's Law.
Again if you have information regarding electric flux through any closed surface then total charge enclosed
by that surface can also be easily calculated using Gauss's Law.
Surface on which Gauss's Law is applied is known as Gaussian surface which need not be a real surface.
Gaussian surface can be an imaginary geometrical surface which might be empty space or it could be
partially or fully embedded in a solid body.
Again consider equation 11
In left hand side of above equation E·da is scalar product of two vectors namely electric field vector E and
area vector da. Area vector da is defined as the vector of magnitude |da| whose direction is that of outward
normal to area element da. So, da=nˆda where nˆ is unit vector along outward normal to da.
From above discussion we can conclude that,
(1) If both E and surface area da at each points are perpendicular to each other and has same magnitude at
all points of the surface then vector E has same direction as that of area vector as shown below in the
figure.
since E is perpendicular to the surface
(2) If E is parallel to the surface as shown below in the figure
E at all points on the surface.
APPLICATIONS OF GAUSS'S LAW
* Derivation of Coulomb’s Law.
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Coulomb’s law can be derived from Gauss's law.
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Consider electric field of a single isolated positive charge of magnitude q as shown below in the figure.
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Field of a positive charge is in radially outward direction everywhere and magnitude of electric field
intensity is same for all points at a distance r from the charge.
We can assume Gaussian surface to be a sphere of radius r enclosing the charge q.
From Gauss's law
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
since E is constant at all points on the surface therefore,
surface area of the sphere is A=4πr2
thus,
Now force acting on point charge q' at distance r from point charge q is
This is mathematical statement of Coulomb’s law.
* Electric field due to line charge
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Consider a long thin uniformly charged wire and we have to find the electric field intensity due to the wire
at any point at perpendicular distance from the wire.
If the wire is very long and we are at point far away from both its ends then field lines outside the wire are
radial and would lie on a plane perpendicular to the wire.
Electric field intensity has same magnitude at all points which are at same distance from the line charge.

We can assume Gaussian surface to be a right circular cylinder of radius r and length l with its ends
perpendicular to the wire as shown below in the figure.

λ is the charge per unit length on the wire. Direction of E is perpandicular to the wire and components
of E normal to end faces of cylinder makes no contribution to electric flux. Thus from Gauss's law

Now consider left hand side of Gauss's law
since at all points on the curved surface E is constant. Surface area of cylinder of radius r and length l is
A=2πrl therefore,

Charge enclosed in cylinder is q=linear charge density x length l of cylinder,
or, q=λl
From Gauss's law
Thus electric field intensity of a long positively charged wire does not depends on length of the wire but on
the radial distance r of points from the wire.
Electric field due to charged solid sphere
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We'll now apply Gauss's law to find the field outside uniformly charged solid sphere of radius R and total
charge q.
In this case Gaussian surface would be a sphere of radius r>R concentric with the charged solid sphere
shown below in the figure. From Gauss's law


where q is the charge enclosed.
Charge is distributed uniformly over the surface of the sphere. Symmetry allows us to extract E out of
the integral sign as magnitude of electric field intensity is same for all points at distance r>R.
Since electric field points radially outwards we have
also as discussed magnitude of E is constant over Gaussian surface so,
where 4πr2 is the surface area of the sphere.
Again from Gauss's law we have
Thus we see that magnitude of field outside the sphere is exactly the same as it would have been as if all
the charge were concentrated at its center.
* Electric field due to an infinite plane sheet of charge
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Consider a thin infinite plane sheet of charge having surface charge density σ (charge per unit area).
We have to find the electric field intensity due to this sheet at any point which is distance r away from the
sheet.
We can draw a rectangular Gaussian pillbox extending equal distance above and below the plane as shown
below in the figure.
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By symmetry we find that Eon either side of sheet must be perpandicular to the plane of the sheet, having
same magnitude at all points equidistant from the sheet.
No field lines crosses the side walls of the Gaussian pillbox i.e., component of E normal to walls of pillbox
is zero.
We now apply Gauss's law to this surface
in this case charge enclosed is
q=σA
where A is the area of end face of Gaussian pillbox.
E points in the direction away from the plane i.e., E points upwards for points above the plane
and downwards for points below the plane. Thus for top and bottom surfaces,
thus
2A|E|=σA/ε0
or,
|E|=σ/2ε0
Here one important thing to note is that magnitude of electric field at any point is independent of the sheet
and does not decrease inversely with the square of the distance. Thus electric field due to an infinite plane
sheet of charge does not falls of at all.
……………………………………………………………….
Electric Potential and Electric Field
Consider a charge
placed in an electric field generated by fixed charges. Let us chose some arbitrary reference
point
in the field. At this point, the electric potential energy of the charge is defined to be zero.
This uniquely specifies the electric potential energy of the charge at every other point in the field. For instance, the
electric potential energy
to
at some point
along any path. As, we can calculate
depends both on the particular charge
is simply the work
done in moving the charge from
. It is clear, from (from previous knowledge of
, that
which we place in the field, and the magnitude and direction of the electric
field along the chosen route between points
and
. However, it is also clear that
is directly
proportional to the magnitude of the charge
is
. Thus, if the electric potential energy of a charge
then the electric potential energy of a charge
at the same point is
at point
. We can exploit this fact to
define a quantity known as the electric potential. The difference in electric potential between two points
and
in an electric field is simply the work done in moving some charge between the two points divided by the
magnitude of the charge. Thus,
(80)
where
denotes the electric potential at point
potential between points
and
, etc. This definition uniquely defines the difference in electric
, but the absolute value of the potential at point
therefore, without loss of generality, set the potential at point
of a charge
at some point
potential
at that point:
remains arbitrary. We can
equal to zero. It follows that the potential energy
is simply the product of the magnitude of the charge and the electric
(81)
The electric potential at point
(relative to point
) is solely a property of the electric field, and is, therefore,
the same for any charge placed at that point. We shall see exactly how the electric potential is related to the electric
field later on.
The dimensions of electric potential are work (or energy) per unit charge. The units of electric potential are,
therefore, joules per coulomb (
). A joule per coulomb is usually referred to as a volt (V): i.e.,
(82)
Thus, the alternative (and more conventional) units of electric potential are volts. The difference in electric potential
between two points in an electric field is usually referred to as the potential difference, or even the difference in
``voltage,'' between the two points.
A battery is a convenient tool for generating a difference in electric potential between two points in space. For
instance, a twelve volt (12V) battery generates an electric field, usually via some chemical process, which is such
that the potential difference
between its positive and negative terminals is twelve volts. This means
that in order to move a positive charge of 1 coulomb from the negative to the positive terminal of the battery we
must do 12 joules of work against the electric field. (This is true irrespective of the route taken between the two
terminals). This implies that the electric field must be directed predominately from the positive to the negative
terminal.
More generally, in order to move a charge
work
through a potential difference
we must do
, and the electric potential energy of the charge increases by an amount
the process. Thus, if we move an electron, for which
minus 1 volt then we must do
an electron volt (eV): i.e.,
in
C, through a potential difference of
joules of work. This amount of work (or energy) is called
(83)
The electron volt is a convenient measure of energy in atomic physics. For instance, the energy
required to break up a hydrogen atom into a free electron and a free proton is
eV.
We have seen that the difference in electric potential between two arbitrary points in space is a function of the
electric field which permeates space, but is independent of the test charge used to measure this difference. Let us
investigate the relationship between electric potential and the electric field.
Consider a charge
which is slowly moved an infinitesimal distance
along the
difference in electric potential between the final and initial positions of the charge is
change
in the charge's electric potential energy is given by
-axis. Suppose that the
. By definition, the
(84)
From Eq. (76), the work
which we perform in moving the charge is
(85)
where
is the local electric field-strength, and
the
-axis. By definition,
Energy conservation demands that
on the charge), or
is the angle subtended between the direction of the field and
, where
is the
-component of the local electric field.
(i.e., the increase in the charge's energy matches the work done
(86)
which reduces to
(87)
We call the quantity
the gradient of the electric potential in the
-direction. It basically measures how
fast the potential
varies as the coordinate
is changed (but the coordinates
and
are held constant).
Thus, the above formula is saying that the
-component of the electric field at a given point in space is equal
to minus the local gradient of the electric potential in the
-direction.
According to Eq. (87), electric field strength has dimensions of potential difference over length. It follows that the
units of electric field are volts per meter (
per coulomb: i.e.,
. Of course, these new units are entirely equivalent to newtons
(88)
Consider the special case of a uniform
-directed electric field
generated by two uniformly charged parallel
planes normal to the
-axis. It is clear, from Eq. (87), that if
is to be constant between the plates then
must vary linearly with
in this region. In fact, it is easily shown that
(89)
where
is an arbitrary constant. According to Eq. (89), the electric potential
decreases continuously as we
move along the direction of the electric field. Since a positive charge is accelerated in this direction, we conclude
that positive charges are accelerated down gradients in the electric potential, in much the same manner as masses fall
down gradients of gravitational potential (which is, of course, proportional to height). Likewise, negative charges are
accelerated up gradients in the electric potential.
According to Eq. (87), the
-component of the electric field is equal to minus the gradient of the electric potential
in the
-direction. Since there is nothing special about the
-direction, analogous rules must exist for the
and -components of the field. These three rules can be combined to give
-
(90)
Here, the
derivative is taken at constant
and
, etc. The above expression shows how the electric
field
, which is a vector field, is related to the electric potential
, which is a scalar field.
We have seen that electric fields are superposable. That is, the electric field generated by a set of charges distributed
in space is simply the vector sum of the electric fields generated by each charge taken separately. Well, if electric
fields are superposable, it follows from Eq. (90) that electric potentials must also be superposable. Thus, the electric
potential generated by a set of charges distributed in space is just the scalar sum of the potentials generated by each
charge taken in isolation. Clearly, it is far easier to determine the potential generated by a set of charges than it is to
determine the electric field, since we can sum the potentials generated by the individual charges algebraically, and
do not have to worry about their directions (since they have no directions).
Equation (90) looks rather forbidding. Fortunately, however, it is possible to rewrite this equation in a more
appealing form. Consider two neighboring points
displacement of point
relative to point
. Let
points. Suppose that we travel from
along the
move from
along the
-axis, and finally moving
to
to
and
. Suppose that
is the vector
be the difference in electric potential between these two
by first moving a distance
along the
as we move along the
-axis, then moving
-axis. The net increase in the electric potential
is simply the sum of the increases
-axis, and
along the
as we move along the
-axis,
as we
as we move
-axis:
(91)
But, according to Eq. (90),
, etc. So, we obtain
(92)
which is equivalent to
(93)
where
is the angle subtended between the vector
and the local electric field
. Note that
attains its
most negative value when
. In other words, the direction of the electric field at point
corresponds to the
direction in which the electric potential
decreases most rapidly. A positive charge placed at point
is
accelerated in this direction. Likewise, a negative charge placed at
is accelerated in the direction in which the
potential increases most rapidly (i.e.,
). Suppose that we move from point
to a neighboring
point
in a direction perpendicular to that of the local electric field (i.e.,
). In this case, it follows
from Eq. (93) that the points
and
lie at the same electric potential (i.e.,
). The locus of all the
points in the vicinity of point
which lie at the same potential as
is a plane perpendicular to the direction of
the local electric field. More generally, the surfaces of constant electric potential, the so-called equipotential
surfaces, exist as a set of non-interlocking surfaces which are everywhere perpendicular to the direction of the
electric field. Figure 14 shows the equipotential surfaces (dashed lines) and electric field-lines (solid lines) generated
by a positive point charge. In this case, the equipotential surfaces are spheres centred on the charge.
Figure 14: The equipotential surfaces (dashed lines) and the electric field-lines (solid lines) of a positive point
charge.
In Sect. 4.3, we found that the electric field immediately above the surface of a conductor is directed perpendicular
to that surface. Thus, it is clear that the surface of a conductor must correspond to an equipotential surface. In fact,
since there is no electric field inside a conductor (and, hence, no gradient in the electric potential), it follows that the
whole conductor (i.e., both the surface and the interior) lies at the same electric potential.
………………………………………………………………………………
MULTIPOLE EXPANSION
A multipole expansion is a series expansion of the effect produced by a given system in terms of an expansion
parameter which becomes small as the distance away from the system increases. Therefore, the leading one or terms
in a multipole expansion are generally the strongest. The first-order behavior of the system at large distances can
therefore be obtained from the first terms of this series, which is generally much easier to compute than the general
solution. Multipole expansions are most commonly used in problems involving the gravitational field of mass
aggregations, the electric and magnetic fields of charge and current distributions, and the propagation of
electromagnetic waves.
Electric Multipole Expansion
For a given charge distribution, we can write down a multipole expansion, which gives the potential as a series in
powers of
, where is the distance from the origin to the observation point.
We know that the potential in general is
In the integral,
where
is the position of charge element
is the angle between
and
. From the law of cosines
. We can rewrite this as
From the theory of Legendre polynomials, it is known that the last factor in this expression is a generating
function for the polynomials. That is, if we write the square root as an power series, we get
The coefficient of
in the series is the Legendre polynomial
. This can be verified for the first few
terms by calculating the Taylor series expansion of the square root term about
. This is tedious to do by
hand, but using Maple, we get, defining
:
It is important to note that the angle
is equivalent to the angle in spherical coordinates only if the observation
point lies on the axis, since that is the only configuration where the angle between the observation vector and a
charge element corresponds to the spherical coordinate angle . (A more general multipole expansion uses spherical
harmonics rather than just Legendre polynomials, but that’s a topic for a more advanced post.)
With this restriction, we can substitute the series expansion back into 1 to get
The first few terms in this series have special names. The
term is
where
is the total charge. This is called the monopole term, and shows that to a first approximation, the potential
of any charge distribution is just the potential of a point charge with the same total charge.
The next term in the series is
This is called the dipole term.
For
Finally, for
, we get the quadrupole term
we get the octopole term
As an example, consider a solid sphere with a charge density
We can use the integrals above to find the first non-zero term in the series, and thus get an approximation for the
potential. Note that we can do this only for points on the z axis.
By direct calculation, we have for the monopole term:
since the integral over
gives zero. Thus the monopole term vanishes, as it always does if the total charge is zero.
For the dipole term, we get
This time, the integral over
For the quadrupole term
gives zero, since the term
is odd relative to the interval
.
The octopole term comes out to zero, since the terms in
are again odd relative to the interval
.
……………………………………………………………………………….
Work and energy – point charges
Since charges exert forces on each other through their electric fields, it will require the expenditure of energy, or
work, to assemble any configuration of charges. Here we’ll have a look at how much energy is required to assemble,
and thus how much energy is stored, in a collection of discrete charges.
The force on a charge due to an electric field
is
. From elementary physics, we know that the work done
when an object is moved against a force is the negative (since we’re opposing the force) of (force) times (distance).
In general, if the force varies as a function of position, we get
where the integral is taken over the path through which the object is moved. The minus sign is an indication that we
are opposing the force ; if we work instead with the force that we must exert to move the object, then the minus
sign is omitted.
For the electric field, then, we get
To get the last line, we’ve used the fact that, in electrostatics, the line integral of the electric field is independent of
the path; it depends only on the endpoints and . We’ve seen earlier that the line integral of the field is the
negative of the potential difference between the two endpoints.
So, in other words, the potential difference between two points is the work per unit charge required to move a charge
between those two points. If we’ve set the reference point for the potential at infinity (that is,
at infinity),
then the work required to bring in a charge from infinity to a point is
We can apply this formula to find out how much energy is required to assemble a collection of point charges. To
place a single charge
at a location
takes no work, since there are no fields to work against. Bringing in a
second charge
requires working against the field due to . The potential due to
is
, so if we want to place
at position
the work required is
Before we go any further, it’s worth noting that this formula gives rise to a bit of a problem. What if we want to
assemble a point charge itself? That is, suppose we want to build up a point charge of a certain size by bringing
together other point charges and, in effect, gluing them together. This seems to be a valid procedure, since after all,
if a charge is truly a mathematical point, we should be able to pile as many of these point charges on top of each
other as we like without increasing the volume (that is, zero) occupied by the sum of all the charges.
However, if we try that, the above formula says this will require an infinite amount of work (since
). This
is, in fact, a recognized problem in electrodynamics, and the problems don’t go away even in the quantum
mechanical theory. In fact, we can’t even get out of the problem by saying that there is no such thing as a point
charge, since a lot of physicists think that the electron might actually be a point charge (at least its diameter, if it’s
non-zero, is so small that nobody has actually measured it yet).
With that caution in mind, let’s ignore the problem and carry on. If we assume that the existence of point charges is
possible (without taxing our minds as to how they are built), we can continue to add more point charges to our
distribution. Adding a third charge
at location
requires work
The total work required to assemble all three charges is then
The general pattern should be fairly obvious by now. To assemble
charges, the total work is
If we extend the second sum to cover 1 through
), we get
where
(excluding
is the potential due to all the charges in the collection except
.
As an example, suppose we have arranged two charges of
at the ends of a diagonal in a square, and a charge
of
on one of the other two corners of the square. How much work is required to bring in another charge of
from infinity and place it at the remaining corner of the square?
To work this out, we need to find the potential at this corner due to the existing three charges. If the length of each
side of the square is , then we get
Once all four charges have been assembled, the total energy stored in the collection is
In the second line, the potential at the location of one of the
charges is
. Multiplying this
by
gives the first two terms in the square brackets. Similar logic for one of the
charges gives the last two
terms in the brackets. The factor of 2 in the numerator arises from the fact that there are two each of
and
charges.
Work and energy – continuous charge
For discrete point charges, the work required to assemble a collection of
where
is the potential at the location of charge
For a continuous charge density
charges
is
due to all the other charges (excluding
).
we can write this as an integral over the volume containing the charges:
Note, however, that there is a subtle distinction between the discrete and continuous formulas. In the discrete
formula, the potential term in the sum excludes the charge , but in the integral form, the potential is the complete
potential due to the entire charge distribution. In the continuous case, we don’t talk about point charges (unless we
write the density as a sum of delta functions), so in that sense the continuous formula is more accurate. However,
if point charges such as electrons do truly exist, then we can’t either build them or take them apart, so it seems fair
enough to exclude the energies involved in doing so. However, as we’ve seen in the post on discrete charges, the
energy associated with a point charge is actually infinite, so it’s dodgy to just ignore it. This problem plagues both
classical and quantum electrodynamics, but most books just ignore it.
The integral formula can be expressed in terms of the electric field by using a bit of vector calculus. We know
from Gauss’s law for electrostatics that
so we can write the work as
A theorem from vector calculus says
In the last line, we used the relation
.
Therefore, the work becomes
We can now use the divergence theorem on the first term, and convert it from a volume integral to a surface integral
if we select some surface that encloses all the charge (we’re assuming that we’re dealing with realistic systems so
that all the charge is at a finite distance, and not with things like infinite planes of charge). That is, we can say
Since all we require is that the surface encloses all the charges, we can let the surface tend to infinity. In that case,
since the charges are all at finite distances, and we are taking the potential to be zero at infinity, and the electric field
falls off as
, the surface integral will go to zero at infinity. The volume integral is always positive (since we’re
integrating the square of the field, which is always positive), so what happens is that as we include more volume, the
surface integral decreases and the volume integral increases in such a way as to keep the total work constant. That is,
we get
where the integral now covers all space. Note the distinction between 2 and 10: in the first case, we need to integrate
only over that volume where
; in the second case we need to integrate over all space, since in general the
electric field is always non-zero over any finite distance, even for a localized charge distribution.
As an example, we can work out the energy stored in a uniformly charged solid sphere of radius
We’ll do it four different ways to show how each of the above methods works.
and charge
.
Example 1. We can use 2. We found the potential of the sphere earlier (Example 1 in this post). The charge
density in terms of the total charge is
and the potential inside the sphere (all we need here, since we need to integrate only over that volume where
) is
Therefore, we get
Example 2. Same problem, but now we use 10. We worked out the field earlier (Example 2 in this post). In this
case, we will need the field for both inside and outside the sphere, since we must integrate over all space. We have,
for inside:
and for outside:
The energy is then
Example 3. This time we use the formula 9, so the integral is split between a volume integral and a surface integral.
If we use a surface of radius
, then the volume component can be worked out using the same integrals as in
Example 2, but changing the limit on the second integral.
Note that as
, this integral tends to the total energy as worked out in the previous two
examples:
.
The surface integral uses the potential and field at a distance
sphere, which is
The field at
is
Both of these are constants over the bounding sphere, so we get
The total energy is
. This time we need the potential outside the
Example 4. Finally, we can build up the solid sphere by adding successive layers of charge of thickness
have, when the sphere has intermediate radius :
This amount of charge is brought from infinity and added to a spherical volume of charge
from above that the potential of such a sphere at its outer boundary is
and radius
. We
. We know
so the amount of work needed to add this charge to the sphere is
We can then integrate this to find the total work:
…………………………………………………………..
Dielectric Constant?
It is also known as Relative permittivity. If two charges q 1 and q 2 are separated from each other by a small
distance r. Then by using the coulombs law of forces the equation formed will be
———— 2.6
In the above equation
is the electrical permittivity or you can say it, Dielectric constant.
If we repeat the above case with only one change i.e. only change in the separation medium between the charges.
Here some material medium must be used. Then the equation formed will be.
——2.7
Now after division of equation 2.6 with equation 2.7 we get…
——2.8
In the above figure
is the Relative Permittivity. Again one thing to notice is that the dielectric constant is
represented by the symbol(K) but permittivity by the symbol
Now after the division third equation is created. Its we describe this equation then the new definition can be made
from this. i.e. It is the ratio of the force of communication between the two point charges separated from each by a
distance using air/Vacuum as a medium to the force of communication between the same charges placed at the same
distance and using some material medium. Faraday has done many experiments on this topic. He found that When
we insert some insulating material in the space present between the plates of a capacitor , having its plates fully
charged then the capacitance of the capacitor will increase definitely.
Values of dielectric constant varies with the change in materials. Some of the examples of materials is shown below
in the table
Values of K
Material
Air( 1 atm)
1.0006
Amber
2.8
Alcohol
25
Water
81
Hydrogen
1.00026
Germanium
16
Vaccum/Air
1
Metals
∞
Polar and Non Polar Dielectrics?
Those materials which have the ability to transfer the electric effects without conducting.
Dielectrics exists basically in two types.
1.Polar Dielectrics
2. Non polar Dielectrics
Polar Dielectrics: Polar dielectrics are those in which the possibility of center coinciding of the positive as well as
negative charge is almost zero i.e. they don’t coincide with each other. The reason behind this is their shape. They
all are of asymmetric shape. Some of the examples of the polar dielectrics is NH3, HCL, water etc.
Non Polar dielectrics: In case of non polar dielectrics the centres of both positive as well as negative charges
coincide. Dipole moment of each molecule in non polar system is zero. All those molecules which belong to this
category are symmetric in nature. Examples of non polar dielectrics are: methane , benzene etc.
……………………………
What is Dielectric Polarization?
In this case a non polar dielectric like methane is taken and placed in some external electric field. Center of positive
charge of individual molecules is pulled automatically in the same direction as that of electric field towards the plate
having negative charge. Similarly, the centre of the negative charged electrons is dragged in the opposite direction of
the electric field, towards the plate having positive charge. So the centres of positive as well as negative charges are
set apart. Due to this the molecules are deformed from their original shape and hence separate at last. So, due to the
above process each molecule gets some dipole moment . After sometime these molecules will get polarized when
the forces of attraction between the centres of positive and negative charges and the force due to electric field will
reach at some stable state.
Now the individual molecules will exist as separate tiny dipoles.
————–------------------------------------ 2.22
In the above equation alpha is proportionality constant. Its is also called as Atomic polarizability.
By using the equation 2.22
= Cm / (C2 N-1m-2 ) * (NC-1)
=m3
Now from the above result it is cleared that the Alpha i.e. atomic polarizability has similar dimensions as that of
volume. Mostly values of alpha vary from 10-29 to 10-30.
We are going to made some experiments on a rectangular slab. Let’s name it as ABCD. Keep it in the influence of
the electric field present between the two plates. For this we have to think that all the atoms must be polarized in a
uniform manner. Let the displacement between the charges be x. So the equation will be:
p=qx
If the total number of atoms present per unit volume be taken as N. But N is equal to the total dipole moment density
.
New equation formed will be:
P=Np
Or we can write it as
P=Nqx
Here P is the density of the dipole. Its other name is Electric polarization. Its units are C -m-2
Lets us think about the interior of the slab. The cancellation of the charges due to their equal magnitude of positive
and negative charge, the volume charge density will be Zero. Some amount of positive charge will generate in the
surface CD.It can be seen in the figure below. The equation of Effective electric field present inside a dielectric
which is polarized is given by:
E=Eo – E p ---------------------------------------—-2.23
Eo=
= p/
i
/
= Qi/a *
------------------------------------—-2.24
The nature of the dielectric slab will affect the value of E. Now it is proved from the definition and the equation
given below. The charge in external electric field divided by the reduced value of electric field gives us the constant
i.e. k , which is the constant for material of the rectangular slab ABCD.
/E=K
----------------------------------------------—2.25
(b) Electric Susceptibility: It states that the reduced value of electric field effects the electric polarization P due to a
relation of direct proportion between them.
P is directly proportional to E.
So
P=
*
*E
--------------------------------——2.26
In the above equation
is the electric susceptibility. The main purpose of
to make the
dimensionless. Its
main purpose is to tell about the complete electric manner of acting of a dielectric. Value of Susceptibility varies
with the variation of dielectrics. The value of Electric susceptibility for vacuum is equal to zero.
E= Eo – P /
–
*
*E/
By using the equation no 2.25
Or E = Eo –
E
Or E o= E+E=E (1+
K=1+
)
---------------------------------——– 2.27
Dielectric constant and the susceptibility are related to each other according to the equation 2.27
In polar dielectrics individual molecules have their individual dipole moments. But when the External electric field’s
influence remains inactive then the net dipole moment of the elements of dielectric becomes zero.
Lets consider the case in which external electric fields influence is present. Due to this value of dipole moment in
the individual elements increase by a little amount. Then another affect will take place, a twisting force will be
generated which will try to change the alignment of electric dipoles parallel to that of the electric field. The
alignment depends upon the amount of the electric field passed. Stronger the electric field higher will be the
alignment. This also depends upon the temperature. The alignment of the dipoles increases if we increase the
temperature.
…………………….
See pdf file “CM & Multipole”