2. static electric fields

Static Electric Fields
In this chapter we will discuss on the followings:
• Coulomb's Law
• Electric Field & Electric Flux Density
• Gauss's Law with Application
• Electrostatic Potential•
Boundary Conditions for Static Electric Fields
• Capacitance and Capacitors
• Electrostatic Energy
• Laplace's and Poisson's Equations
Introduction
Electric charge is a fundamental property of matter and charge exist in integral multiple of
electronic charge. Electrostatics can be defined as the study of electric charges at rest.
Electric fields have their sources in electric charges.
In this chapter we first study two fundamental laws governing the electrostatic fields, viz,
(1) Coulomb's Law and
(2) Gauss's Law.
Both these laws have experimental basis. Coulomb's law is applicable in finding electric
field due to any charge distribution, Gauss's law is easier to use when the distribution is
symmetrical.
Coulomb's Law
Coulomb's Law states that the force between two point charges Q1 and Q2 is directly
proportional to the product of the charges and inversely proportional to the square of the
distance between them.
Point charge is a hypothetical charge located at a single point in space. It is an idealised
model of a particle having an electric charge.
Mathematically, ,where k is the proportionality constant.

In SI units, Q1 and Q2 are expressed in Coulombs(C) and R is in meters.
Force F is in Newtons (N) and , is called the permittivity of free space.
(We are assuming the charges are in free space. If the charges are any other dielectric
medium, we will use instead where is called the relative permittivity or the
dielectric constant of the medium).
Therefore .......................(2.1)
Vector Force:
As shown in the Figure 2.1 let the position vectors of the point charges Q1and Q2 are
given by and . Let represent the force on Q1 due to charge Q2.
Fig 2.1: Coulomb's Law
The charges are separated by a distance of . We define the unit
vectors as
and ..................................(2.2)
can be defined as . Similarly the force on Q2 due to
charge Q1 can be calculated and if represents this force then we can write

Force due to a number of point charges:
When we have a number of point charges, to determine the force on a particular charge
due to all other charges, we apply principle of superposition. If we have N number of
charges Q1,Q2,.........QN located respectively at the points represented by the position
vectors , ,...... , the force experienced by a charge Q located at is given by,
.................................(2.3)
Electric Field
The electric field intensity or the electric field strength at a point is defined as the force per
unit charge. That is
or, .......................................(2.4)
The electric field intensity E at a point r (observation point) due a point charge Q located
at (source point) is given by:
..........................................(2.5)
For a collection of N point charges Q1 ,Q2 ,.........QN located at , ,...... , the electric
field intensity at point is obtained as
........................................(2.6)
E for continuous charge distributions:
The expression (2.6) can be modified suitably to compute the electric filed due to a
continuous distribution of charges.
In figure 2.2 we consider a continuous volume distribution of charge (t) in the region
denoted as the source region.
For an elementary charge , i.e. considering this charge as point charge, we
can write the field expression as:

.............(2.7)
Fig 2.2: Continuous Volume Distribution of Charge
When this expression is integrated over the source region, we get the electric field at the
point P due to this distribution of charges. Thus the expression for the electric field at P can
be written as:
..........................................(2.8)
Similar technique can be adopted when the charge distribution is in the form of a line charge
density or a surface charge density.
........................................(2.9)
........................................(2.10)
Electricfluxdensity:
As stated earlier electric field intensity or simply ‘Electric field' gives the strength of the
field at a particular point. The electric field depends on the material media in which the field is
being considered. The flux density vector is defined to be independent of the material media
(as we'll see that it relates to the charge that is producing it).

For a linear, isotropic medium under consideration; the flux density vector is defined as:
................................................(2.11)
We define the electric flux  as
.....................................(2.12)
Gauss's Law: Gauss's law is one of the fundamental laws of electromagnetism and it states
that the total electric flux through a closed surface is equal to the total charge enclosed by
the surface.
Fig 2.3: Gauss's Law
Let us consider a point charge Q located in an isotropic homogeneous medium of dielectric
constant . The flux density at a distance r on a surface enclosing the charge is given by
...............................................(2.13)
If we consider an elementary area ds, the amount of flux passing through the elementary
area is given by
.....................................(2.14)
But , is the elementary solid angle subtended by the area at the location
of Q. Therefore we can write

For a closed surface enclosing the charge, we can write
which can seen to be same as what we have stated in the definition of Gauss's Law.
Application of Gauss's Law
Gauss's law is particularly useful in computing or where the charge distribution has
some symmetry. We shall illustrate the application of Gauss's Law with some examples.
1.An infinite line charge
As the first example of illustration of use of Gauss's law, let consider the problem of
determination of the electric field produced by an infinite line charge of density LC/m. Let us
consider a line charge positioned along the z-axis as shown in Fig. 2.4(a) (next slide). Since
the line charge is assumed to be infinitely long, the electric field will be of the form as shown
in Fig. 2.4(b) (next slide).
If we consider a close cylindrical surface as shown in Fig. 2.4(a), using Gauss's theorm
we can write,
.....................................(2.15)
Considering the fact that the unit normal vector to areas S1 and S3 are perpendicular to the
electric field, the surface integrals for the top and bottom surfaces evaluates to zero. Hence
we can write,

Fig 2.4: Infinite Line Charge
.....................................(2.16)
2. Infinite Sheet of Charge
As a second example of application of Gauss's
theorem, we consider an infinite charged sheet
covering the x-z plane as shown in figure 2.5.
Assuming a surface charge density of for the
infinite surface charge, if we consider a
cylindrical volume having sides placed
symmetrically as shown in figure 5, we can write:
..............(2.17)
Fig 2.5: Infinite Sheet of Charge
It may be noted that the electric field strength is independent of distance. This is true for the infinite plane of
charge; electric lines of force on either side of the charge will be perpendicular to the sheet and extend to
infinity as parallel lines. As number of lines of force per unit area gives the strength of the field, the field
becomes independent of distance. For a finite charge sheet, the field will be a function of distance.

3. Uniformly Charged Sphere
Let us consider a sphere of radius r0 having
a uniform volume charge density of v C/m3.
To determine everywhere, inside and
outside the sphere, we construct Gaussian
surfaces of radius r < r0 and r > r0 as shown
in Fig. 2.6 (a) and Fig. 2.6(b).
For the region ; the total enclosed
charge will be
.........................(2.18)
Fig 2.6: Uniformly Charged Sphere
By applying Gauss's theorem,
...............(2.19)
Therefore
...............................................(2.20)
For the region ; the total enclosed charge will be
....................................................................(2.21)
By applying Gauss's theorem,
.....................................................(2.22)

Electrostatic Potential and Equipotential Surfaces
In the previous sections we have seen how the electric
field intensity due to a charge or a charge distribution
can be found using Coulomb's law or Gauss's law. Since
a charge placed in the vicinity of another charge (or in
other words in the field of other charge) experiences a
force, the movement of the charge represents energy
exchange. Electrostatic potential is related to the
work done in carrying a charge from one point to the
other in the presence of an electric field.
Let us suppose that we wish to move a positive test
charge from a point P to another point Q as shown in
the Fig. 2.8.
The force at any point along its path would cause the
particle to accelerate and move it out of the region if
unconstrained. Since we are dealing with an
electrostatic case, a force equal to the negative of that
acting on the charge is to be applied while moves
from P to Q. The work done by this external agent in
moving the charge by a distance is given by:
Fig 2.8: Movement of Test Charge in Electric
Field
.............................(2.23)
The negative sign accounts for the fact that work is done on the system by the external
agent.
.....................................(2.24)
The potential difference between two points P and Q , VPQ, is defined as the work done per
unit charge, i.e.
...............................(2.25)
It may be noted that in moving a charge from the initial point to the final point if the potential
difference is positive, there is a gain in potential energy in the movement, external agent
performs the work against the field. If the sign of the potential difference is negative, work is
done by the field.
We will see that the electrostatic system is conservative in that no net energy is exchanged if
the test charge is moved about a closed path, i.e. returning to its initial position. Further, the
potential difference between two points in an electrostatic field is a point function; it is
independent of the path taken. The potential difference is measured in Joules/Coulomb
which is referred to as Volts.

Let us consider a point charge Q as shown in the Fig. 2.9.
Fig 2.9: Electrostatic Potential calculation for a point charge
Further consider the two points A and B as shown in the Fig. 2.9. Considering the movement
of a unit positive test charge from B to A , we can write an expression for the potential
difference as:
..................................(2.26)
It is customary to choose the potential to be zero at infinity. Thus potential at any point
( rA = r) due to a point charge Q can be written as the amount of work done in bringing a unit
positive charge from infinity to that point (i.e. rB = 0).
..................................(2.27)
Or, in other words,
..................................(2.28)

Let us now consider a situation where the point charge Q is not located at the origin as shown in Fig. 2.10.
Fig 2.10: Electrostatic Potential due a Displaced Charge
The potential at a point P becomes
..................................(2.29)
So far we have considered the potential due to point charges only. As any other type of
charge distribution can be considered to be consisting of point charges, the same basic
ideas now can be extended to other types of charge distribution also.
Let us first consider N point charges Q1, Q2,.....QN located at points with position vectors
, ,...... . The potential at a point having position vector can be written as:
..................................(2.30a)
or, ...........................................................(2.30b)
For continuous charge distribution, we replace point charges Qn by corresponding charge
elements or or depending on whether the charge distribution is linear,
surface or a volume charge distribution and the summation is replaced by an integral. With
these modifications we can write:

For line charge, ..................................(2.31)
For surface charge, .................................(2.32)
For volume charge, .................................(2.33)
It may be noted here that the primed coordinates represent the source coordinates and the
unprimed coordinates represent field point.
Further, in our discussion so far we have used the reference or zero potential at infinity. If
any other point is chosen as reference, we can write:
.................................(2.34)
where C is a constant. In the same manner when potential is computed from a known electric
field we can write:
.................................(2.35)
The potential difference is however independent of the choice of reference.
.......................(2.36)
We have mentioned that electrostatic field is a conservative field; the work done in moving a
charge from one point to the other is independent of the path. Let us consider moving a
charge from point P1 to P2 in one path and then from point P2 back to P1 over a different path.
If the work done on the two paths were different, a net positive or negative amount of work
would have been done when the body returns to its original position P1. In a conservative
field there is no mechanism for dissipating energy corresponding to any positive work neither
any source is present from which energy could be absorbed in the case of negative work.
Hence the question of different works in two paths is untenable, the work must have to be
independent of path and depends on the initial and final positions.
Since the potential difference is independent of the paths taken, VAB = - VBA , and over a
closed path,
.................................(2.37)

Applying Stokes's theorem, we can write:
............................(2.38)
from which it follows that for electrostatic field,
........................................(2.39)
Any vector field that satisfies is called an irrotational field.
From our definition of potential, we can write
.................................(2.40)
from which we obtain,
..........................................(2.41)
From the foregoing discussions we observe that the electric field strength at any point is the
negative of the potential gradient at any point, negative sign shows that is directed from
higher to lower values of . This gives us another method of computing the electric field, i.
e. if we know the potential function, the electric field may be computed. We may note here
that that one scalar function contain all the information that three components of carry,
the same is possible because of the fact that three components of are interrelated by the
relation .
Boundary conditions for Electrostatic fields
In our discussions so far we have considered the existence of electric field in the
homogeneous medium. Practical electromagnetic problems often involve media with different
physical properties. Determination of electric field for such problems requires the knowledge
of the relations of field quantities at an interface between two media. The conditions that the
fields must satisfy at the interface of two different media are referred to as boundary
conditions .
In order to discuss the boundary conditions, we first consider the field behavior in some
common material media.

In general, based on the electric properties, materials can be classified into three
categories: conductors, semiconductors and insulators (dielectrics). In conductor , electrons
in the outermost shells of the atoms are very loosely held and they migrate easily from one
atom to the other. Most metals belong to this group. The electrons in the atoms
of insulators or dielectrics remain confined to their orbits and under normal circumstances
they are not liberated under the influence of an externally applied field. The electrical
properties of semiconductors fall between those of conductors and insulators since
semiconductors have very few numbers of free charges.
The parameter conductivity is used characterizes the macroscopic electrical property of a
material medium. The notion of conductivity is more important in dealing with the current flow
and hence the same will be considered in detail later on.
If some free charge is introduced inside a conductor, the charges will experience a force
due to mutual repulsion and owing to the fact that they are free to move, the charges will
appear on the surface. The charges will redistribute themselves in such a manner that the
field within the conductor is zero. Therefore, under steady condition, inside a
conductor .
From Gauss's theorem it follows that
= 0 .......................(2.51)
The surface charge distribution on a conductor depends on the shape of the conductor.
The charges on the surface of the conductor will not be in equilibrium if there is a tangential
component of the electric field is present, which would produce movement of the charges.
Hence under static field conditions, tangential component of the electric field on the
conductor surface is zero. The electric field on the surface of the conductor is normal
everywhere to the surface . Since the tangential component of electric field is zero,
the conductor surface is an equipotential surface. As = 0 inside the conductor, the
conductor as a whole has the same potential. We may further note that charges require a
finite time to redistribute in a conductor. However, this time is very small sec for good
conductor like copper.
Fig 2.14: Boundary Conditions for at the surface of a Conductor
Let us now consider an interface between a conductor and free space as shown in the
figure 2.14.

Let us consider the closed path pqrsp for which we can write,
.................................(2.52)
For and noting that inside the conductor is zero, we can write
=0.......................................(2.53)
Et is the tangential component of the field. Therefore we find that
Et = 0 ...........................................(2.54)
In order to determine the normal component En, the normal component of , at the surface of the conductor, we
consider a small cylindrical Gaussian surface as shown in the Fig.12. Let represent the area of the top and bottom
faces and represents the height of the cylinder. Once again, as , we approach the surface of the
conductor. Since = 0 inside the conductor is zero,
.............(2.55)
..................(2.56)
Therefore, we can summarize the boundary conditions at the surface of a conductor as:
Et = 0 ........................(2.57)
.....................(2.58)
Behavior of dielectrics in static electric field: Polarization of dielectric
The electric flux density
When the dielectric properties of the medium are linear and isotropic, polarisation is directly
proportional to the applied field strength and

........................(2.75)
is the electric susceptibility of the dielectric. Therefore,
.......................(2.76)
is called relative permeability or the dielectric constant of the medium. is
called the absolute permittivity.
A dielectric medium is said to be linear when is independent of and the medium is
homogeneous if is also independent of space coordinates. A linear homogeneous and
isotropic medium is called a simple medium and for such medium the relative permittivity is
a constant.
Dielectric constant may be a function of space coordinates. For anistropic materials, the
dielectric constant is different in different directions of the electric field, D and E are related
by a permittivity tensor which may be written as:
.......................(2.77)
For crystals, the reference coordinates can be chosen along the principal axes, which make
off diagonal elements of the permittivity matrix zero. Therefore, we have
.......................(2.78)
Media exhibiting such characteristics are called biaxial. Further, if then the medium is
called uniaxial. It may be noted that for isotropic media, .
Lossy dielectric materials are represented by a complex dielectric constant, the imaginary
part of which provides the power loss in the medium and this is in general dependant on
frequency.
Another phenomenon is of importance is dielectric breakdown. We observed that the
applied electric field causes small displacement of bound charges in a dielectric material that
results into polarization. Strong field can pull electrons completely out of the molecules.
These electrons being accelerated under influence of electric field will collide with molecular
lattice structure causing damage or distortion of material. For very strong fields, avalanche
breakdown may also occur. The dielectric under such condition will become conducting.

The maximum electric field intensity a dielectric can withstand without breakdown is
referred to as the dielectric strength of the material.
Boundary Conditions for Electrostatic Fields:
Let us consider the relationship among the field components that exist at the interface
between two dielectrics as shown in the figure 2.17. The permittivity of the medium 1 and
medium 2 are and respectively and the interface may also have a net charge
density Coulomb/m.
Fig 2.17: Boundary Conditions at the interface between two dielectrics
We can express the electric field in terms of the tangential and normal
components ..........(2.79)
where Et and En are the tangential and normal components of the electric field respectively.
Let us assume that the closed path is very small so that over the elemental path length the
variation of E can be neglected. Moreover very near to the interface, . Therefore
.......................(2.80)
Thus, we have,
or i.e. the tangential component of an electric field is continuous
across the interface.

For relating the flux density vectors on two sides of the interface we apply Gauss’s law to a
small pillbox volume as shown in the figure. Once again as , we can write
..................(2.81a)
i.e., .................................................(2.81b)
i.e., .......................(2.81c)
Thus we find that the normal component of the flux density vector D is discontinuous
across an interface by an amount of discontinuity equal to the surface charge density
at the interface.
Example
Two further illustrate these points; let us consider an example, which involves the refraction
of D or E at a charge free dielectric interface as shown in the figure 2.18.
Using the relationships we have just derived, we can write
.......................(2.82a)
.......................(2.82b)
In terms of flux density vectors,
.......................(2.83a)
.......................(2.83b)
Therefore, .......................(2.84)

Fig 2.18: Refraction of D or E at a Charge Free Dielectric Interface
Capacitance and Capacitors
We have already stated that a conductor in an electrostatic field is an Equipotential body
and any charge given to such conductor will distribute themselves in such a manner that
electric field inside the conductor vanishes. If an additional amount of charge is supplied to
an isolated conductor at a given potential, this additional charge will increase the surface
charge density . Since the potential of the conductor is given by , the
potential of the conductor will also increase maintaining the ratio same. Thus we can
write where the constant of proportionality C is called the capacitance of the isolated
conductor. SI unit of capacitance is Coulomb/ Volt also called Farad denoted by F. It can It
can be seen that if V=1, C = Q. Thus capacity of an isolated conductor can also be defined
as the amount of charge in Coulomb required to raise the potential of the conductor by 1
Volt.
Of considerable interest in practice is a capacitor that consists of two (or more) conductors
carrying equal and opposite charges and separated by some dielectric media or free space.
The conductors may have arbitrary shapes. A two-conductor capacitor is shown in
figure 2.19.

Fig 2.19: Capacitance and Capacitors
When a d-c voltage source is connected between the conductors, a charge transfer occurs which results into
a positive charge on one conductor and negative charge on the other conductor. The conductors are
equipotential surfaces and the field lines are perpendicular to the conductor surface. If V is the mean
potential difference between the conductors, the capacitance is given by . Capacitance of a capacitor
depends on the geometry of the conductor and the permittivity of the medium between them and does not
depend on the charge or potential difference between conductors. The capacitance can be computed by
assuming Q(at the same time -Q on the other conductor), first determining using Gauss’s theorem and
then determining . We illustrate this procedure by taking the example of a parallel plate
capacitor.
Example: Parallel plate capacitor

Fig 2.20: Parallel Plate Capacitor
For the parallel plate capacitor shown in the figure 2.20, let each plate has area A and a distance h separates the
plates. A dielectric of permittivity fills the region between the plates. The electric field lines are confined betwee
plates. We ignore the flux fringing at the edges of the plates and charges are assumed to be uniformly distributed
the conducting plates with densities and - , .
By Gauss’s theorem we can write, .......................(2.85)
As we have assumed to be uniform and fringing of field is neglected, we see that E is
constant in the region between the plates and therefore, we can write . Thus,
for a parallel plate capacitor we have,
........................(2.86)
Series and parallel Connection of capacitors
Capacitors are connected in various manners in electrical circuits; series and parallel
connections are the two basic ways of connecting capacitors. We compute the equivalent
capacitance for such connections.
Series Case: Series connection of two capacitors is shown in the figure 2.21. For this case
we can write,
.......................(2.87)

Fig 2.21: Series Connection of Capacitors
Fig 2.22: Parallel Connection of Capacitors
The same approach may be extended to more than two capacitors connected in series.
Parallel Case: For the parallel case, the voltages across the capacitors are the same.
The total charge
Therefore, .......................(2.88)

Electrostatic Energy and Energy Density
We have stated that the electric potential at a point in an electric field is the amount of work
required to bring a unit positive charge from infinity (reference of zero potential) to that point.
To determine the energy that is present in an assembly of charges, let us first determine the
amount of work required to assemble them. Let us consider a number of discrete
charges Q1, Q2,......., QN are brought from infinity to their present position one by one. Since
initially there is no field present, the amount of work done in bring Q1 is zero. Q2 is brought in
the presence of the field of Q1, the work done W1= Q2V21 where V21 is the potential at the
location of Q2 due to Q1. Proceeding in this manner, we can write, the total work done
.................................................(2.89)
Had the charges been brought in the reverse order,
.................(2.90)
Therefore,
................(2.91)
Here VIJ represent voltage at the Ith charge location due to Jth charge. Therefore,
Or, ................(2.92)
If instead of discrete charges, we now have a distribution of charges over a volume v then we
can write,
................(2.93)
where is the volume charge density and V represents the potential function.
Since, , we can write

.......................................(2.94)
Using the vector identity,
, we can write
................(2.95)
In the expression , for point charges, since V varies as and D varies as ,
the term V varies as while the area varies as r2. Hence the integral term varies at least
as and the as surface becomes large (i.e. ) the integral term tends to zero.
Thus the equation for W reduces to
................(2.96)
, is called the energy density in the electrostatic field.
Poisson’s and Laplace’s Equations
For electrostatic field, we have seen that
..........................................................................................(2.97)
Form the above two equations we can write
..................................................................(2.98)
Using vector identity we can write, ................(2.99)

For a simple homogeneous medium, is constant and . Therefore,
................(2.100)
This equation is known as Poisson’s equation. Here we have introduced a new
operator, ( del square), called the Laplacian operator. In Cartesian coordinates,
...............(2.101)
Therefore, in Cartesian coordinates, Poisson equation can be written as:
...............(2.102)
In cylindrical coordinates,
...............(2.103)
In spherical polar coordinate system,
...............(2.104)
At points in simple media, where no free charge is present, Poisson’s equation reduces to
...................................(2.105)
which is known as Laplace’s equation.
Laplace’s and Poisson’s equation are very useful for solving many practical electrostatic field
problems where only the electrostatic conditions (potential and charge) at some boundaries
are known and solution of electric field and potential is to be found throughout the volume.
We shall consider such applications in the section where we deal with boundary value
problems.

Electrostatic boundary value problem
In this section we consider the solution for field and potential in a region where the
electrostatic conditions are known only at the boundaries. Finding solution to such problems
requires solving Laplace’s or Poisson’s equation satisfying the specified boundary condition.
These types of problems are usually referred to as boundary value problem. Boundary value
problems for potential functions can be classified as:
• Dirichlet problem where the potential is specified everywhere in the boundary
• Neumann problem in which the normal derivatives of the potential function are specified
everywhere in the boundary
• Mixed boundary value problem where the potential is specified over some boundaries and
the normal derivative of the potential is specified over the remaining ones.
Usually the boundary value problems are solved using the method of separation of variables,
which is illustrated, for different types of coordinate systems.
Example 3: Potential distribution in a coaxial
conductor
Let us consider a very long coaxial conductor as
shown in the figure 2.33, the inner conductor of
which is maintained at potential V0 and outer
conductor is grounded.
Fig 2.33: Potential Distribution in a Coaxial
Conductor
From the symmetry of the problem, we observe that the potential is independent of
variation. Similarly, we assume the potential is not a function of z. Thus the Laplace’s
equation can be written as
...................................................(2.132)
Integrating the above equation two times we can write
.....................................................(2.133)
where C1 and C2 are constants, which can be evaluated from the boundary conditions

............................................................(2.134)
Evaluating these constants we can write the expression for the potential as
.......................(2.135)
Furthure, .......................(2.136)
Therefore, ........................(2.137)
If we consider a length L of the coaxial conductor, the capacitance
....................................(2.138)
Example-4: Concentric conducting spherical shells
As shown in the figure 2.34, let us consider a pair of concentric spherical conduction shells,
outer one being grounded and the inner one maintained at potential V0.

Considering the symmetry, we find that the potential is function
of ronly. The Laplace’s equation can be written as:
............(2.139)
From the above equation, by direct integration we can write,
...............................(2.140)
Applying boundary conditions
................................(2.141)
we can find C1 and C2 and the expression for the potential can
be written as : ...........................(2.142)
The electric field and the capacitor can be obtained using the
same procedure described in Example -3.
Fig 2.34: Concentric conducting spherical
shells

2. static electric fields

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