Chapter 2

Electrostatic Fields

The main objective of this chapter is to provide detailed concepts of electrostatics. They include:

• applications of electrostatics
• charge distributions
• Coulomb’s law, applications and limitations
• Gauss’s law, applications and limitation
• potential functions
• Poisson’s and Laplace’s equations
• conductors, dielectrics and capacitors
• energy stored in electrostatic fields and capacitors
• fields in dielectrics and boundary conditions
• solved problems, points/formulae to remember, objective and multiple choice questions and exercise problems.

2.1 INTRODUCTION

Electrostatic fields are also called static electric fields or steady electric fields. These fields are not variant with time. They are produced by static charges or charge distributions. These fields have a wide range of applications.

In this chapter, the important laws, theorems and equations, with their applications are presented to provide a thorough understanding of electrostatics. The important areas are Coulomb’s and Gauss’s laws, Stoke’s and divergence theorems and Poisson’s and Laplace’s equations.

2.2 APPLICATIONS OF ELECTROSTATIC FIELDS

Electrostatic fields are used

1. in cathode ray oscilloscopes to obtain the electron beam deflection
2. in ink-jet printers to obtain speed of printing and quality of print
3. to sort out minerals in ore separators, to sort out seeds in agriculture and for spraying plants and trees
4. in electrostatic generators
5. to produce potential
6. to produce force on charges for their mobility
7. in electric power transmission
8. in lightning protection
9. to measure moisture content
10. to spin cotton
11. in field-effect transistors
12. in capacitors
13. in LCDs
14. in touch pads
15. in capacitance keyboards
16. in fast baking of bread
17. in ECGs, EEGs, ERG, EMG, EOG in medical applications
18. in spray painting
19. in electrodeposition
20. in electrochemical machining
21. X-ray machines
22. computer peripherals

2.3 DIFFERENT TYPES OF CHARGE DISTRIBUTIONS

Charges at rest produce electrostatic field. The charges are basically of two types: positive and negative.

Properties and Functions of Charges

1. Charge is conserved. It can neither be created nor destroyed.
2. Charge is quantisised that is, it comes in discrete quantities and in integer multiples of the basic unit of charge. Electron charge or proton charge is the basic charge.
3. When a charge is accelerated, electromagnetic field is produced. A fraction of the field is detached and propagates at the speed of light. The detached field carries energy, momentum and angular momentum. This is called electromagnetic radiation.
4. Charges are surrounded by electric and magnetic fields.
5. A charge experiences a force in the presence of a field.
6. Charges mediate the integration of fields.

Charge density and charge distribution indicate the same property. In practice, there are four types of charge distributions.

 

(i)	Point charges, Q (Coulomb)    These are the charges which do not occupy any space, that is, the volume of the point charge is zero. For example, an electron is considered to be a point charge and has a charge of 1.6 × 10–19 Coulombs (C).
(ii)	Line charge distribution, ρL (c/m)    This is a charge distribution in which the charge is distributed along a line like a filament, that is, this has only length but no width or breadth. ρL is defined as the charge per unit length, that is,

where ΔQ is small charge, ΔL is small length, dQ is differential charge and dL is differential length.

Sometimes, ρ_L is simply considered as Q/L.

An example is the electron beam in CRT.

Problem 2.1    If there is a charge of 10μC over a filament length of 0.5 m, find its line charge density.

Solution    Here,         Q = 10μC = 10 × 10^–6 C

                                         L = 0.5m

Line charge density,    ρ_L = Q/L = 10 / 0.5 = 20 μC / m

 

(iii)

Surface charge distribution, ρs (c/m2) When a charge is confined to the surface of a conductor, it is said to be surface charge distribution. Such a surface has both length and width but no breadth.

Surface density is defined as the charge per unit area, that is,

where ΔS is small area and dS is differential area.

Sometimes, ρ_s is simply considered as Q/S.

An example is the conductor surface of a capacitor.

Problem 2.2    If there is a total charge of 10 pC over a surface area of 0.2 m², find the surface charge density.

Solution    Here,          Q = 10 pC = 10 × 10^–12C

                                           S = 0.2 m²

Surface charge density, ρ_s = Q/ S = (10 × 10^–12)/0.2

                                               = 50pC/m²

 

(iv)	Volume charge distribution, ρv (c/m3) Volume charge density is defined as the charge per unit volume, that is,

where dv is differential volume.

Sometimes, ρ_v is simply considered as

Examples are ionospheric region, electron cloud in vacuum tube.

Problem 2.3    If there exists a total charge of 12 nC in a spherical volume of 0.1 m³, find the volume charge density.

Solution     Here       Q = 12nC = 12 × 10 ^–9C

                                       v = 0.1 m³

The volume charge density,

                    ρ_v = Q / v

                        = 12 / 0.1

                        = 120 nC / m³

2.4 COULOMB’S LAW

Charles Augustin De Coulomb (1736–1806) was a French physicist. After conducting several experiments on charged bodies, he concluded that there exists a force between them. He formulated a law known as Coulomb’s law.

The law appears in the following forms:

1. Gaussian form

   Force,   

2. SI form

   Force,   

3. Heavside-Lorentz form

   Force,   

Law statement    Coulomb’s law states that there exists a force between charged bodies and it is:

1. Proportional to the product of the two charges,
2. Inversely proportional to the square of the distance between the charges. The force also depends on the medium in which the charges are located.

The force is a vector quantity and it is attractive if the charges are unlike and repulsive if the charges are alike. It acts along the straight line joining the two charges.

Mathematically, Coulomb’s law is given by,

where	k = a constant of proportionality
	(metre/farad-radian)

Q₁, Q₂ are two point charges (C)

 

	=	force between the charges (Newton)
r	=	distance between the charges (m)
∈	=	permittivity of the medium in which the charges are located (F/m)
	=	∈0∈r
∈0	=	permittivity of vacuum or free space
	=
∈r	=	relative permittivity of the medium with respect to free space (has no unit)
	=	1 for free space

k in free space

ar	= unit vector along the line joining the two charges

The force on Q₂ due to Q₁ in free space is written in the form of

where,

∈₀ = permittivity of free space

The force on Q₁ due to Q₂ is

Problem 2.4    If Q₁ and Q₂ are two point charges and are located at r₁ and r₂ respectively in free space, find (a) the force on Q₂ due to Q₁, (b) the force on Q₁ due to Q₂.

 

Solution	Q1 = first charge
	Q2 = second charge
	r1 = location of Q1
	r2 = location of Q2

Distance vector along the line joining two charges,

 

The magnitude of r₂₁ is

 

The direction of         

1. The force on Q₂ due to Q₁
   
2. The force on Q₁ due to Q₂,
   

Here	r12 = r1 − r2 = −(r2 − r1) = −r21
	r12 = r21 = |r1 − r2| = |r2 − r1|
	a12 = −a21
	F1 = −F2

Problem 2.5    A charge of Q₁ = –1.0μC is placed at the origin of a rectangular coordinate system and a second charge, Q₂ = –10 mC is placed on the x-axis at a distance of 50 cm from the origin. Find the force on Q₁ due to Q₂ if they are in free space.

 

Solution	Q1 = –1.0μC at (0, 0, 0)
	Q2 = –10 mC at (0.5, 0, 0)

By Coulomb’s law,

Problem 2.6    A point charge, Q₁ = 2 μC is at (2, 3, 6) and another charge, Q₂ = 5 μC is at (0, 0, 0) in free space. Find the force on Q₁ due to Q₂.

Solution    We have

Problem 2.7    Three equal point charges of 2μC are in free space at (0, 0, 0), (2, 0, 0) and (0, 2, 0), respectively. Find net force on Q₄ = 5μC at (2, 2, 0).

Solution

	Q1	=	Q2 = Q3 = 2μC =Q
	Q4	=	5μC
	r1	=	(0, 0, 0)
	r2	=	(2, 0, 0) = 2ax
	r3	=	(0, 2, 0)= 2ay
	r4	=	(2, 2, 0)= 2ax + 2ay
	r41	=	r4 − r1 = 2ax + 2ay
	r42	=	r4 − r2 = 2ax + 2ay − 2ax = 2ay
	r43	=	r4 − r3 = 2ax + 2ay − 2ay = 2ax
	r41	=
	r42	=	2, r43 = 2

The net force on Q₄ due to Q₁, Q₂, Q₃ is

Problem 2.8    Two charges Q₁ = 2μC and Q₂ = 5μC are located at (–3, 7, –4) and (2, 4, –1), respectively. Determine the force on Q₂ due to Q₁ and the force on Q₁ due to Q₂.

Solution    The force on Q₂ due to Q₁ is given by

where

Now the force on Q₁ due to Q₂

2.5 APPLICATIONS OF COULOMB’S LAW

Coulomb’s law is used to:

1. find the force between a pair of charges
2. find the potential at a point due to a fixed charge
3. find the electric field at a point due to a fixed charge
4. find the displacement flux density indirectly
5. find the potential and electric field due to any type of charge distribution
6. find the charge if the force and the electric field are known.

2.6 LIMITATION OF COULOMB’S LAW

It is difficult to apply the law when charges are of arbitrary shape. Here, the distance, r cannot be determined accurately as the centres of arbitrarily shaped charged bodies cannot be identified accurately.

2.7 ELECTRIC FIELD STRENGTH DUE TO POINT CHARGE

Electrostatic field is produced by a charge at rest. It is defined by Coulomb’s law. Electric field strength or electric field intensity or electric field indicate the same property in this book.

Definition 1    Electric field due to a charge is defined as the Coulomb’s force per unit charge. It is a vector and has the unit of Newton per Coulomb or volt per metre, that is,

where F is Coulomb’s force, Newton and Q is charge, Coulomb.

Definition 2    Electric field is also defined as a negative gradient of a potential due to a charge, that is,

 

where V is potential due to the charge, volts.

 

Let	Qf = fixed point charge, C
	Qt = test point charge, C
	rf = location of fixed charge
	rt = location of test charge

Then the force on Q_t due to a fixed charge in free space, Q_f is given by

The electric field, E at the location of Q_t due to Q_f is defined as the ratio of force on Q_t due to Q_f and the test charge, Q_t, that is,

If there are N point charges, the field at a point is the vectorial sum of fields due to N charges, that is,

 

2.8 SALIENT FEATURES OF ELECTRIC INTENSITY

1. It has units of Newton/Coulomb or volts/metre.
2. It is a vector.
3. It has both direction and magnitude.
4. Its direction is the same as that of Coulomb’s force.
5. Its magnitude depends on the magnitude of Coulomb’s force and the charge on which the force is acting.
6. It depends on the medium.
7. It depends on the permittivity of the medium.
8. It depends on the distance of the charge from another charge which produces Coulomb’s force.
9. It depends on the location of the charges.
10. It originates from a positive charge and terminates on a negative charge.
11. When a unit charge at a distance is moved around a fixed charge, the field lines and force appear as in Fig. 2.2.

Problem 2.9    If Coulomb’s force, F = 2a_x + a_y + a_z N, is acting on a charge of 10C, find the electric field intensity, its magnitude and direction.

Solution    Force,

The magnitude of E is

The direction of E is

Problem 2.10    If a charge of 2μC is located at P₁ (1, 0, 0) and another charge of 1 μC is located at P₂ (0, 1, 0) in free space, find the electric field at P₂.

Solution    Let

Electric field strength at P₂ is

As

Problem 2.11    There are three charges which are given by Q₁ = 1μC, Q₂ = 2μC and Q₃ = 3μC. The field due to each charge at a point, P in free space is a_x + 2a_y – a_z, a_y + 3a_z and 2a_x – a_y N/C. Find the total field at the point, P due to all the three charges.

Solution    The field, E₁ at P due to Q₁ (1μC)

 

The field, E₂ due to Q₂ (2μC)

 

The field, E₃ due to Q₃(3μC)

 

The total field at P,

Problem 2.12    A charge, Q₁ = –10nC is at the origin in free space. If the x-component of E is to be zero at the point (3, 1, 1), what charge, Q_t should be kept at the point (2, 0, 0)?

Solution    Consider Fig. 2.3.

E at (3, 1, 1)

If E_x is to be zero, we should have

Problem 2.13    Two point charges Q₁ = 5.0C and Q₂ = 1.0nC are located at (–1, 1, – 3) m and (3, 1, 0) m, respectively. Determine the electric field at Q₁.

 

Solution	Q1 = 5.0 C is at (–1, 1, –3) m
	Q2 = 1.0 nC is at (3, 1, 0) m

Electric field, E at Q₁ (–1, 1, –3) m is

2.9 ELECTRIC FIELD DUE TO LINE CHARGE DENSITY

By definition, line charge density is given by

Here, Q is the total charge.

But electric field due to Q at a distance of r is given by

Problem 2.14    If a fine filament carries a uniform charge distribution of ρ_L c/m, find the electric field at a point, P₂ (x₁, y₁, z₁).

Solution    Consider a differential length, dL on the line carrying the uniform charge. Let it be at a point P₁ (x₂, y₂, z₂) on the line.

The position            P₁ = r₁ = x₁a_x + y₁a_y + z₁a_z

The position            P₂ = r₂ = x₂ a_x + y₂ a_y + z₂a_z

 

r	=	r2 − r1
	=	(x2 − x1) ax + (y2 − y1) ay + (z2 − z1) az
|r|	=	r =|r2 − r1|
ar	=

The electric field strength at P₂ is

Problem 2.15    Find out electric field due to a uniformly charged short line (refer Fig. 2.4).

Solution    It may be noted that there exists symmetry with the line charge along z-axis. Assume that the origin is at the centre of the line.

Consider dL at a distance of L from the origin. The magnitude of the charge, dQ of dL length is given by

 

 

Now
dE

or,

Let

and               

and

2.10 ELECTRIC FIELD STRENGTH DUE TO INFINITE LINE CHARGE

The electric field due to a uniform infinite line charge is given by

Proof    For a line charge extending from –∞ to ∞ along z-axis, the electric field does not vary with z when ϕ and ρ are constants. It also does not vary with ϕ when z and ρ are constants due to symmetry.

The position of infinite line charge is shown in Fig. 2.5.

Consider a differential length, dL along the line charge. Let it be at a distance of L from the origin. By definition,

The charge dQ of dL = ρ_L dL.

Consider a point A on y-axis at a distance of ρ from the origin. Let the field at A due to dQ be dE.

From Coulomb’s law

Here,                  

So,

Put	L = ρ tan θ
	dL = ρ sec2 θ dθ

Now

As E_z = 0 due to symmetry, the field strength is given by

Problem 2.16    A charge density of ρ_L = 10 PC/m is uniformly distributed along an infinite line. Determine electric field at a point .

Solution    The field at is

where

But                    

or                        

So,

Problem 2.17    Find E at (2, 0, 2) if a line charge of 10 PC/m lies along the y-axis.

Solution    We have   

where

or,

Problem 2.18    Three parallel line charges, ρ_L1 = 5 nC/m, ρ_L2 = 4 nC/m and ρ_L3 = –6 nC/m are located at (0, 0), (3, 0) and (0, 4) m, respectively. Find D and E at (3, 4).

Solution    We have

                    ρ_L1 = 5 nC/m at (0,0)

                    ρ_L2 = 4 nC/m at (3, 0)

                    ρ_L3 = –6 nC/m at (0, 4)

                      D = D₁ + D₂ + D₃

where

or,

or,

Problem 2.19    Three infinitely long lines charged uniformly are parallel to the z-axis. They are separated by a distance of b m. The charge density of each is ρ_L = 2.0 PC/m. Find the electric field E at a point P on the y-axis at y = a m. If a = b = 1 m, what is the electric field, E?

Solution    Consider Fig. 2.6.

Expression for the electric field at a distance of P due to an infinitely long charged line, E is

E_y1 at the point P due to ρ_L1

E_y3 at the point P due to ρ_L3

E_y2 at the point P due to ρ_L2

As                ρ_L1 = ρ_L2 = ρ_L3 = ρ_L

Total

If a = b = 1m, ρ_L = 2.0 PC/m

Problem 2.20    An infinite length of uniform line charge has ρ_L = 10 PC/m and it lies along the z-axis. Determine the electric field E at (4, 3, 3) m.

Solution    In cylindrical coordinates,

Due to symmetry, E is constant with z. Hence

or,

Problem 2.21    An infinitely long uniform line distribution of ρ_L = 3.0 nC/m is at y = 3, z = 5. Determine E at (a) (0, 0, 0), (b) (0, 2, 1), (c) (3, 2, 1).

Solution    ρ_L = 3.0 × 10^–9 C/m is at (x, 3, 5).

Let us take a general point (x, y, z). Then the radial vector from the location of ρ_L, that is, (x, 3, 5) to (x, y, z) is

E at (x, y, z)

1. E at (0, 0, 0)
   
2. E at (0, 2, 1)
   
3. E at (2, 3, 1)
   

2.11 FIELD DUE TO SURFACE CHARGE DENSITY, ρ_s (C/m²)

Field due to surface charge density,

Consider an infinitely charged sheet lying in x-y plane. Assume that the sheet is uniformly charged.

Consider a strip of a differential width of dx as in Fig. 2.7.

The sheet extends from – ∞ to ∞ in both x and y directions. It is obvious that the field does not vary with x or y due to symmetry. Now, there is only z-component.

By definition,            

or,	dQ = ρs dS
	= ρs dx dy
that is,

For a differential strip of width dx, we have

 

The field at a point, A on z-axis is given by

Here,

that is,
or,

If the surface charge sheet lies in y-z plane, the field at a point on x-axis is

Similarly, if it is in x-z plane, the field at a point on y-axis is

In general, the field at a point on the axis normal to the plane of the sheet is given by

Problem 2.22    An infinite sheet in x-y plane extending from – ∞ to ∞ in both directions has a uniform charge density of 10 nC/m². Find the electric field at z = 1.0 cm.

Solution                 ρ_s = 10 nC/m²

For a sheet of charge lying in x-y plane, the field at any point on z-axis is given by

Problem 2.23    A sheet of charge lies in y-z plane at x = 0 and has uniform surface charge density of 5.0 PC/m². Find the electric field at a point P (– 5, 0, 0) on x-axis.

 

Solution	ρs = 5.0 PC/m2
	= 5.0 × 10–12 C/m2

E at P(–5, 0, 0)

Problem 2.24    A point charge, Q is at the centre of a neutral spherical conducting shell. Find the surface charge density at the inner surface and at the outer surface (Fig. 2.8).

Solution    The surface density at the inner surface is

The surface density at the outer surface is

Problem 2.25    A plane z = 1.0 m has a uniform charge density of ρ_s = 2.0 PC/m². Find the electric field E above the plane.

Solution                  ρ_s = 2.0 × 10^–12 C/m²

or,                      

Problem 2.26    Determine the force on a point charge of 5 nC at (0, 0, 5) m due to uniformly distributed charge of 5 mC over a circular disc of radius r ≤ 1 m in z = 0 plane.

Solution    Consider Fig. 2.9.

The surface charge density, ρ_s is

From Fig. 2.9

 

The differential charge on dS is

 

Then the differential force due to differential charge is

as

From Fig. 2.9, it is obvious that the radial components will be cancelled out due to symmetry.

The force on Q_t is

2.12 FIELD DUE TO VOLUME CHARGE DENSITY, ρ_v (C/m³)

Volume charge density is defined as

or,

Here, determination of field due to volume charge density simply involves the estimation of total charge, Q from ρ_v.

As

Problem 2.27    A sphere of volume 0.1 m³ has a charge density of 8.0 PC/m³. Find the electric field at a point (2, 0, 0) if the centre of the sphere is at (0, 0, 0).

Solution    The field, E due to volume charge density is

Problem 2.28    Find the electric field at a point on the axis of a charged disc (Fig. 2.10).

Solution    Consider a differential area shown by the dotted line.

The differential area          = 2πr dr

The field at P               

The vertical component of E is

at	r = 0, θ = 0
at	r = a, θ = α
Also	r = h tan θ
that is,	dr = h sec2 θ dθ

where            

Problem 2.29    Determine the charge enclosed in a cylinder (Fig. 2.11) when the volume charge density is ρ_v = 1.0e^–z (x² + y²)^–1/4 C/m³.

Solution         In cylindrical coordinates,

 

Hence the total charge

Problem 2.30    In a spherical region, the electric displacement is given by D = 10r² a_r mC/m². Find the total charge enclosed by the volume specified by r = 40 cm, θ = π/4 and ϕ = 2π.

Solution    The point form of Gauss’s law is

         ∇.D = ρ_v

and

But in the present problem, D is only a function of r, or, D_ϕ = 0, D_θ = 0.

So,

2.13 POTENTIAL

A charge at rest produces potential at a specified point. It is a scalar quantity.

 

The potential, V at a point due to a fixed charge, Q_f is given by

Simply            

2.14 POTENTIAL AT A POINT

The potential at a point due to a point charge is given by

Proof    Consider a fixed charge, Q and 1 C of charge at an infinite distance (Fig. 2.12). There exists a force on 1 C due to Q. If 1 C of charge is moved against the force of repulsion, some work has to be done.

If 1 C is at point B which is at a distance of x from Q, then force on 1 C due to Q is

If 1 C is moved through a distance of dx in the opposite direction of a_x, then the differential work done is

Total work done in moving 1 C from ∞ to r is

or,

This work done is the potential. The potential at a distance of r from Q is

Problem 2.31    A charge of 10 PC is at rest in free space. Find the potential at a point, A 10 cm away from the charge.

Solution     The potential at A is

Problem 2.32    The potential at a point A is 10 volts and at B it is 15 volts. If a charge, Q = 10μC is moved from A to B, what is the work required to be done?

 

Solution	VA = 10 V
	VB = 15 V
	VBA = 15 – 10 = 5 volt

By definition
	Q = 10μC
Work done	= VBA × Q
	= 5 × 10μC
	W = 50μJ

Problem 2.33    Two point charges Q₁ = 2 nC and Q₂ = 4 nC are located at (1, 1, 1) and (1, 0, 0) respectively. Determine the potential at P (1, 1, 0) due to the point charge.

Solution

                     Q₁ = 2 nC

                       r₁ = (1, 1, 1) = a_x + a_y + a_z

                       r_p = (1, 1, 0) = a_x + a_y

                r_p – r₁ = a_x + a_y – a_x – a_y – a_z

                          = –a_z

                |r_p – r₁| = 1.0

The potential at P due to Q₁ is

The potential at P due to Q₂ is

The total potential is    V = V₁ + V₂ = 18 + 36

Problem 2.34    An electric field is given by E = 10ya_x + 10xa_y, V/m. Find the potential function, V. Assume V = 0 at the origin.

Solution    We have

At x = 0 and y = 0, V = 0

2.15 POTENTIAL DIFFERENCE

The work done,         

or,               

where Q = charge that is being moved from A to B.

Potential difference between A and B is also defined as the difference between the potentials at A and B.

Let V_A be the potential at A. V_B be the potential at B. Then

2.16 SALIENT FEATURES OF POTENTIAL DIFFERENCE

1. Potential difference depends only on the initial and final points.
2. It does not depend on the path between the points.
3. It is zero around a closed path,

   that is,         

   or,            

4. Negative value of V_AB represents loss in potential energy in moving Q from A to B.
5. Positive value of V_AB represents gain in potential energy.
6. V_AB has the units of Joules/Coulombs or Volt.
7. V_AB depends on the distance between A and B.
8. V_AB is created by a fixed charge.
9. If B is reference at ∞ from the fixed charge, V_AB is the potential of A itself.
10. The potential at ∞ is zero.

Problem 2.35    A point charge, Q = 10 nC is at the origin. Determine the potential difference at A (1, 0, 0) with respect to B (2, 0, 0).

 

Solution	Q = 10 nC = 10 × 10–9 C
	rA = 1m
	rB = 2m

The potential difference,

2.17 POTENTIAL GRADIENT

2.18 SALIENT FEATURES OF POTENTIAL GRADIENT

1. Potential gradient is a vector.
2. It is always normal to equi-potential surfaces everywhere.
3. It lies in the direction of maximum increase of potential.
4. Negative potential gradient gives the electric field, that is,

    

   E = −∇V

    

5. The electric field and potential gradient are in opposite directions.
6. The potential gradient in different coordinate systems is given by
   

Problem 2.36    If the potential function, V is given by V = x³y – xy² + 3z, find the potential gradient.

Solution             V = x³y – xy² + 3z

The potential gradient is given by

2.19 EQUIPOTENTIAL SURFACE

2.20 POTENTIAL DUE TO ELECTRIC DIPOLE

Potential due to a dipole

Here,	Q = charge
	d= distance between two charges
	r = distance of the point from the centre of the pair of charges
	p = dipole moment = Qd (C-m)

Electric field at a point due to a dipole

Proof    Fig. 2.13 shows a dipole (A pair of charges)

Let P be a point at which the potential and electric field are to be determined. Let r₁ be the distance of P from (−) ve charge and be the distance of P from (+) ve charge and r₂ be the distance from the centre of the dipole.

From Fig. 2.13, we have

Potential at P due to Q and –Q is

If

The expression for the potential becomes

where p = Qd = electric dipole moment.

If d cos θ is written as

 

Potential V becomes

where p ≡ Qd is known as dipole moment. It is a vector and has the direction of d. Here d is the distance vector from –Q to Q. If the centre of the dipole is at r′ instead of at the origin, potential is given by

Problem 2.37    An electric dipole represented by 0.1a_y nC-m is at the origin. Find the potential at a point P (0, 10, 0).

Solution    We have

Problem 2.38    An electric dipole, 1.0a_y nC-m is located at (0, 0, 0). Find the potential at

Solution    The potential due to an electric dipole is given by

where

Hence      

2.21 ELECTRIC FIELD DUE TO DIPOLE

In spherical coordinates, the gradient of the potential is given by

where      

It is evident from the above expression that V is not a function of ϕ. Therefore,

E is only a function of r and θ.

or in terms of electric dipole moment, E is expressed as

Problem 2.39    An electric dipole, 1.0a_z nC-m is at (0, 0, 0). Find the electric field at (0, 0, 1).

Solution    We have

Problem 2.40    If an electric dipole located at the origin is represented by 0.1a_z nC-m, find E at

Solution    We have

2.22 ELECTRIC FLUX

Electric flux is also known as Electric Displacement Flux.

Definition 1    Electric flux is defined as the displaced charge, that is,

Electric flux, Ψ ≡ Q, Coulomb.

Definition 2    Electric flux is defined as the surface integral of electric flux density, that is,

Electric flux, Ψ ≡

2.23 SALIENT FEATURES OF ELECTRIC FLUX

1. It is independent of the medium.
2. The electric field creates a force on a charge and hence the charge moves along a certain path. This path is called the flux line.
3. The force between two charges acts along a certain path. This path is also called the flux line.
4. Magnitude of flux depends only on the charge from which it originates.
5. The flux lines are equal to the charge in Coulombs.
6. Flux line is only an imaginary line.
7. Its direction is the same as that of the electric field.
8. The flux lines from a point charge are shown in Fig. 2.14.
   

   Fig. 2.14 Field from an isolated charge

9. The flux lines between a (+)ve and a (–)ve point charges are shown in Fig. 2.15.
   

   Fig. 2.15 The field in a system of equal but opposite charges

10. It is a scalar quantity.
11. The flux lines from a pair of (+)ve charges are shown in Fig. 2.16.

2.24 FARADAY’S EXPERIMENT TO DEFINE FLUX

Apparatus used    One small metallic sphere, two hemispheres forming a full sphere larger than the small sphere, if held together over a dielectric material.

Procedure    See the steps given below:

1. Inner sphere is charged to Q Coulombs.
2. Dielectric material is pasted radially over the sphere with uniform thickness.
3. Two hemispheres are held together on the dielectric strip so that they form an outer sphere.
4. Outer sphere is momentarily grounded to discharge its charge.
5. The two hemispheres are removed with insulated tools without disturbing the induced charge.
6. The charge on each sphere is measured.
7. The total induced charge on the outer sphere is found to be negative.
8. The magnitude of the induced charge is equal to that of the inner sphere.
9. The displacement took place from inner to outer sphere through the dielectric material.
10. This displacement is known as displacement flux.
11. The displacement flux is also called electric flux.
12. This flux is independent of the type of dielectric material.
13. It is independent of the separation between inner and outer spheres.
14. The final result of Faraday’s experiment is,

2.25 ELECTRIC FLUX DENSITY

This is also known as Displacement Electric Flux Density.

Definition 1    Electric flux density, D is defined as

where ψ is the electric flux crossing the differential area, dS. The direction of dS is always outward, normal to dS, that is, dS = dS a_n.

Definition 2    Electric flux density, D is also defined as

 

 

where	∈0 = permittivity of free space, F/m
	E = electric field strength, V/m

2.26 SALIENT FEATURES OF ELECTRIC FLUX DENSITY, D

1. The unit of electric flux density is C/m².
2. It is a vector.
3. It is inversely proportional to r², r being the radius of the sphere.
4. In free space, D is in the direction of E.
5. D in a Gaussian surface is determined from Gauss’s law.
6. D is independent of the medium.
7. D is given by
8. D in a general medium is given by

    

               D = ∈ E

    

9. D in a dielectric medium is given by

             D = ∈₀ E + P

             P = polarisation of medium

Problem 2.41    If an electric field in free space is given by

             E = a_x + 2a_y + 5a_z V/m.

find the electric flux density.

Solution    Electric field, E = a_x + 2a _y + 5a _z

Problem 2.42    A point charge, Q = 10 nC is at the origin in free space. Find the electric field at P (1, 0, 1). Also find the electric flux density at P.

Solution

                Q = 10 nC = 10 × 10⁻⁹ C

                P = (1, 0, 1)

                r = (1, 0, 1) − (0, 0, 0)

                   = a_x + a_z

Electric field,

The electric flux density,

Problem 2.43    What is the electric flux, ψ that passes the surface shown in Fig. 2.17, if the displacement flux density is D = ya_x + xa_y mC/m².

Solution    Electric flux density is given by

            D = ya _x + xa _y (mC/m²)

The differential area, dS is given by

             dS = dx dz a _y

The differential flux, dψ passing through dS is

2.27 GAUSS’S LAW AND APPLICATIONS

Generalised Faraday’s law is Gauss’s law.

 

This is known as Gauss’s law in integral form. Gauss’s law is applicable only on Gaussian surfaces.

Proof    Consider a spherical surface which encloses a charge Q at its centre (Fig. 2.18).

The differential area, ds is on the surface of the sphere whose direction is a_n. Let r be the radius of the sphere.

The electric field, E at the spherical surface is given by

But

Taking dot product with ds on both sides, we get

For the spherical surface under consideration, a_r and a_n are in the same direction.

Taking surface integral on both sides, we get

But S = 4πr² for a sphere. So,

2.28 PROOF OF GAUSS’S LAW (ON ARBITRARY SURFACE)

Consider an arbitrary surface of Fig. 2.19.

Consider a sphere of radius one metre within the surface, S which encloses a charge, Q at its centre.

Electric flux density, D at ds on the arbitrary surface is

Taking dot product with ds on both sides

Here, if a_r and a_n make an angle of θ, then

Now taking surface integral on both sides,

On the right hand side, the integrand consists of The numerator dscosθ represents the projection of area ds on the spherical surface whose radius is r. Hence will be the projection of ds on the spherical surface of radius equal to unity. This is the solid angle subtended by an area ds at the location of the point charge. Therefore, is the total solid angle subtended by s at the point charge. It is the sum of the projections of all ds on the spherical surface of radius unity and centered at Q. This is equal to the area of the spherical surface of unity radius. Hence

2.29 GAUSS’S LAW IN POINT FORM

Gauss’s law in point form states that the divergence of electric flux density is equal to the volume charge density, that is,

 

Proof    Consider a differential parallelepiped as shown in Fig. 2.20.

Assumptions

1. The differential parallelepiped has the dimensions ∆x, ∆y and ∆z.
2. The point, P is in the centre of the element.
3. The flux density D at the centre is given by

    

   D = D_c = D_cx a_x + D_cy a_y + D_cz a_z

The integral form of Gauss’s law is

where

Face 1 represents the face ABCD

Face 2 represents the face EFGH

Face 3 represents the face ABFE

Face 4 represents the face DCGH

Face 5 represents the face ADHE

Face 6 represents the face BCGF

D₁, D₂, D ₃, D₄, D₅ and D₆ are the flux densities on the faces 1, 2, 3, 4, 5 and 6 respectively.

dS₁, dS₂, dS₃, dS₄, dS₅ and dS₆ are differential areas of the faces 1, 2, 3, 4, 5 and 6, respectively.

As the flux density at the centre is known, it is found on each face of the parallelepiped by considering the first two terms of Taylor’s theorem.

Taylor’s theorem states that if f (x) has continuous derivatives in the neighbourhood of a point x = a, then

If (x − a) is very small, we have

Accordingly, we can simplify D₁, D₂, D₃, D₄, D₅ and D₆

Similarly

If

This becomes exact if Δv → 0.

But

2.30 DIVERGENCE OF A VECTOR, ELECTRIC FLUX DENSITY

This means that the divergence of flux density is the outflow of electric flux from a closed surface per unit volume as the volume shrinks to zero.

The point form of Gauss’s law in different coordinate systems is

2.31 APPLICATIONS OF GAUSS’S LAW

1. It is useful to find flux, ψ or flux density, D from the knowledge of enclosed charge and surface.
2. It is useful to find the electric field from the knowledge of enclosed charge and surface.
3. It is useful to find the enclosed charge from the knowledge of either D or E.

2.32 LIMITATIONS OF GAUSS’S LAW

1. It cannot be applied on Non-Gaussian surfaces.
2. It can be applied only if the surface encloses the volume completely.

2.33 SALIENT FEATURES OF GAUSS’S LAW

1. It relates volume integral to surface integral.
2. It is applicable only if the surface completely encloses the volume.
3. It does not specify any particular shape for the closed surface.
4. It does not require knowledge of the distance from the points on the surface.
5. It cannot be applied if the charges are outside the surface.
6. It is useful to find D or E from the knowledge of the charge enclosed and the surface.
7. It is applicable only on Gaussian surfaces.
8. It is useful to find the outward flow of flux from a closed surface from the enclosed charge.

Problem 2.44    If electric flux density, D is given by,

       D = [(2y² + z) a_x + 4xy a_y + xa_z)] μC/m²

find the volume charge density at (0, 0, 0) and (−1, 0, 4).

Solution

But             ρ_v = volume charge density = ∇.D

2.34 POISSON’S AND LAPLACE’S EQUATIONS

Proof    The point form of Gauss’s law is

 

	∇.D = ρv
But	D = ∈ E
and	E = −∇V
	∇.D = ∇.∈ E = ∇.∈ (−∇V) = ρv

or,               

[as ∇ . ∇ = ∇²]

where ∇² is a scalar operator and is called Laplace’s operator.

In the regions where ρ_v = 0, Poisson’s equation becomes

Laplace’s equation in one dimensional form is given by

Its solution is in the form of

 

Laplace’s equation in two dimensional form is given by

Its solution is given by

where r is the radius of a circle about a point (x, y).

Laplace’s equation in three dimensional form is given by

The value of V at point p is the average value of V over a spherical surface of radius r centered at p and it is given by

Poisson’s equation in different coordinates

2.35 APPLICATIONS OF POISSON’S AND LAPLACE’S EQUATIONS

1. They are useful in determining surface charge densities and electric field in the regions of interest.
2. They can be used to find the capacitance of different structures by appropriate application of boundary conditions.

2.36 UNIQUENESS THEOREM

Proof    Assume that V₁ and V₂ are the solutions of Laplace’s equation and V_1b and V_2b are the potentials on the boundaries. By hypothesis,

 

	∇2 V1 = 0
and	∇2 V2 = 0
	∇2 (V1 − V2) = 0

As V_1b and V_2b are the values on the boundary we have

 

	V1b = V2b = Vb
or,	V1b − V2b = 0

Consider a vector identity, namely

 

where ψ is a scalar and A is a vector quantity.

This is valid for any scalar, ψ and any vector, A.

In the present case, let ψ = V₁ – V₂, A = ∇ (V₁ – V₂). Then, the above identities become

 

Take volume integral on both sides

Applying divergence theorem to the left hand side, volume integral is replaced by surface integral,

that is,

But on the boundary, this becomes

Right hand side is zero because V_1b = V_2b. By hypothesis

So the remaining integral is

This is possible if

1. Integrand is zero.
2. Integrand is positive in some region and negative in some other region.

The second condition cannot be true as it is a square term.

So the integrand is zero,

that is,       [∇(V₁ − ∇₂)]² = 0 or ∇ (V₁ − V₂) = 0

If the gradient of a scalar is zero everywhere, then (V₁ – V₂) cannot change with any coordinate. Hence (V₁ – V₂) = constant.

The constant is easily evaluated by considering a point on the boundary. Here V₁ – V₂ = V₁b – V₂b = 0,

that is,                    Hence proved.

Problem 2.45    Consider concentric spherical shells in free space in which V = 0 volts at r = 10 cm and V = 10 volts at r = 20 cm. Find E and D.

Solution    Here V is a function of only r and not θ, and ϕ. Then Laplace’s equation

Integrating twice, we get

The boundary conditions are

V = 0 V at r = 10 cm = 0.1 m and

V = 10 V at r = 20 cm = 0.2 m,

that is,

So,          

But            

or,

or,            

Problem 2.46    There exists a potential of V = –2.5 V on a conductor at 0.02 m and V = 15.0 V at r = 0.35 m. A dielectric material whose ∈_r = 3.0 exists between the conductors. Determine the surface charge densities on the conductors.

Solution    As V is a function of only r, Laplace’s equation is given by

Integrating twice, we get

A and B can be found by the use of boundary conditions,

that is,

Solving, we get

                    A = 37.12 × 10⁻² , V-m

                    B = 16.06

and                

But

and

or,            

On the conductor surfaces,

 

	Dn = ρs = surfacedensity
At	r = 0.02 m
	ρs = −2.4649 × 10−8 C/m2

or,                     

At

or,                     

Problem 2.47    In what manner does permittivity vary to satisfy Laplace’s equation in a non-homogeneous, charge-free space?

Solution    For charge-free space,

 

	ρv = 0
	∇.D = 0
	∇.∈ E = 0	[as D = ∈ E]
	∇.(−∈ ∇V) = 0	[as E = − ∇V]
	∇.(∈ ∇V) = 0

If ∈ varies spatially, then

 

	∇.(∈∇V) = ∇V.∇∈ + ∇2 V = 0
As	∇2 V = 0
	(∇V).(∇∈) = 0

This is true only when ∇V and ∇∈ are perpendicular to each other. Hence permittivity should vary so that its gradient is perpendicular to the electric field.

Problem 2.48    If a potential V = x² yz + Ay³ z, (a) find A so that Laplace’s equation is satisfied (b) with the value of A, determine electric field at (2, 1, −1).

Solution    (a) Laplace’s equation is

 

or,         

As             V = x² yz + Ay³ z

The above equation becomes

 

So,            

(b) With

But

2.37 BOUNDARY CONDITIONS ON E AND D

1. The tangential component of E is continuous across any boundary, that is,
   or

   The tangential component of E in medium 1 is the same as that of E in medium 2 at any boundary.

2. The normal component of D is continuous across any boundary except at the surface of the conductor. In general,
   

   ρ_s = surface charge density, (C/m²). For any point other than the conductor boundary, D_n1 = D_n2.

2.38 PROOF OF BOUNDARY CONDITIONS

Consider the rectangular loop on the boundary of two media (Fig. 2.21).

It is well known that electric field is conservative and hence the line integral of E .dL is zero around a closed path,

that is,               

From the figure shown above, LHS is written as

As Δy → 0, we get

Thus,                   E_x1 = E_x2

It is obvious that E_x1 and E_x2 are the tangential components of E in medium 1 and 2 respectively.

So,             E_tan1 = E_tan2

Now consider a cylinder across the media 1 and 2 (Fig. 2.22).

According to Gauss’s law,

Applying this to the cylindrical surface on the boundary spreading over medium 1 and medium 2, we get Δh → 0

Problem 2.49    The region y < 0 contains a dielectric material for which ∈_r1 = 2.0 and the region y > 0 contains a dielectric material for which ∈_r2 = 4.0. If E₁ = –3.0a_x + 5.0a_y + 7.0a_z V/m, find the electric field, E₂ and D₂ in medium 2.

Solution    As y < 0 belongs to medium 1 and y > 0 belongs to medium 2

                E_tan1 = −3.0a_x + 7.0a_z V/m

                  E_n1 = 5.0a_y V/m

                  ∈_r1 = 2

                  ∈_r2 = 4

The boundary condition on tangential component of E is

                E_tan1 = E_tan2

                E_tan2 = −3.0a_x + 7.0a_z V/m

and

Problem 2.50    An electric field in medium 1 of ∈_r1 = 7 passes into a medium 2 of ∈_r2 = 2. When the field, E makes an angle of 60° as shown in Fig. 2.23 with the axis normal to the boundary line, find the angle made by the field with the normal in medium 2.

Solution    From Fig. 2.23

According to the boundary conditions,

and

2.39 CONDUCTORS IN ELECTRIC FIELD

Conductors

Definition 1    Conductors are materials which have very low resistance. Examples: Copper, Silver and Aluminium.

Definition 2    Conductors are materials for which no forbidden gap exists between valance band and conduction band.

Definition 3    A material is defined as a conductor

Conductors have a variety of applications in all fields of life.

2.40 PROPERTIES OF CONDUCTORS

1. Charge density is zero within a conductor.
2. The surface charge density resides on the exterior surface of a conductor.
3. In static conductors, current flow is zero.
4. Electric field is zero within a conductor.
5. Conductivity is very large.
6. Resistivity is small.
7. Magnetic field is zero inside a conductor.
8. Good conductors reflect electric and magnetic fields completely.
9. A conductor consists of a large number of free electrons which constitute conduction current with the application of an electric field.
10. A conductor is an equipotential body.
11. The potential is same everywhere in the conductor.
12. E = –∇V = 0 in a conductor.
13. In a perfect conductor, conductivity is infinity.
14. When an external field is applied to a conductor, the positive charges move in the direction E and the negative charges move in the opposite direction. This happens very quickly.
15. Free charges are confined to the surface of the conductor and hence surface charge density, J_s is induced. These charges create internal induced electric field. This field cancels the external field.

    It is interesting to note that copper and silver are not super conductors but aluminium is a superconductor for temperature below 1.14 K.

2.41 ELECTRIC CURRENT

The current through a given medium is defined as charge passing through the medium per unit time. It is a scalar, that is,

Current is of three types.

1. Convection current
2. Conduction current
3. Displacement current

1. Convection current    It is defined as the current produced by a beam of electrons flowing through an insulating medium. This does not obey Ohm’s law. For example, current through a vacuum, liquid and so on is convection current.
2. Conduction current    It is defined as the current produced due to flow of electrons in a conductor. This obeys Ohm’s law. For example, current in a conductor like copper is conduction current.
3. Displacement current    It is defined as the current which flows as a result of time-varying electric field in a dielectric material. For example, current through a capacitor when a time-varying voltage is applied is displacement current.

2.42 CURRENT DENSITIES

In electromagnetic field theory, it is of interest to describe the events at a point instead of in a large region. This is the reason why current densities are considered. Current densities are vector quantities.

 

Current densities are of three types:

1. Convection current density
2. Conduction current density
3. Displacement current density

1. Convection current density (A/m²)    It is defined as the convection current at a given point through a unit normal area at that point, that is,

   Convection current density

   

   where	dI = differential convection current
   	dS = differential area
   	= dSan
   	an = outward unit normal to dS

   As convection current density is confined to specific media, it is not of much interest in this book.

2. Conduction current density, J_c (A/m²)    It is defined as the conduction current at a given point through a unit normal area at that point,

    

   that is,	Jc ≡ σE
   and

   Conduction current density exists in the case of conductors when an electric field is applied.

3. Displacement current density, J_d (A/m²)    It is defined as the rate of displacement electric flux density with time, that is,
   

   If I_d is the displacement current in a dielectric due to applied electric field, displacement current density is defined as

   

   As

   

   In fact, displacement current density exists due to displacement of bound charges in a dielectric by the applied electric field.

2.43 EQUATION OF CONTINUITY

Equation of continuity in integral form is

            I = outward flow of current (A)

            J = conduction current density (A/m²)

where               

Proof    If Q_i is the charge inside a closed surface, the rate of decrease of charge due to the outward flow of current is given by

From the principle of conservation of charge, we have

From divergence theorem, we have

So,

Two volume integrals are equal if the integrands are equal. So,

In the above equation the derivative became a partial derivative as the surface is kept constant.

2.44 RELAXATION TIME (T_r)

It is also called rearrangement time.

Relaxation time is defined as the time taken by a charge placed in a material to reach 36.8 per cent of its initial value. It is given by

where ∈ = permittivity (F/m), σ = conductivity (mho/m).

Problem 2.51    Find the relaxation time of sea water whose ∈_r = 81 and σ = 5 mho/m.

Solution    Relaxation time of sea water

Problem 2.52    Find the relaxation time of porcelain whose σ = 10^–10 mho/m, ∈_r = 6.

Solution    Relaxation time of porcelain

2.45 RELATION BETWEEN CURRENT DENSITY AND VOLUME CHARGE DENSITY

            J = conduction current density, A/m²

            V = velocity of the charge (m/s)

Proof    We know that   

Consider an element charge (Fig. 2.24)

 

Assume that the charge element is oriented parallel to the coordinate axes. Let there be only an x-component of velocity. It moves a distance of Δx in a time Δt as in the figure. Therefore,

 

The resultant current is

where V_x = x -component of velocity of the charge.

Similarly, if the charge moves in y and z-directions, we get

 

	Jy = ρv Vy
	Jz = ρv Vz
	J = ρv (Vx ax + Vy ay + Vz az)
As	V = Vx ax + Vy ay + Vz az
	Hence proved.

Problem 2.53    If the current density, find the current passing through a sphere radius of 1.0 m.

Solution      

where

where

that is,

Problem 2.54    Find the electric flux density in free space if the electric field, E = 6a_x – 2a_y + 3a_z , V/m.

Solution    Electric field in free space,

              E = 6a_x − 2a_y + 3a_z , V/m

            ∈₀ = 8.854 × 10⁻¹²

The electric flux density,

2.46 DIELECTRIC MATERIALS IN ELECTRIC FIELD

Definition 1    An ideal dielectric material is one which does not contain free electrons.

Definition 2    An ideal dielectric material is one in which the charges are well bounded and cannot be set in motion easily.

Definition 3    An ideal dielectric material is one for which there exists a large forbidden gap between valance band and conduction band.

Definition 4    A material is defined as dielectric material if

Definition 5    A material is defined as dielectric material if it does not conduct electric current and opposes the flow of current.

2.47 PROPERTIES OF DIELECTRIC MATERIALS

1. Conductivity is zero.
2. Volume charge density, ρ_ν = 0.
3. Electric and magnetic fields exist in a dielectric material.
4. Resistivity is ∞.
5. Electric and magnetic fields penetrate the dielectric material freely.
6. There exists no free electrons.

Dielectrics in Electric Field

An atom of a dielectric consists of a nucleus and a bunch of electrons. Similarly, a molecule of a dielectric consists of nuclei and a set of electron bunches. The charge of the nucleus is positive and the charge of the electron bunch is negative. The atoms and molecules are electrically neutral as they contain an equal number of negative and positive charges.

Dielectrics are classified into polar and non-polar type of materials.

Polar type of dielectrics

The centres of positive and negative charges of a molecule of polar type of dielectric material are separated by a small distance. Each pair acts as a dipole and there exists a dipole moment. However, such pairs are randomly distributed in a dielectric material. Hence the overall dipole moment is zero.

If such a material is kept in an electric field, all the positive charges move in the direction of the electric field and all the negative charges move in the opposite direction. As a result, dipole moment is induced by the electric field. Under these conditions, the material is said to be under a state of polarisation.

A polar type of molecule is shown in Fig. 2.25.

Examples of polar dielectrics are water, hydrochloric acid, sulphur dioxide and others.

Non-polar type of dielectrics

The centres of positive and negative charges of a molecule of non-polar type of dielectric material coincide as in Fig. 2.26, that is, there is no separation between them and hence dipole moment is zero.

However, when such a material is placed in an electric field, the centres of positive and negative charges are displaced and there exists a distance between them, that is, dipole moment is induced. Under these conditions, the material is said to be under a state of polarisation.

Examples of non-polar dielectrics are oxygen, hydrogen, nitrogen and so on.

The important conclusion is that dielectric materials are not polarised in the absence of electric field and they are polarised in the presence of an electric field. As a result, the electric flux density is greater than that in free space conditions with the same field intensity. The intensity of polarisation is described in terms of dipole moment and polarisation.

2.48 DIPOLE MOMENT, p

that is,	p ≡ Qd, (C-m)
where	p = dipolemoment
	Q = charge magnitude
	d = distance vector from(−)ve to(+)ve charges of the dipole

If there are N dipoles in a dielectric material of volume Δν, the total dipole moment with the application of electric field is

2.49 POLARISATION, P

that is,

In some dielectrics, the polarisation, P is defined as

 

	P ≡ χe ∈0 E
where	χe = electric susceptibility of the dielectric

The charges are bound in a dielectric and the total positive bound charge on a surface, S enclosing the dielectric is given by

On the other hand, the charge that remains inside the surface, S is –Q_b and it is given by

If there is some free charge in the dielectric, the free charge volume charge density is ρ_ν. If ρ_b is the bound volume charge density, then the total volume charge density is given by

                ρ_t = ρ_v + ρ_b

                    = ∇.∈ E

or,

As P ≡ χ_e ∈₀ E, we have

                D = ∈₀ E + χ_e ∈₀ E

                    = ∈₀ E (1 + χ_e)

                D = ∈₀ ∈_r E = ∈ E

where        

or,            

In summary, we have

Problem 2.55    A pair of negative and positive charges of 10 μC each are separated by a distance of 0.1 m along the x-axis. Find the dipole moment.

 

Solution	Q1 = −10μC
	Q2 = 10μC
	d = 0.1ax
	Q = 10μC

The dipole moment is

Problem 2.56     If a dielectric material of ∈_r = 4.0 is kept in an electric field E = 3.0a _x + 2.0a _y + a _z, V/m, find the polarisation.

Solution

          ∈_r = 4.0

          E = 3.0a_x + 2.0a_y + a_z

Polarisation in the dielectric,

       P = χ_e ∈₀ E

       χ_e = ∈_r −1

            = 4 − 1 = 3

       P = 3 × 8.854 × 10⁻¹² × (3a_x + 2a_y + a_z)

          = (79.68a_x + 53.12a_y + 26.56a_z) PC/m²

Problem 2.57     What are the magnitudes of electric flux densities and polarisation for a dielectric material in which E = 150 kV/m? Electric susceptibility of the dielectric material is 4.75.

Solution     We have

Polarisation is

Problem 2.58     Find the polarisation, P in a homogeneous and isotropic dielectric material whose ∈_r = 3.0 when D = 3.0a_r μC/m².

Solution     Polarisation,

 

		P = χe ∈0 E
	and	D = ∈0 ∈r E, χe = ∈r − 1
		D = ∈0 E + P
	or,	P = D−∈0 E

Problem 2.59     If the polarisation P = 3a _x nC/m² in a homogeneous and isotropic dielectric material whose χ_e = 4.5, find E in the material.

Solution     We have

       P = 3a_x nC/m²

          = χ_e ∈₀ E

or,

Problem 2.60     A dielectric slab (∈_r = 2) is placed under the influence of electric flux density = 10a_x C/m² . The slab has a volume of 0.1 cm³ . Determine the polarisation in the slab and total dipole moment.

Solution     The electric flux density,

 

	D = ∈0 E + P
or,	P = D −∈0 E

Dipole moment,	p = polarisation × volume
Slab volume	= 0.1 cm3 = 0.1 × (10−2)3 = 10−7 m3
Dipole moment,

Problem 2.61     What are the magnitudes of P and D for a dielectric material in which E = 1.0 V/m and χ _e = 5.0?

Solution

2.50 CAPACITANCE OF DIFFERENT CONFIGURATIONS

Any two conducting bodies separated by free space or a dielectric material exhibit a capacitance between them.

 

that is,             

Capacitance depends only on the geometry of the system and properties of the dielectrics involved. It does not depend on Q and V.

The capacitance of a capacitor is the ability to store electric charge. It opposes sudden changes in voltage. Its reactance is ohms. Here f is the frequency and C is capacitance in Farads.

Capacitance of a parallel plate capacitor It is given by

where	= permittivity of the dielectric between the conductors (F/m)
	A = area of the conductor
	d = distance between the conductors

Proof     Consider Fig. 2.27.

Let Q be the absolute charge on one of the plates, V be the potential difference between them, d be the distance between the plates, A be the area of the plate and ∈ be the permittivity of the medium between the plates.

Then the electric flux density is

Potential difference between the plates is

Capacitance of parallel plate capacitor of n dielectric slabs

It is given by          

 

where	di = width of the ith slab
	∈i = permittivity of the ith slab

Proof     Let ∈₁, ∈₂, ∈₃,…, ∈_n be the permittivity of dielectric materials between the plates; d₁, d₂, d₃,…, d_n respectively be their thickness. (Fig. 2.28).

The potential difference across the capacitor is

If Q is the charge on the plates, then the electric flux density, D is

Then

or,

Hence         Hence proved.

Capacitance between two concentric spheres

It is given by

where	r1 = radius of the inner sphere
	r2 = radius of the outer sphere

Proof     Refer Fig. 2.29.

The radial electric field is given

Potential difference between the two spheres is given by

If r₂ is ∞, the structure will result in an isolated sphere.

The capacitance of an isolated sphere of radius r,₁ = r is

 

Capacitance of a coaxial cable

It is given by

where	l = length of the cable
	ρ1, ρ2 = radius of inner and outer conductors
	∈ = permittivity of the dielectric between the conductors

Proof     Refer Fig. 2.30.

Let ρ₁, and ρ₂ be the radii of the inner and outer conductors of the coaxial cable. Let ρ_L be the line charge density on the inner conductor and -ρ _L be that on the outer conductor. Then the electric field in the radial direction is

The potential difference between the cylinders is

or,

The capacitance of a cable of length l metres is

Capacitance of parallel wires

It is given by

where	d = distance between the wires
	r = radius of each conducting wire

Proof     Refer Fig. 2.31.

Let ρ_L and -ρ_L be the line charge densities of the lines A and B respectively. Let d be the distance between the wires and r be the radius of each wire.

Electric field between the wires at the point P due to ρ_L is

Electric field at P due to -ρ_L is

The potential difference, V

that is,

In all practical cases, d >> r. As a result

So,

The capacitance,

or capacitance of a pair of wires of length l metres is, therefore, given by

Problem 2.62     Find the capacitance of an isolated sphere of radius 1 cm.

Solution     The expression for the capacitance of an isolated sphere is

 

	C = 4π ∈0 r
Here	r = 1 cm = 0.01m
	C = 4π ∈0 × 0.01 m

Problem 2.63     A parallel plate capacitor has conducting plates of area equal to 0.04 m². The plates are separated by a dielectric material whose ∈_r = 2 with the plate separation of 1 cm. Find (a) its capacitance value (b) the charge on the plates when a potential difference of 10 V is applied (c) the energy stored.

Solution     (a) The capacitance of parallel plate capacitor, C

 

where	∈ = ∈0∈r = 2 ∈0
	A = 0.04 m2
	d = 1 cm = 0.01 m

(b) We have

(c) Energy stored

Problem 2.64     Find the capacitance for a 10 km long coaxial cable shown in Fig. 2.32.

Solution     For the coaxial cable shown, the inner radius, a = 1 cm, and outer radius, b = 1.6 cm.

The capacitance of the cable is,

Problem 2.65     The cable shown in Fig. 2.33 is 10 km long. If r₁ = 10 mm, r₂ = 15 mm, r₃ = 20 mm, ∈_r1 = 2.0, _r2 = 4.0, find the capacitance of the cable.

Solution     The capacitance of the two inner conductors

The capacitance of the two outer conductors

As these two are in series, the resultant capacitance, C

But

2.51 ENERGY STORED IN AN ELECTROSTATIC FIELD

 

When a positive charge is brought from a distance of infinity to a point in a field of another positive charge, work is done by an external source. Energy spent in doing so represents the potential energy.

If the external source is removed, the charge that is brought moves back, acquiring kinetic energy of its own and it is capable of doing some work.

If V is the potential at a point due to some fixed charge,

Work done    = potential energy

that is,        W_E = QV

where Q is the charge brought by an external source,

V is the potential at the point due to a fixed charge.

Let us consider two charges Q1 and Q₂ separated by a distance of infinity. If Q₁ is fixed, work done on bringing Q₂ towards Q₁ is given by

where                

Similarly, consider another charge, Q₃ which is at infinity from Q₁ and Q₂. Now work done in bringing Q₃ towards Q₁ and Q₂

This is because there exists force due to Q₁ and Q₂ after Q₂ is brought toQ₁.

In the above equation, are the potentials at Q₃ due to Q₁ and Q₂ respectively. Therefore, total work done in bringing Q₂ and Q₃ is

 

In a similar fashion, consider n charges. Then we have

that is,             

where is the potential of Q_j at the location of Q_i. Note that

W_t may be written as

Adding the above two equations and simplifying, we get

	= Q1 × (potential at Q1 due to all other charges) + Q2 × (potential at Q2 due to all other charges) + Qn × (potential at Qn due to all other charges)
	= Q1 V1 + Q2 V2 + …Qn Vn

or,

This equation represents the potential energy stored in a system of n point charges.

 

If
But	∇.D = ρυ or ∇.E = ρυ/∈0

From standard vector identity,

 

this becomes

Applying divergence theorem, first term on the right hand side can be written as

However, viewing from a surface bounding complete space, the charge distribution of finite volume appears as a point charge, say Q. We know that

and

From the expressions of E and V, we get

Hence W_t becomes

or,

This expression is the total energy stored in an electrostatic field. Therefore, the integrand represents energy density, that is,

The energy density,

2.52 ENERGY IN A CAPACITOR

Proof     Method 1 We know that energy stored in the electric field of a capacitor is given by

If the space between the conductors is occupied by a dielectric material whose relative permittivity is ∈_r, then

As the plates are assumed to be separated in x-direction,

and

Method 2     When a capacitor is charged, energy is stored in the electrostatic field which is set up in the dielectric medium.

Assume that a capacitor C is charged to a voltage, V. If the potential difference across the plates at any instant of charging is V, this is equal to the work done in shifting one coulomb of charge from one plate to another. If dQ is the charge transferred, work done is

 

As per the definition of C,

 

	Q = CV
	dQ = C dV
or,	dWc = CV dV

Total work done in producing a potential of V is

Energy stored in a capacitor

 

1. Coulomb’s force depends on the medium in which the charges are placed.	(Yes/No)
2. Constant of proportionality in Coulomb’s law has units.	(Yes/No)
3. The directions of electric field and Coulomb’s force are the same.	(Yes/No)
4. For a thin filament extending from –∞ to ∞ along the z-axis, field varies with z and ϕ coordinates.	(Yes/No)
5. For a line charge extending from –∞ to ∞ along the z-axis, field at a point varies with ρ only.	(Yes/No)
6. For N point charges, the electric field at a point is the vectorial sum of the fields due to each of N charges.	(Yes/No)
7. Coulomb’s law can be applied to find electric field at a point.	(Yes/No)
8. Charge distributions produce electric field.	(Yes/No)
9. Electric field at a point due to a point charge is inversely proportional to r	(Yes/No)
10. Electric field at a point due to a point charge is proportional to 1/r2.	(Yes/No)
11. A surface of a conductor is an equipotential surface.	(Yes/No)
12. Gradient of the potential and equipotential surface are orthogonal to each other.	(Yes/No)
13. Electric dipole is a pair of two positive point charges.	(Yes/No)
14. Electric dipole is a pair of a positive charge and a negative charge.	(Yes/No)
15. Displacement electric flux is equal to the charge enclosed.	(Yes/No)
16. Faraday’s experiment says ψ = Q.	(Yes/No)
17. Gauss’s law is applicable on all surfaces.	(Yes/No)
18. Gauss’s law is applicable only on Gaussian surfaces.	(Yes/No)
19. Gauss’s law is applicable even if the charge is outside the closed surface.	(Yes/No)
20. Displacement flux depends on the dielectric material.	(Yes/No)
21. Laplace’s equation has several solutions.	(Yes/No)
22. Etan1 = Etan2	(Yes/No)
23. Dn1 = Dn2	(Yes/No)
24. Aluminium is a better conductor than silver at very low temperatures.	(Yes/No)
25. For a good conductor,	(Yes/No)
26. Convection and conduction current densities are identical.	(Yes/No)
27. Convection current obeys Ohm’s law.	(Yes/No)
28. Conduction current obeys Ohm’s law.	(Yes/No)
29. Displacement current in a conductor is greater than conduction current,	(Yes/No)
30. Displacement current in dielectrics is greater than conduction current.	(Yes/No)
31. The current density, J = ρν V.	(Yes/No)
32.	(Yes/No)
33.	(Yes/No)
34. Electric dipole moment is a vector.	(Yes/No)
35. The polarisation, P = χe∈0 E.	(Yes/No)
36. Electric susceptibility has the unit of permittivity.	(Yes/No)
37. Electric susceptibility has no units.	(Yes/No)
38. Polarisation is a result of free electrons in dielectrics.	(Yes/No)
39. Polarisation is a result of bound electrons in dielectrics.	(Yes/No)
40. Electric flux density in dielectrics is greater than that in free space for field.	(Yes/No)
41. Capacitance depends on dielectric material between the conductors.	(Yes/No)
42. Gauss’s law can be applied on arbitrary Gaussian surfaces.	(Yes/No)
43. Potential is inversely proportional to r.	(Yes/No)
44. The unit of potential is Joule/Coulomb.	(Yes/No)
45. Potential obeys the superposition principle.	(Yes/No)
46. Electric field obeys the superposition principle.	(Yes/No)
47. Electrostatic energy is quadratic in the fields.	(Yes/No)
48. Electrostatic energy does not obey superposition principle.	(Yes/No)
49. is the same as ∇ × E = 0.	(Yes/No)
50. E is perpendicular to the surface just outside a conductor.	(Yes/No)
51. ρν = 0 inside a conductor.	(Yes/No)
52. Charge resides on the surface of a conductor.	(Yes/No)
53. Direction of dipole moment is in the direction of applied electric field.	(Yes/No)
54. Coulomb’s force has the unit of _____.
55. The unit of constant of proportionality in Coulomb’s law is _____.
56. The unit of line charge density is _____.
57. The unit of surface charge density is _____.
58. The unit of volume charge density is _____.
59. Electric field is defined as _____.
60. The unit of electric field is _____.
61. The unit of electric flux is _____.
62. Laplace’s equation is _____.
63. Relaxation time in dielectrics is _____.
64. The unit of electric dipole moment is _____.
65. The unit of polarisation of dielectric is _____.
66. If ∈r = 3, electric susceptibility is _____.
67. Application of electric field to dielectric material produces _____.
68. Energy density in electric field is _____.
69. Energy stored in a capacitor is _____.
70. If the charge is doubled everywhere, the total energy is _____.
71. Atomic polarisability has unit of _____.

1. Coulomb’s force is proportional to
   1. r
   2. r²
   3. 
   4. 
2. The proportionality constant in Coulomb’s law has unit of
   1. Farads
   2. Farads/metre
   3. Newton
   4. metre/Farad
3. The value of proportionality constant in Coulomb’s law is
   1. 9 × 10⁹
   2. 9 × 10^–9
   3. 8.854 × 10^–12
   4. 
4. The unit of electric field is
   1. Newton
   2. Coulomb/Newton
   3. Newton/Coulomb
   4. Coulomb/metre
5. If the direction of Coulomb’s force on a unit charge is a_x, the direction of electric field is
   1. –a_x
   2. a_y
   3. a_x
   4. a_z
6. The unit of electric flux is
   1. Coulomb
   2. Coulomb/metre
   3. Weber
   4. Weber/m²
7. The electric field on x-axis due to a line charge extending from –∞ to ∞ is
   1. 
   2. 
   3. 
   4. 
8. Electrostatic field due to a dipole consists of
   1. 
   2. 
   3. 
   4. r terms
9. Potential at all the points on the surface of a conductor is
   1. the same
   2. not the same
   3. zero
   4. infinity
10. Gradient of the potential and an equipotential surface
    1. have the same direction
    2. have opposite directions
    3. are orthogonal to each other
    4. have no directional relation
11. The potential at a point due to electric dipole consists of
    1. r terms
    2. 
    3. 
    4. 
12. The unit of electric dipole moment is
    1. C/m
    2. C-m
    3. C/m²
    4. C-m²
13. The unit of polarisation in dielectric is
    1. C/m²
    2. C/m
    3. C/m³
    4. C-m²
14. Point form of Gauss’s law is
    1. ∇.D = ρ_v
    2. ∇.D = ρ_s
    3. ∇.D = ρ_v/∈₀
    4. ∇.D = Q
15. The Laplacian operator, ∇²
    1. has unit of m²
    2. is a vector operator
    3. has unit of 1/m²
    4. has no unit
16. Laplace’s equation has
    1. two solutions
    2. infinite solutions
    3. no solution
    4. only one solution
17. The surface charge density in a good dielectric is
    1. zero
    2. ρ_s
    3. infinity
    4. – ρ_s
18. Relaxation time of a medium with ∈_r = 3.0 and σ = 3.0 Mho/m is
    1. 8.854 picosecond
    2. 9 picosecond
    3. 7.9686 picosecond
    4. 1 second
19. The force magnitude between Q₁ = 1C and Q₂ = 1C when they are separated by 1 m in free space is
    1. 9 × 10⁹ N
    2. 8.854 × 10^–12 N
    3. 
    4. 9 × 1^–9 N
20. When the force on 2 C due to fixed charge of 4 C is 2 N, the electric field at the charge of 2 C is
    1. 1 N/C
    2. 4 N/C
    3. 8 N/C
    4. 16 N/C
21. If ∈_r = 2 for a dielectric medium, its electric susceptibility is
    1. 1
    2. 2
    3. 3
    4. 2_∈0
22. If a pair of (+)ve and (-)ve charges of 1 C each are separated by a distance of 1.0 μm, the magnitude of dipole moment is
    1. 2 C-μm
    2. 1 C-μm
    3. 0 C-μm
    4. 1 C-μm
23. If dipole moment of 1 C-m in a dielectric material of volume 0.1 m³ exists, the polarisation is
    1. 10 C/m²
    2. 0.1 C/m²
    3. 10 C/m
    4. 0.1 C/m
24. If a charge element, whose volume charge density is 2.0 C/m², is moving with a velocity of 3a_x m/s, the current density is
    1. 6a_x A/m²
    2. 6a_x A/m
    3. 1.5a_x A/m²
    4. 1.5a_x A/m
25. If E_tan1 = a_x and E_n1 = 0, the electric field E₂ in a dielectric medium 2 is
    1. a_x
    2. 2a_x
    3. ^a_y
    4. a_z
26. Two point charges Q₁ = 1 C and Q₂ = 3 C are separated by 1.0 m. The force on Q₁ is
    1. zero
    2. repulsive
    3. attractive
    4. increasing linearly
27. A charge density of 10 nC/m² is distributed on a plane z = 10 m, the electric field intensity at the origin is
    1. – 180π a_z V/m
    2. – 10π a_z V/m
    3. – 360π a_z V/m
    4. – 18π a_z V/m
28. If a force, F = 4a_x + a_y + 2a_z moves 1μC charge through a displacement of 4a_x + 2a_y – 6a_z, the resultant work done is
    1. 6 μ J
    2. 12 μ J
    3. 18 μ J
    4. 24 μ J
29. If a potential of 1 V is applied across a capacitor of 10 PF, the energy stored is
    1. 5 PJ
    2. 10 PJ
    3. 100 PJ
    4. 0.01 PJ
30. Example of non-polar type of dielectric is
    1. water
    2. hydrochloric acid
    3. sulphur dioxide
    4. oxygen
31. Example of polar type of dielectric is
    1. oxygen
    2. water
    3. hydrogen
    4. nitrogen
32. If the voltage applied across a capacitor is increased, the capacitance value
    1. increases
    2. decreases
    3. remains constant
    4. becomes infinity
33. If the electric field intensity is 1 V/m in free space, the energy density is
    1. 4.427 PJ/m³
    2. 8.854 PJ/m³
    3. 4.427 PJ
    4. 8.854 PJ
34. If electric susceptibility of a dielectric is 4, its relative permittivity is
    1. 5
    2. 4
    3. 3
    4. 2
35. The unit of electric flux is
    1. Coulomb
    2. Coulomb/m
    3. Weber
    4. Tesla
36. Gauss’s law is
    1. 
    2. 
    3. 
    4. 
37. Gauss’s law in point form is
    1. ∇.D = ρ_v
    2. ∇.D = ρ_s
    3. ∇.D = Q
    4. ∇D = ρ_v
38. Equation of continuity is
    1. 
    2. 
    3. 
    4. 
39. Relaxation time is
    1. 
    2. 
    3. 
    4. 
40. In dielectrics,
    1. 
    2. 
    3. 
    4. 
41. Potential has the unit of
    1. Joules/Coulomb
    2. Joules
    3. Joules/m³
    4. Joules/m²
42. If a total charge of 10 coulombs is uniformly distributed along a filament of length 10 m, the line charge density is
    1. 1 C/m
    2. 100 C-m
    3. 100 C/m
    4. 1 C-m
43. If a charge of 10 coulombs is uniformly distributed on the surface of a conductor of area 10 m², the surface charge density is
    1. 1 C/m²
    2. 100 C/m²
    3. 1 C-m²
    4. 100 C-m²
44. If a total charge of 1 C is contained in a tiny sphere of volume 0.1 m³, the volume charge density is
    1. 10 C/m³
    2. 0.1 C/m³
    3. 10 C-m^{3 Q}
    4. 10 C/m^{2 Q}
45. Electric flux density is
    1. 
    2. 
    3. 
    4. 
46. Poisson’s equation is
    1. ∇² V = ρ_v / ∈
    2. ∇² V = −ρ_v / ∈
    3. ∇² V = −ρ_v
    4. 
47. Boundary condition for the normal component of E on the boundary of a dielectric is
    1. E_n1 = E_n2
    2. E_n1 – E_n2 = ρ_s
    3. 
    4. E_n1 = 0
48. Electric flux lines
    1. originate at (+)ve charge
    2. originate at (–)ve charge
    3. are closed loops
    4. originate at (+)ve charge and also terminate at (+)ve charge
49. Unit of electric flux is
    1. Coulomb
    2. Weber
    3. Tesla
    4. Weber/m
50. Potential due to a charge at a point situated at ∞ is
    1. zero
    2. ∞
    3. –∞
    4. 1

1. An infinite line charge, ρ_L = 10 nC/m parallel to z-axis is at x = 3, y = 4 in free space. Find E at
   1. (0, 0, 0)
   2. (0, 1, 2)
   3. (1, 1, 1)
2. Determine the force on a point charge of 5 nC at (0, 0, 5) m due to uniformly distributed charge of 5 mC over a circular disc of radius r ≤ 1 m in z = 0 plane.
3. Find the total charge in the volume specified by 0 ≤ x ≤ 1 m, 0 ≤ y ≤ 1 m and 0 ≤ z ≤ 1 m when ρ_v = 30x² y (nC/m³).
4. Three infinitely long lines charged uniformly are parallel to the z-axis. They are separated by a distance of b m. The charge density of each is ρ_L = 2.0 PC/m. Find the electric field E at a point P on the y-axis at y = a m. If a = b = 1 m, what is the electric field, E?
5. A dipole consists of two charges Q and – Q separated by 2b m as shown in Fig. 2.34. Determine the electric field at point P.
   

   Fig. 2.34

6. An infinite length of uniform line charge has ρ_L = 10 PC/m and it lies along the z-axis. Determine electric field E at (4, 3, 3) m.
7. When six equal charges Q = 10 PC are located at (2, 0, 0), (3, 0, 0), (4, 0, 0), (5, 0, 0), (6, 0, 0) and (7, 0, 0), find the potential at the origin.
8. If the electric potential is as shown in Fig. 2.35, sketch the respective electric field distribution in a specified region.
   

   Fig. 2.35

9. 1. What is the force magnitude between two charges Q₁ = 4 nC and Q₂ = 6 nC which are separated by 0.1 m in free space.
   2. What will happen if they are kept in ice whose ∈_r = 4.2?
10. Consider the Fig. 2.36.
    

    Fig. 2.36

    If Q₁ = 4 nC, Q₂ = 4 nC, Q₃ = 4 nC, find the electric field at point S. Assume that the charges are in free space.

11. Two point charges Q₁ = 5.0 nC and Q₂ = 1.0 nC are located at (– 1, 1, –3) m and (3, 1, 0) m respectively. Determine the electric field at Q₁ and Q₂.
12. An electric field is given by E = 10ya_x + 10xa_y, V/m. Find the potential function, V. Assume V = 0 at the origin.
13. Consider concentric spherical shells in free space in which V = 0 at r = 10 cm and V = 10 volts at r = 20 cm. Find E and D.
14. If electric flux density, D is given by D = [(2y² + z)a_x + 4xya_y + xa_z)] mC/m², find the volume charge density at (0, 0, 0) and (–1, 0, 4).
15. A sphere whose radius is 0.5 m contains a charge density of ρ_v = (5 – 2r)mC/m³. Find D at a distance 10 m away from the centre of the sphere.