emtl unit-1

VENKATA REDDY BOLLA GNITS Page 0

2013

ELECTROMAGNETICTHEORY ANDTRANSMISSION LINESVENKATA REDDY BOLLA

V E N K A T A R E D D Y B O L L A , G N I T S


UNIT-1

Co-ordinate System....

In order to describe the spatial variations of the quantities, appropriate coordinate system

is required.

- A point or vector can be represented in a curvilinear coordinate system that may be orthogonal

or non-orthogonal.

- A way of addressing the points in the space ,,,,

- An orthogonal system is one in which the coordinates are mutually perpendicular to each

other.

- The different co-ordinate system available are:

- Cartesian or Rectangular.

- Circular cylindrical.

- Spherical.

- Elliptical Cylindrical.

- Hyperbolic Cylindrical.

- Parabolic Cylindrical.

The choice depends on the geometry of the application

The frequently used and hence discussed herein are

- Rectangular Co-ordinate system.(Example: Cube, Cuboid)

- Cylindrical Co-ordinate system.(Example : Cylinder)

- Spherical Co-ordinate system.(Example : Sphere)

- A set of 3 scalar values that define position and a set of unit vectors that define direction form

a co-ordinate system.

- The 3 scalar values used to define position are called co-ordinates. All coordinates are defined


with respect to an arbitrary point called the origin.

Cartesian Co-ordinate System / Rectangular Co-ordinate

System (X, Y, Z)

A Vector in Cartesian system is represented as

(Ax, Ay, Az)

Or

A = Axax+ Ayay+ Azaz

Where ax, ay and az are the unit vectors in x, y, z direction respectively.

Range of the variables:

It defines the minimum and the maximum value that x, y and z can have in Cartesian system.

-∞ < x,y,z < ∞

Differential Displacement / Differential Length (dl):

It is given as dl = dxax + dyay + dzaz

Differential length for a surface is given as:

dl = dxax + dyay --- ( For XY Plane or Z Constant Plane).

dl = dyay + dzaz, ---( For YZ Plane or X Constant Plane).

dl = dxax + dzaz ---( For XZ Plane or Y Constant Plane).

Differential length for a line parallel to x, y and z axis

are respectively given as:

dl = dxax ---( For a line parallel to x-axis).

dl = dyay ---( For a line Parallel to y-axis).

dl = dzaz, ---( For a line parallel to z-axis).

Differential Normal Surface (ds):


The differential surface (area element) is defined as

ds = ds an,

where an is the unit vector perpendicular to the surface.

For the 1st figure, ds = dydz ax

2nd figure, ds = dxdz ay

3rd figure, ds = dxdy az

Differential surface is basically a cross product between two parameters of the surface. For

example, consider the first figure. The surface has two differential lengths, one is dy and dz. The

differential surface (ds) is hence given as:

dS = dy * dz

= |dy| |dz| sinAB an

=|dy| |dz| an

Where an is the unit vector normal to both dy and dz

i.e. an = ay * az = ax

In other words the differential surface element (ds) has an area equal to product dydz, and a

normal vector that points in ax direction.

Differential Volume element (dv)

The differential volume element (dv) can be expressed in terms of the triple product.

dv = dx . (dy * dz)

Consider a cubical surface having dimension x * y * z. The differential volume (dv) of the

cubical surface is given as the triple product of the dimensions.

dv = dx . (dy * dz)

= dx ax . (dy dz a xsin θ AB )

= dx ax . (dy dz a x)

= dx dy dz

Where dy and dz are mutually perpendicular to each other. Therefore the angle between them is

90o.


ax.ax= 1 ax.ay= 0 ax.az= 0

ay.ax= 0 ay.ay= 1 ay.az= 0

az.ax= 0 az.ay= 0 az.az= 1

One thing to remember is that, the three parameters of Cartesian

coordinate system i.e. X, Y, Z are all mutually perpendicular to

each other.

Therefore ax, ay and az are all mutually perpendicular to each other.

(Circular) Cylindrical Co-ordinate System...

A Vector in Cylindrical system is represented as

(Aρ, Aφ, Az)

or

A = Aρaρ+ Aφaφ+ Azaz

Where aρ, aφand az are the unit vectors in ρ, φ and z direction respectively.

The physical significance of each parameter of cylindrical coordinates:

- The value ρ indicates the distance of the point from the z-axis. It is the radius of the cylinder.

- The value φ, also called the azimuthal angle, indicates the rotation angle around the z-axis. It

is basically measured from the x axis in the x-y plane. It is measured anti-clockwise.

- The value z indicates the distance of the point from z-axis. It is the same as in the Cartesian

system. In short, it is the height of the cylinder.

Range of the variables:It defines the minimum and the maximum value that ρ, φ and z can have in Cartesian system.0 ≤ ρ < ∞0 ≤ φ < 2π

-∞ < z < ∞


Cylindrical System - Unit vectors:

Since the co-ordinate system is orthogonal, the unit vectors aρ, aφ and az are mutually

perpendicular to each other.

- aρ points in the direction of increasing ρ, i.e aρ points away from the z-axis.

- aφ points in the direction of increasing φ (anticlockwise).

- az points in the direction of increasing z.

Relationship between Cylindrical and Cartesian Co-ordinate System

Consider the parallelogram ABOC,

X = ρcosφ.

Y = ρsinφ.

Z = Z.

From the above equations we have,


Relationship between (ax, ay and az) and (aρ , aφand az) :

Since az is common between the two coordinate system, our main focus is to find out the relation

between ax, ay and aρ , aφ

We know φ is the angle from the x-axis on the x-y plane.

From the above figures two equations can be deduced,

ax= aρ cos - asin

ay= aρ sin + acos

az= az

Transformation of vector A from (Ax, Ay, Az) to (Aρ, Aφ, Az) i.e. transformation of Vector A

from Cartesian to Cylindrical can be obtained as


Transformation of vector A from (Aρ, Aφ, Az) to (Ax, Ay, Az) i.e. transformation of Vector A

from Cylindrical to Cartesian can be obtained as

Differential Analysis - Cylindrical Co-ordinate System...

Differential Length(dl):

In General the differential length is given as

dl = dρ aρ + ρdφ aφ+ dz az

Differential Length for a surface is given as:

- dl = dρ aρ+ ρdφ aφ ---(For ρ-φ Plane or Z constant Plane)

- dl = ρdΦ aφ+ dz az ---(For φ-z Plane or ρ Constant Plane)

- dl = dρ aρ+ dz az ---(For ρ-z Plane or φ Constant Plane)

Differential length for a line parallel to ρ, φ and z axis are respectively given as:

- dl = dρ aρ ---(For a line parallel to ρ axis)

- dl = ρdφ aφ ---(For a line parallel to φ axis)

∫ dl = ∫o2π ρdφ = ρ( 2π - 1) = 2πρ

This resembles the circumference of a circle. Hence if φ varies with ρ and z constant, then


the length is the circumference of the circle.

dl = dzaz ---(For a line parallel to z axis)

Differential Surface (ds):

- ds = ρdρ dφ az

This surface describes a circular disc. Always remember- To define a circular disk we need

two parameter one distance measure and one angular measure. An angular parameter will always

give a curved line or an arc.

In this case dΦ is measured in terms of change in arc.

Arc is given as:

Arc= radius * angle

Therefore, whenever there is a change in angle the radius always remains constant. Hence ρ

always assist dφ.

∫ ds = ∫oρ∫o2π ρ dρ dφ = (ρ2/2) (2π -1) = π ρ2

This answer describes the surface area of a circle. Hence the surface is a circular disc.

ds = ρdφ dz aρ

This surface describes the curved surface of the cylinder. We can also say that this surface

defines a hollow cylinder.

Suppose the height of the cylinder varies from 0 to h.

∫ ds = ∫oh∫o2π ρ dφ dz = ρ (h - 0) (2π - 0) = 2 πρh

This answer describes the surface area of a cylinder. Hence the surface is a hollow cylinder.

- ds = dρ dz aφ

This surface describes a simple ρ-z plane which is along the direction of φ.

Differential Volume (dv):

dv = ρdρ dφ dz ---(Scalar Quantity)


∫ dv = ∫oρ ∫o2π ∫oh ρ dρ dφ dz = (ρ2/2) (2π -0) (h -0) = π ρ2 h

This answer describes the volume of a cylinder.

1 Express the following points in Cartesian co-ordinate system.

a) P1 (2, 30o, 5)

b) P2 (4, 30o, 60o)

Ans.

a) P1 (ρ, φ, z) → P1 (x, y, z)

x = ρ cosφ = 1cos 60o = 0.5

y = ρ sinφ = 1sin 60o = 0.87

z = z = 2

Therefore P1 = (0.5, 0.87, 2)

b) P2 (r, θ, φ) → P2 (x, y, z)

x = r sinθ cosφ = 4sin 30o cos 60o = 1

y = r cosθ cosφ = 4cos 30o cos 60o= 1.73

z = r cosθ = 4cos 60o = 3.46

Therefore P2 = (1, 1.73, 3.46)


Q.2. Express the point P (1, -4, -3) in cylindrical and spherical co-ordinates?

Ans.

P1 (x, y, z) → P1 (ρ, φ, z)

P2 (x, y, z) → P2 (r, θ,φ)

Q. 3

a) If V = XZ – XY +YZ, express V in Cylindrical co-ordinate system.

b) If U = X2 + 2Y2 +3z2, express U in Spherical coordinates System.

Ans.

a) Since the equation given is a scalar equation, hence we just need to substitute the values of x,

y and z in terms of ρ, φ and z.

We know x = ρ cosφ and y = ρ sinφ

V= xz – xy – yz

= (ρ cosφ) z – (ρ cosφ) (ρ sinφ) – (ρ sinφ)z

= ρz cosφ – ρ2 cosφ sinφ – ρz sinφ

b) We know x = r sinθcosφ; y = r sinθsinφ & z = rcosθ

U = x2 + 2y2 +3z2


= (r sinθcosφ)2 +2(r sinθsinφ)2 +3(r cosθ)2

= r2 + r2 sin2 θsin2 φ + 2 r2cos2θ

Q.4 Transform the vector E = (y2 – x2) ax + xyz ay + (x2 – z2) ax to cylindrical and spherical

system?

Ans.

Eρ = (y2 – x2) cosφ + xyz sinφ

= ρ2 (sin2 φ - cos2 φ) + ρ2 z cosφ sin2 φ

= - ρ2 cos2φ cosφ + ρ2 z cosφ sin2 φ

E φ = - (y2 – x2) sinφ + xyz cosφ

= ρ2 cos2φ cosφ + ρ2 z cos2 φ sin φ

Ez = x2 – z2 = ρ2 cos2 φ - z2

E = ρ2 cosφ (z sin2 φ - cos2φ)aρ + ρ2 sinφ (z cos2 φ + cos2φ)aφ

+ (ρ2 cos2 φ - z2)az

Now E in spherical co-ordinate system is given as ,

Er = (y2 – x2) sinθ cosφ + xyz sinθ sinφ + (x2 – z2) cosθ

= (r2 sin2 θsin2 φ – r2 sin2 θcos2 φ) sinθ cosφ +(r sinθcosφ)(r sinθsinφ)(r cosθ) sinθ sinφ + ( r2 sin2

θcos2 φ – r2 cos2 θ) cosθ

= r2 sin3 θ (sin2 φ - cos2 φ) cosφ + r3 sin3 θsin2 φ cosθcosφ + ( r2 sin2 θcos2 φ – r2 cos2 θ) cosθ


Similarly,

Eθ = - r2 sin2 θ (cos2φ) cosθ cosφ + r2 sin2 θsin2 φ cos2 θcosφ - (r2 sin2 θcos2 φ – r2 cos2 θ) sinθ

Eφ = -(r2 sin2 θsin2 φ – r2 sin2 θcos2 φ) sinφ + r3 sin2 θsinφ cosθ cos2 φ

E = Er ar + Eθ aθ + Eφ aφ

Q.5 Express the vector A = ρ (z2 + 1)aρ - ρz cosφaφ in Cartesian co-ordinate system?

.

Ax = ρ (z2 + 1) cosφ + ρz cosφ sinφ

Ay = ρ (z2 + 1) sinφ - ρz cos2φ


Az = 0

Hence, A = Axax + Ayay + Azaz

Q.6 Express the vector E =2r sinθ cosφar + r cosθ cosφaθ – r sinφaφ in Cartesian co-ordinate

system?

Ans.

Ex = 2r sin2 θcos2 φ + r cos2 θcos2 φ + r sin2 φ

Substituting the above values, we get


VECTOR ANALYSIS

SCALAR -

- A scalar quantity is defined as a quantity that has magnitude only.

- A scalar quantity does not point to any direction i.e. a scalar quantity has no directional

component.

example - when we say, the temperature of the room is 30o C, we don’t specify the direction.

- Hence examples of scalar quantities are mass, temperature, volume, speed etc.

- A scalar quantity is represented simply by a letter – A, B, T, V, S.

VECTOR :

- A Vector has both a magnitude and a direction. Hence a vector quantity is a quantity that has

both magnitude and direction.

- Examples of vector quantities are force, displacement, velocity, etc.

A vector quantity is represented by a letter with an arrow over it.

UNIT VECTORS (aA):

- When a simple vector is divided by its own magnitude, a new vector is created known as the

unit vector


NOTE:

1. Unit vector has a magnitude of one. Hence the name “unit vector”.

2. . A unit vector is always used to describe the direction of respective vector.

Rearranging the terms, we have

Hence any vector can be written as the product of its magnitude and its unit vector.

- Unit Vectors along the co-ordinate directions are referred to as the base vectors.

For example unit vectors along X, Y and Z directions are ax, ay and az respectively.

Position Vector / Radius Vector ( r ):

- A Position Vector (rQ)/ Radius vector defines the position of a point in space relative to the

origin.

rQ = xax + yay +zaz

- If the coordinates of some point is given as x =1, y =2 and z =3, then the position vector is

defined as

r = ax + 2ay +3az.

Hence Position vector is another way to denote a point in space.

- Position vector for Cartesian system in general is written as

r = xax + yay +zaz

But we cannot say the position vector for cylindrical and spherical coordinate system to be

r = ρaρ + φaφ + zaz

r = rar +θaθ + φaφ

because θ and φ are not a unit of distance.

- Hence the correct position vector for cylindrical and spherical system is given as:

r = ρcosφax + ρsinφay + zaz

r = rsinθcosφax +rsinθsinφay + rcosθaz


- A position vector should always be expressed using Cartesian base vectors (ax, ay, az).

- Displacement Vector is the displacement or the shortest distance from one point to another.

Vector Multiplication:

When two vectors are multiplied the result is either a scalar or a vector depending on how they

are multiplied.

The two important types of vector multiplication are:

- Dot Product/Scalar Product (A.B)

- Cross product (A x B)

DOT PRODUCT (A . B):

- Dot product of two vectors A and B is given as:

A . B = |A| |B| cosθAB

Where θAB is the angle formed between A and B.

Also θ ranges from 0 to π i.e. 0 ≤ θAB ≤ π

- The result of A.B is a scalar, hence dot product is also known as Scalar Product.

If A = (Ax, Ay, Az) and B = (Bx, By, Bz) then

A.B = AxBx + AyBy + AzBz

If A.B = |A| |B|, then obviously cosθAB =1 which means θAB = 0o

This shows that A and B are in the same direction or we can also say that A and B are parallel

to each other.

If A.B = - |A| |B|, then obviously cosθAB = -1 which means θAB = 180o.

This shows that A and B are in the opposite direction or we can also say that A and B are

antiparallel to each other.

Similarly if A.B = 0, then cosθAB =0 which means θAB =90o.

This shows that A and B are orthogonal or perpendicular to each other.

Since we know the Cartesian base vectors are mutually perpendicular to each other, we have


ax . ax = ay . ay = az . az = 1

ax . ay = ay . az = az . ax = 0

CROSS PRODUCT (A x B):

- Cross Product of two vectors A and B is given as:

A x B = |A| |B| sinθAB an

Where θAB is the angle formed between A and B and an is a unit vector normal to both A and B.

Also θ ranges from 0 to π i.e. 0 ≤ θAB ≤ π

- The cross product is an operation between two vectors and the output is also a vector.

If A = (Ax, Ay, Az) and B = (Bx, By, Bz) then,

- The resultant vector is always normal to both the vector A and B. - If A x B = 0, then sin θAB =

0 which means θAB = 0o or 180o;

This shows that A and B are either parallel or antiparallel to each other.

-Since we know the Cartesian base vectors are mutually perpendicular to each other, we have

ax x ax = ay x ay = az x az =0

ax x ay = az

ay x az = ax

az x ax = ay


SCALAR TRIPLE PRODUCT:

- A . (B x C) = B . (C x A) = C . (A x B)

Volume of a parallelogram having A, B and C as edges is given by the Scalar Triple Product.

VECTOR TRIPLE PRODUCT:

A x (B x C) = B (A . C) - C (A .B)

COMPONENT OF A VECTOR:

Scalar Component AB of vector A along vector B

AB = AcosθAB = A |aB|cosθAB = A.aB

Vector Component AB of vector A along vector B

is the scalar product multiplied by the unit vector along B

i.e. AB = (A.aB) aB

1. Prove that (A * B) . A = 0.

A * B = ax ( Ay Bz - Az By ) - ay ( Ax Bz - Az Bx ) + az ( Ax By - Ay Bx )


A = Ax ax + Ay ay + Az az

(A * B ) . A = Ax Ay Bz - Ax Az By - Ax Ay Bz + Ay Az Bx + Ax Az By - Ay Az Bx = 0.

a3A + B = 3A + B/ |3A + B| = (8ax + 2ay + 3az)/ (77)1/2

Q.2 Given Points P (1, -3, 5), Q (2, 4, 6) and R (0, 3, 8) find

a) Position vectors of P and R.

b) Distance vector rQR.

c) Distance between Q and R.

Ans:

a) rp = ax – 3ay + 5az.

rR = 3ay +8az.

b) rQR = rR -rQ

= 3ay +8az – (2ax + 4ay + 6az)

= - 2ax – ay + 2az.

c) |rQR| = (22 + 12 + 22)1/2 = 3.

Q. 3 Find the unit vector along the line joining point (2, 4, 4) to point (-3, 2, 2)?

Ans:

Ā = (-3, 2, 2) – (2, 4, 4)

= (-5, -2, -2)

Q. 4 Given that A = 3ax + 5ay – 7az and B = ax – 2ay + az ; find

a) | 2B + 0.4A |

b) A.B - | B |2

c) A x B

Ans:


a) 2B + 0.4A

= 2ax – 4ay +2az + 0.4 (3ax + 5ay – 7az)

= 3.2ax + 6ay - 0.8az

| 2B + 0.4A | = (3.22 + 62 + 0.82)1/2 = 6.846

b) A.B - | B |2

= (3ax + 5ay -7az) . (ax -2ay + az) – (12 + 22 +12)1/2

= - 14 – (6)1/2

= - 16.4494

c) A x B

Q.5 Given Vectors T = 2ax – 6ay + 3az and S = ax + 2ay +az; find

a) the scalar projection of T on S.

b) the vector projection of S on T.

c) the smaller angle between T and S.

Ans:

a)

b)


= -0.286 ax + 0.857ay – 0.43az

Sin θTS = 0.9129 => θTS = 65.91o

Q.6 Let E = 3ay + 4az and F = 4ax – 10ay + 5az

a) Find the component of E along F.

b) Determine a Unit vector perpendicular to both E and F.

Ans:

a)

= - 0.28ax +0.71ay - 0.35az

b)

= ± (0.94, 0.27, -0.21)


Q.7 E and F are vector fields given by E = 2xax + ay +yzaz. and

F = xyax – y2ay +xyzaz. Determine:

a) | E| at (1, 2, 3)

b) Component of E along F at (1, 2, 3)

c) A vector perpendicular to both E and F at (0,1, -3) whose magnitude is unity?

Ans:

a) At (1, 2, 3), E = (2, 1, 6)

b) At (1, 2, 3), F = (2, -4, 6)

= 1.29 ax – 2.57 ay + 3.86 az

c) At (0, 1, -3), E = (0, 1, -3) and F = (0, -1, 0)

LINE, SURFACE AND VOLUME INTEGRAL...

- The Line integral of Vector A along a path L is given as

∫L A .dl

- The line integral is the dot product of a vector with a specified curve C.

- We can also say that line integral is the integral of the tangential component of vector A along

the curve C.


- If the path of integration is a closed path, the line integral becomes a closed line integral and is

called the circulation of A around C.

Line Integral is useful in finding the electric field intensity along a path L.

- The surface integral of a vector B across a surface S is defined as

∫s B .ds

- When the surface S is closed, the surface integral becomes the net outward flux of B across S,

i.e.

- Surface integral is useful in finding the magnetic flux through a surface S.

- The volume integral of a scalar T over a volume v is given as

∫v T . dv

Q.1 Calculate the circulation of A = ρ cosφ aρ+ z sinφ az around the edge L of the wedge

defined by 0 < ρ < 2, 0 < φ < 60o, z = 0 as shown.


A = ρcosφ aρ+z sinφ az

∫1 A .dl = ∫02 (ρcosφ aρ+zsinφ az) dρaρ = ρ2 cosφ / 2 = 4 / 2 = 2 (since φ = 0o)

∫2 A .dl = ∫0π/3 (ρcosφ aρ + zsinφ az) ρdφaφ = 0

∫3 A .dl = ∫20(ρcosφ aρ+zsinφ az) dρaρ = - 4 cosφ / 2

= -1 (since φ = 60o)

Q.2 given that H = x2 ax + y2 ay, evaluate ∫L H .dl where L is along the curve y =x2 from (0,

0) to (1, 1).

Ans:

∫L H .dl = ∫( x2 ax + y2 ay ) . (dx ax + dy ay + dz az)

= ∫( x2 dx+ y2 dy )

But on L, y = x2 hence dy = 2x dx


Therefore

∫L H .dl = ∫01[x2 dx+ x4 (2xdx) ] = ∫0

1( x2 dx+ 2x5 dx )

= | x3/3 |01 + 2 | x6 /6 |0

1 = 0.667

Q.3 Given that ρs = x2 + xy, calculate ∫s ρsds over the region y ≤ x2, 0< x< 1.

Ans:

∫sρsds = ∫ ∫ x2 dxdy + ∫ ∫ xy dx dy

= ∫01x2 dx ∫dy + ∫01x dx ∫ydy

= ∫01x2 dx | y | + ∫01x dx | y2 / 2 |

= ∫01x4 dx + ∫01 ( x5 / 2) dx = 1/5 + 1/12 = 0.2833


ELECTROSTATICS

Introduction to Electrostatics and Coulomb's Law...

- Electrostatics is a branch of science that involves the study of various phenomena caused by

electric charges that are slow-moving or even stationary.

- Electric charge is a fundamental property of matter and charge exist in integral multiple of

electronic charge.

Electrostatics as the study of electric charges at rest.

- The two important laws of electrostatics are

- Coulomb’s Law.

- Gauss’s Law.

- Both these laws are used to find the electric field due to charge configurations.

- Coulomb’s law is applicable in finding electric field due to any charge configurations where

as Gauss’s law is applicable only when the charge distribution is symmetrical.

COULOMB’S LAW:

- Coulomb’s law is the “Law of Action” which describes the force between two point charges.


- A point charge is a charge that occupies a region of space which is negligibly small compared

to the distance between the point charge and any other object.

- Coulombs law states that

“The force of attraction or repulsion between two point charges Q1 and Q2 is:

- Proportional to the charges Q1 and Q2.

- Varies inversely as the square of distance between them.

- Acts along the line joining the two point charges.

- Mathematically, Coulomb’s law is expressed as:

Where k is the proportionality constant and is defined as

εo -is known as the permittivity of free space and εr is called the relative permittivity of any

dielectric material.

In Free Space, εr = 1, hence

Hence equation of force in free space becomes:


- Here the diagram represents the coulomb

vector force on Point charges Q1 and Q2.

Force F12 on Q2 due to Q1 is given as:

- Similarly F21 on Q1 due to Q2 is given as:

F21 = |F12 | aR21 = |F12 | ( - aR12 )

= - F12 since [aR21 = - aR12 ]

- The force on Q1 due to Q2 is equal in magnitude but opposite in direction to the force on Q2

due to Q1.

- If there are more than two point charges, then the principle of Superposition can be used to


determine the force on a particular charge.

-If there are N numbers of charges Q1, Q2, Q3…..Qn located respectively at points with position

vectors r1, r2, r3….rn, the force experienced by a charge Q located at position vector r is given as:

Electric Field Strength / Electric field Intensity (E)...

- Electric field due to a charge is the space around the unit charge in which it experiences a

force.

- Electric field intensity or the electric field strength at a point is defined as the force per unit

charge.

Mathematically,

E = F / Q

OR

F = E Q

- The force on charge Q is the product of a charge (which is a scalar) and the value of the

electric field (which is a vector) at the point where the charge is located.

- Hence force will be either parallel or anti-parallel to the Electric field intensity.

(i.e. Q > 0) the force F points in the same direction as the electric field E.

- If the charge is negative (i.e. Q < 0) the force F points in the opposite direction as the

electric field E.


- Electric field intensity (E) at point r due to a point charge Q located at a point with position

vector r1 is given as:

Similarly for N point charges Q1, Q2 ….Qn located at points with position vectors r1, r2,….rn,

the electric field intensity at point r is given as:

Continuous Charge Distributions

- Charges can occur as point charge, line charge, surface charge and volume charge.

The charge element dQ and the total charge due to different charge distribution is given as:

dQ=ρldl → Q = ∫L ρldl (Line Charge)

dQ = ρsds → Q = ∫S ρsds (Surface Charge)

dQ = ρvdv → Q = ∫V ρvdv (Volume Charge)

- Electric field intensity due to different charge distribution is hence given as:


Electric Field Intensity Due To a Finite Line Charge...

(for this refer ur class notes )

- Hence the electric field intensity due to an infinite line charge is given as:


Electric Field Intensity Due To a Infinite Sheet Charge....

(for this refer ur class notes)

- Consider an infinite sheet of charge in the y-z plane having a uniform charge density of ρs

C/m2.

- In general, Electric field intensity for an infinite sheet of charge is given as:

Where an is a unit vector normal to the sheet.

NOTE :

1. The electric field intensity (E) points away from the plane if ρs is positive and towards the

plane if ρs is negative.

2. The magnitude of the electric field is a constant – the magnitude is independent of the

distance from the infinite plane.

3. This is because no matter how far the point is from the infinite sheet, the distance becomes

incomparable with the dimensions of the plane. Hence it seems the point is very close to the

infinite plane.


4. In a parallel plate capacitor the electric field intensity between the two plates having equal

and opposite charge is given by:

The first –ve sign denotes –ve charge on one plate and the second –ve sign denotes opposite

direction.

Electric Field Intensity (E) Due To a Circular Ring Charge...

- Consider a circular ring of radius 'a' which carries a uniform line charge density ρL as

shown in figure.

- We need to find out electric field at a point P (0, 0, h) on the z axis (z > 0).

- Electric field intensity (E) due to any line charge (ρL) in

general is given as:

In this case,

dl = ρ dφ

(Where radius ρ = a)


dl = a dφ

Consider the triangle shown in the above figure

a aρ + R = h az

R = - a aρ + h az

aR = R / | R |

R2 = a2 + h2

Substituting all these values in the above equations, the electric field

intensity E becomes:

For every element dl there is a corresponding element diametrically opposite that gives an

equal but opposite dEρ so that the two contributions cancel each other.


Hence contribution along aρ due to symmetry adds up to zero.

Therefore the final electric field intensity at point (0, 0, h) has only z component.

Hence Electric field intensity (E) due to a circular ring (of radius a carrying a uniform

charge ρL) placed on the x-y plane and if the point of interest is any point on z axis, then it is

given as:

Similarly if the ring is placed on x-z plane and point of interest is any point on y- axis, then

E is given as:


Electric Field Intensity (E) Due To a Circular Disk Charge...

Consider a circular disc of radius ‘a’ which carries a uniform surface charge

density ρs ,C /m2.

Say the disk lies on x-y plane (or z = 0 plane) with its axis along the z axis as

shown in the figure.

We need to find out electric field (E) due to a circular disk at a point P (0, 0, h) on the

z axis (z > 0).

Electric field intensity (E) at a point due to any surface charge (ρs) is given as:

Consider the triangle shown in figure(Since it’s a disc, the varying

terms are radius ρ and angle φ)

As per the vector law of addition,


ρ aρ + R = h az

R = - ρ aρ + h az

| R | = (ρ2 + h2)1/2

aR = R / | R |

aR = - ρ aρ + h az / (ρ2 + h2)1/2

Substituting all these values in the above equations, the electric field intensity E becomes

Contribution along aρ due to symmetry adds up to zero.

Therefore the final electric field intensity at point (0, 0, h) has only z component.


As a → 0, the electric field intensity (E) also tends to zero i.e. E → 0

Hence electric field intensity (E) at point (0, 0, h) is given as:


Problems on Electric field Intensity (E) and Force (F)

Q.1 Point charges Q1 = 5 μC and Q2 = -4 μC are placed at (3, 2, 1) and (-4, 0, 6)

respectively. Determine the force on Q1.

Ans:

Force on Q1 due to Q2 is given as :

= -5.74 ax – 1.642 ay + 4.104 az mN

Q. 2 Five Identical 15 μC point charges are located at the center and corners of a square

defined by -1 < x, y > 1 and z = 0.

Find the force on the 10 μC point charge at (0, 0, 2).

Therefore,

F / 1.35 = [(0, 0, 2) / 8] + [(-1, -1, 2) / 63/2] + [(1, -1, 2) / 63/2]

+ [(-1, 1, 2) / 63/2] + [(1, 1, 2) / 63/2]

F = 1.35 [(0.25 + (8 / 63/2)] az

= 1.072az N


Q.3 Point Charges Q1 and Q2 are respectively located at (4, 0, -3) and (2, 0, 1). If Q2 = 4nC,

find Q1 such that

a) The E at (5, 0, 6) has no z component.

b) The force on a test charge at (5, 0, 6) has no x component.

a) Electric field intensity at point (5, 0, 6) is given as:

Hence Q1 = - 3.463 nC.

b) F (5, 0, 6) = qE (5, 0, 6)

If Ex = 0


Hence, Q1 = -18.7 nC.

Q. 4 Determine the total charge

a) On line 0 < x < 5 m if ρL = 2x2 mC/m

Ans:

Q = ∫L ρl dl

= ∫05 (2 x2)dx

= | 4x3 |05 = 0.5 C

b) On the cylinder ρ = 3, 0 < z < 4, if ρs = ρz2 nC/m2

∫S ρsds = ∫S ρz2 ds

= ∫S ρz2 (ρ dφ dz)

= ∫(φ =0)2π ∫(z =0)

4 ρz2 (ρ dφ dz)

= (ρ3 / 3) (2π) | z3 / 3 |04

= 9 x (2π) x (43 / 3)

= 1.206 μC


c) Within the sphere r = 4m, if ρv = 10 / (r sinθ ) C / m3

Ans:

Q = ∫V ρvdv

= ∫V [ 10 / (r sinθ ) ] (r2 sinθ dr dθ dφ)

= ∫(r=0)4 ∫(θ=0)

π ∫(φ=0)2π (10 rdr dθ dφ)

= 10 | r2 / 2 |04 (2π) (π)

= 157.91 C

E due to Line, Surface and Mixed Charge Configuration...

Q.1 An infinite long uniform line charge is located at y = 3, z = 5. If ρL = 30 nC/m. Calculate

the electric field at point (5, 6, 1).

Ans:

The electric field intensity (E) due to an infinite line charge is given as:

In this case, ρ = (6 – 3) ay + ρ(1 – 5)az = 3ay – 4az

Hence aρ = (3ay - 4az) / 5

Therefore E is,


Q.2 A uniform line charge density 20 nC lies on z axis between z = 1 and z = 3.

Find electric field intensity at point (4, 0, 0).

Ans:

Electric field due to a line charge density is given as:

In this case

R = 4 ax – z az

ar = (4 ax – z az) / (42 + z2)

Hence we have,


Q. 3 Two identical charges are located on the y axis at y = 3 and y = 9. At what point in

space is the net electric field zero?

Ans:

As the both charges are on the y axis, the point at which the fields due to the two charges can

cancel has to lie on the y-axis also. Since the two charges are identical, the point at which the

net electric field will be zero is midway between them, i.e. at point (0, 6, 0)

Q.4 A point charge 100 pC is located at (4, 1, -3) while the x-axis carries charge 2 nC/m. If

the plane z = 3 also carries charge 5nC / m2, find E at (1, 1, 1).

Ans:

Electric field at point (1, 1, 1) is given as:


= (- 0.0216, 0, 0.0288) + (0, 18, 18) – 264.7 (0, 0, 1)

= - 0.0216 ax + 18ay – 264.7 az V/m

Q.5 Plane x + 2y = 5 carries charge ρs = 6 nC/m2. Determine E at (-1, 0, 1)?

Ans:

Let f (x, y) = x + 2y – 5∇ f = ax + 2ay

A unit normal vector to any surface or plane is given as:

Now since (-1, 0, 1) lies below the plane, therefore Electric field is given as:

ELECTRIC LINES OF FORCES:

- Electric line of force is a pictorial representation of the electric field.

- Electric line of force (also called Electric Flux lines or Streamlines) is an imaginary straight

or curved path along which a unit positive charge tends to move in an electric field.


PROPERTIES OF ELECTRIC LINES OF FORCE:

1. Lines of force start from positive charge and terminate either at negative charge or move

to infinity.

2.if the negative charge is absent , then the flux lines terminate at infinity

3. Lines of force never intersect i.e. they do not cross each other.

4. Tangent to a line of force at any point gives the direction of the electric field E at that

point.

5. Lines are dense close to a source of the electric field and become sparse when one moves

away.

6. The lines of force are independent of the medium in which charges are placed.

7. if the charge on a body is Q , then the total number of the lines originating or terminating on

it is also Q .

8 .The number of lines per unit area, through a plane at right angles to the lines, is proportional to

the magnitude of E. This means that, where the lines of force are close together, E is large and

where they are far apart E is small.

9. If there is no charge in a volume, then each field line which enters it must also leave it.

If there is a positive charge in a volume then more field lines leave it than enter it.


If there is a negative charge in a volume then more field lines enter it than leave it.

Hence we say

“Positive charges are sources and Negative charges are sinks of the field.”

ELECTRIC FLUX DENSITY(D = εE):

- Line of Force may be termed as ‘Electric Flux’ represented by ψ and unit is coulomb (C).

- The density of electric flux is the electric (displacement) flux density, D.

- It is the measure of cluster of ‘electric lines of force’. It is the number of lines of force per

unit area of cross section.

D = ψ / S → ψ = ∫S (D . ds)

Gauss Law (Theory) & Application To A Point Charge...

- Gauss law is one of the fundamental laws of Electrostatics.

- It states that

“The net electric flux emanating or coming from a close surface S is equal to the total

charge contained within the volume V bounded by that surface.”

We know,

Therefore

Ψ = Qenc


Hence Gauss law also states that

“The total electric flux Ψ through any closed surface is equal to the enclosed electric

charge.”

- Charge contained in a volume is given as:

dQ = ρv dv

Q = ∫V ρv dv

Hence we have,

Applying Divergence theorem we have,

Comparing the above two equations, we have

This isMAXWELL`s IEQUATION

THIS IS FINALFORM OFGAUSS`S LAW


This equation is called the 1st Maxwell's equation of electrostatics.

- Gauss law is an easy way of finding electric field for some symmetric problems in

electrostatics.

- Gauss law relates the electric field at points on a closed Gaussian surface to the net charge

enclosed by that surface.

- One thing to remember is Qenc contains charges which are enclosed within the volume.

Charges outside the volume, no matter how large or how close it may be, are not included in

the term Qenc.

Procedure to Apply Gauss Law:

Gaussian surface:

The surface over which the gauss`s law is applied is called Gaussian surface

Conditions of Gaussian surface

1. The surface must be closed

2. At each point of the surface D is either normal or tangential to the surface

When D is normal to the Gaussian surface

D .dS =DdS cos = DdS

When D is tangential to the surface

D .dS =DdS cos = 0


Steps to follow

1. Identify the type of charge distribution ( i.e either point , line , surface ,or volume

charge)

2. Check out whether symmetric charge distribution exists or not.

3. If yes, drew a hypothetical surface called a Gaussian surface

( Gaussian surface is a hypothetical (any imaginary) closed surface enclosing

the charge configuration.)

4. Find the flux

5. Find the Qenc

Then finally, according to Gauss`s law Ψ = Qenc

Application Of Gauss Law

1. A Point Charge:

- Consider a point charge Q at the origin, say at point P, the electric flux density (D) is to be

evaluated.

1. Type of charge – point

2. symmetry - spherical

3. Draw the Gaussian surface

4. Electric flux density is everywhere normal to the Gaussian

surface i.e. D = Dr ar


-Since the Gaussian surface is a hollow sphere, hence the variable terms are θ and φ. Thus the

differential surface for a hollow sphere is given as:

ds = ∫(φ=0)2π ∫(θ=0)

π (r2 sinθ dθ dφ) ar = 4πr2 ar

5. The total charge enclosed by the Gaussian surface, Qenc = Q.

- Hence applying gauss law, we have

Qenc = Dr 4πr2

Hence

Again we know that, D = ε E.

Therefore

2. An Infinite Line

- Consider an infinite line charge carrying a charge per unit

length of ρL along the z axis.

- Gaussian surface selected for a symmetric line charge is a

hollow cylinder of radius ρ and length l as shown in the figure

- A cylinder has basically three surfaces: top, bottom and the

curved cylindrical surface.


From the diagram its clear that

- Electric flux density (D) is parallel to the top and bottom Gaussian surface.

- Electric flux density (D) is normal to the curved cylindrical Gaussian surface.

- Differential surface (ds) is always normal to a surface.

D and ds are normal to each other for the top and bottom Gaussian surface.

Hence

∫ (D . ds) = 0

Hence it’s clear that D and ds are parallel to each other only for the curvilinear Gaussian

surface.

- Since the Gaussian surface is a hollow cylinder, hence the variable terms are φ and z. Thus

the differential surface for a hollow cylinder is given as:

ds = ∫(φ=0)2π ∫0l (ρ dφ dz) aρ → S = 2πρl aρ

Hence applying gauss law, we have

Qenc = Dρ 2πρl = ρL l


3. INFINITE SHEET OF CHARGE

Consider an infinite sheet charge carrying a charge per unit surface of ρs on a x-y or z = 0

plane.

Gaussian surface selected for a symmetric sheet charge can be either a cylinder box or a

rectangular box.

The Gaussian surface is placed such that two of its faces are parallel to the sheet.

In this case say the Gaussian surface is a rectangular box.

From the diagram it’s very clear that only two faces of the rectangular box is parallel to the

sheet and also parallel to the z-axis.

Applying Gauss law, we have

The electric field(E) as well as electric flux

density(D) both points away from the plane if ρs

is positive and towards the plane if ρs is

negative.

The magnitude of the electric field is a constant

– the magnitude is independent of the distance


from the infinite plane.

4. UNIFORMLY CHARGED SPHERE...

Consider a sphere of radius “a” having a uniform volume charge density of ρo C/m3.

- Gaussian surface selected for a symmetric sphere charge is a sphere itself.

- Here we consider two cases:

a) A Gaussian surface with a radius r < a.

b) A Gaussian surface with a radius r > a.

CASE 1: ( r < a )

1. Type of the charge – volume

2. Symmetry – spherical


4. Electric flux is given as:

5. If r < a (consider the figure),

the total charge enclosed by the Gaussian surface is:


From Gauss law Ψ = Qenc

Therefore,

Dr 4π r2 = (4 / 3) (ρo πr3)

D = (r / 3) ρo ar (0 < r < a)

CASE II: (r >a )

1. Type of the charge – volume

2. Symmetry – spherical


4. Electric flux is given as


If r > a (consider the above figure), the total charge enclosed by the Gaussian surface is:

5.

From Gauss law Ψ = Qenc

Therefore,

Dr 4π r2 = (4 / 3) (ρo πa3

D = (a3 / 3r2 ) ρo ar (r > a)


Q.1 A point charge of 30 nC is located at the origin while plane y = 3 carries charge 10 nC /

m2 . Find D at (0, 4, 3)?

Ans:

Electric flux density (D) at point (0, 4, 3) due to a point charge and line charge is given as:

D = DQ + Dρ

= 0.019 (0, 4, 3) + 5ay

= 0.76 ay + 0.057 az + 5ay

= 5.076 ay + 0.057 az

Q.2 A charge distribution in free space has ρv = 2r nC / m3 for 0 < r < 10 m and zero

otherwise. Determine E at r = 2m and r = 12m?

Ans:

Gauss’s law states that:

For 0 ≤ r ≤ 10

Dr (4πr2) = ∫ ∫ ∫ 2r (r2 sinθ dr dθ dφ)


Dr (4πr2) = 4π (2r4) / 4

For r ≥ 10

Dr (4πr2) = ∫ ∫ ∫ 2ro (r2 sinθ dr dθ dφ)

Dr (4πr2) = 4π (2ro4) / 4

Dr (4πr2) = 2πro4


Q.3 If D = (2y2 + z)ax + 4xy ay + x az C/m2, find

a) Volume charge density at (-1, 0, 3)

b) The flux through the cube defined by 0 ≤ x ≤ 1,

c) The total charge enclosed by the cube.

Ans:

a) Volume charge density (ρv) is defined as:

ρv = ∇ . D

= 4

At point (-1, 0,3) ρv = 4

b) The total charge (Q) enclosed by the cube

Q = ∫v ρv dv

= ∫(x=0)1 ∫(y=0)

1 ∫(z=0)1 4x (dxdydz)

= 4 | (x2 / 2) |01 (1) (1)

= 2 C

c) The flux through the cube defined by:

Ψ = Qenc = 2C


Q.4 If the volume charge density (ρv) of a given charge distribution is given by ρ = ρo (a / r)

in spherical co-ordinate, determine the electric flux density (D) at any point ?

Ans:

Gauss’s law states that:

Qenc = ∫v ρv dv

= ∫(r=0)r ∫(θ=0)π ∫(φ=0)

2π ρo (a/r) (r2 sinθ dr dθ dφ)

= - ρo a | r2 / 2 |0r (2π) | (cosθ) |0π

= - ρo a (r2 / 2) (2π) ( -2)

= 2 ρo aπr2

Electric flux is given as:

Since,

Ψ = Qenc


Dr (4πr2) = 2 ρo aπr2

D = [ (ρo a) / 2a ] ar

Electric Potential / Electrostatic Potential (V)

- If a charge is placed in the vicinity of another charge (or in the field of another charge), it

experiences a force.

- If a field being acted on by a force is moved from one

point to another, then work is either said to be done

on the system or by the system.

- Say a point charge Q is moved from point A to

point B in an electric field E, then the work done in

moving the point charge is given as:

WA→B = - ∫AB (F . dl) = - Q ∫AB(E . dl)

where the – ve sign indicates that the work is done on the system by an external agent.

- The work done per unit charge in moving a test charge from point A to point B is the

electrostatic potential difference between the two points(VAB).

VAB = WA→B / Q

= - ∫AB(E . dl)


= - ∫InitialFinal (E . dl)

NOTE :

1. If the potential difference is positive, there is a gain in

potential energy in the movement, external agent performs the work against the field.

2. If the sign of the potential difference is negative, work is done by the field.

3. The electrostatic field is conservative i.e. the value of the line integral depends only on end

points and is independent of the path taken.

- Since the electrostatic field is conservative, the electric potential can also be written as:

VAB = - ∫AB (E . dl )

= - ∫APo (E . dl) - ∫ Po

B (E . dl)

= - ∫PoB (E . dl) - (- ∫AP

o(E . dl)

= VB – VA

Thus the potential difference between two points in an electrostatic field is a scalar field that is

defined at every point in space and is independent of the path taken.

- The work done in moving a point charge from point A to point B can be written as:


WA→B = - Q [VB – VA] = - Q ∫AB (E . dl)

Consider a point charge Q at origin O.

Now if a unit test charge is moved from point A to Point B, then the potential difference

between them is given as:

Electrostatic potential or Scalar Electric potential (V) at any point P is given by:

V = - ∫PoP (E . dl)

The reference point Po is where the potential is zero (analogues to ground in a circuit).

The reference is often taken to be at infinity so that the potential of a point in space is defined

as

V = - ∫∞P (E . dl)

- Basically potential is considered to be zero at infinity.

Thus potential at any point ( rB = 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. rA →

∞)


Electric potential (V) at point r due to a point charge Q located at a point with position vector

r1 is given as:

Similarly for N point charges Q1, Q2 ….Qn located at points with position vectors r1, r2,

r3…..rn, the electric potential (V) at point r is given as:

The charge element dQ and the total charge due to different charge distribution is given as:

dQ = ρldl → Q = ∫L (ρldl) → (Line Charge)

dQ = ρsds → Q = ∫S (ρsds) → (Surface Charge)

dQ = ρvdv → Q = ∫V (ρvdv) → (Volume Charge)


Relationship Between Electric Field Intensity (E) and Electric Potential (V)...

The work done per unit charge in moving a test charge from point A to point B is the

electrostatic potential difference between the two points(VAB).

VAB = VB - VA

Similarly,

VBA = VA – VB

Hence it’s clear that potential difference is independent of the path taken.

Therefore

VAB = - VBA

VAB + VBA = 0

- ∫AB (E . dl) + [ - ∫BA (E . dl) ] = 0

- The above equation shows that the line integral of Electric field intensity (E) along a closed

path is equal to zero.

In simple words,

“No work is done in moving a charge along a closed path in an electrostatic field”.


Applying Stokes’ Theorem to the above Equation, we have:

If the Curl of any vector field is equal to zero, then such a vector field is called an Irrotational

or Conservative Field.

Hence an electrostatic field is also called a conservative field.

Since Electric potential is a scalar quantity, hence dV (as a function of x, y and z variables)

can be written as:

Hence the Electric field intensity (E) is the negative gradient of Electric potential (V).

The negative sign shows that E is directed from higher to lower values of V i.e. E is opposite

to the direction in which V increases.

EQUIPOTENTIAL SURFACE:

An equipotential surface refers to a surface where the potential is constant.

The intersection of an equipotential surface and a plane results into a path called an

equipotential line.


No work is done in moving a charge from one point to the other along an equipotential line

or surface i.e. VA – VB = 0

Hence,

From the above equation, it’s clear that the electric flux lines and the equipotential surface

and normal to each other.

Because the electric field is the negative gradient of electric potential, the electric field

lines are everywhere normal to the equipotential surface and points in the direction of

decreasing potential.

The equipotential lines for a positive point charge. The solid lines show the flux lines

or electric lines of force.

Electric Potential (V) Due To A Circular Disc...

Consider a circular disc of radius ‘a’ which carries a

uniform surface charge density ρs , C/m2.

Say the disk lies on x - y plane (or z = 0 plane) with its axis

along the z axis as shown in the figure.

We need to find out electric potential (V) due to a circular

disk at a point P (0, 0, h) on the z axis (z > 0).


Electric potential (V) at a point due to any surface charge (ρs) is given as:

In this case,

ds = ρ dρ dφ

(Since it’s a disc, the varying terms are radius ρ and angle φ)

R = (ρ2 + h2)1/2

- Hence electric potential (V) is given as:

- On solving further the equation becomes


- As a → 0, electric potential (V) also tends to zero i.e. V → 0.

- Hence the electric potential at point (0, 0, h) is given as:

Electric Dipole

An electric dipole consists of two point charges of equal magnitude but of opposite sign and

separated by a small distance.

Consider an electric dipole centered at origin and placed

in z – axis as shown in the figure:

The potential (V) at point P is given as:

- If the distance between the charges (d) is very small as


compared to the distance of the point P from the origin i.e.

If r >> d,

r2 – r1 ≅ d cosθ ; r1 ≅ r2 = r ; r1r2 ≅ r2

Substituting the values in the above equation, the potential at point P becomes:

- Electric field intensity (E) is the negative gradient of Electric Potential (V).

Hence,


The expressions for electric potential (V) and field intensity (E) above are only valid for a

dipole centered at the origin and aligned with the z-axis.

To determine the fields produced by any arbitrary location and alignment, we first need to

define a new quantity p, called the Dipole Moment.

p = Q d

Since the distance d is a vector quantity, the dipole moment p is also a vector quantity.

Dipole moment p is a measure of the strength of the dipole and indicates its direction.

Vector d is a directed distance that extends from negative charge (- Q) to positive charge (+

Q). This directed distance vector d thus describes the distance between the dipole charges, as

well as the orientation of the charges.

Therefore

d = | d | ad

Where | d | is the distance between the charges and ad defines the orientation or direction of

the dipole.

Say a dipole is aligned along z – axis, then directed distance d is given as:

d = | d | az

From the above diagram it’s clear that:

az . ar = cosθ


Hence the expression can be written as:

Q d cosθ = Q | d | az . ar = Q d. ar = p . ar

Hence electric potential (V) due to a electric dipole centered at origin and aligned with the z

axis is rewritten as:

The above expression no doubt is applicable for all and any dipole moments p, but is valid

for dipoles centered at origin.

Electric potential (V) at point P with a position vector r due to a dipole centered at a point

with position vector r1 is given as:

Q.1 Determine the electric field due to the following potential:

a) V = x2 + 2y2 + 4z2

Ans:

= - (2x ax + 4y ay + 8z az )

= -2x ax - 4y ay - 8z az V/m

b) V = ρ2 (z + 1) sinφ


= - [ 2ρ (z + 1)sinφ aρ + ρ(z + 1) cosφ aφ + ρ2 sinφ az ]

= - 2ρ (z + 1)sinφ aρ - ρ(z + 1) cosφ aφ - ρ2 sinφ az

Q.2 A point charge of 5 nC is located at the origin. If V = 2v at (0, 6, -8), find

a) The potential at A (-3, 2, 6)

b) The potential at B (1, 5, 7)

c) The potential difference VAB

Ans:

If V (0, 6, -8) = 2 V

In this case

x = 0; y = 6; z = -8

Using the scalar relationship between Cartesian and spherical system, we have

r = (x2 + y2 +z2)1/2 = (100)1/2 =10

Hence

Electric potential at point r due to a point charge Q located at origin


C = - 2.5

a) Electric potential at point r due to a point charge Q located at a point (-3, 2, 6) is given as:

= 3. 929 V

b) Electric potential at point r due to a point charge Q located at a point (1, 5, 7) is given as:

= 2.696 V

c) The potential difference VAB is given as:

VAB = VB – VA = 2.696 – 3.929 = - 1.233 V

Q.3 If point charge 3 μC is located at the origin. Also there are two more charges -4 μC and

5 μC are located at (2, -1, 3) and (0, 4, -2) respectively. Find potential at (-1, 5, 2) ? Assume

zero potential at infinity.

Ans:


For N point charges Q1, Q2 ….Qn located at points with position vectors r1, r2, r3…..rn, the

electric potential at point r is given as:

At V ( ∞ ) = 0, C = 0

| r – r1 | = | (-1, 5, 2) – (2, -1, 3) | = (46)1/2

| r – r2 | = | (-1, 5, 2) – (0, 4, -2) | = (18)1/2

| r – r3 | = | (-1, 5, 2) – (0, 0, 0) | = (30)1/2

Hence electric potential is given as:

= 10.3 kV

Q.4 An electric dipole of 100 az pC.m is located at the origin. Find V and E at points

a) (0, 0, 10)

b) (1, π/3, π/2)

Ans:

a) At point (0, 0, 10)

Cartesian → Spherical

(x, y, z) → (r, θ, φ)

(0, 0, 10) → (10, 0o, 0o)


Electric potential (V) due to a electric dipole centered at origin and aligned with the z axis is

written as:

Electric field intensity (E) is the negative gradient of Electric Potential (V).

b)At point (1, π/3, π/2)

Electric potential (V) due to a electric dipole centered at origin and aligned with the z axis is

written as:


Electric field intensity (E) is the negative gradient of Electric Potential (V).

Energy Density In Electrostatic Field / Work Done To Assemble Charges....

In case, if we wish to assemble a number of charges in an empty system, work is required to

do so.

Also electrostatic energy is said to be stored in such a collection.

-Let us build up a system in which we position three point charges Q1, Q2 and Q3 at position

r1, r2 and r3 respectively in an initially empty system.

- Consider a point charge Q1 transferred from infinity to position r1 in the system. It takes no

work to bring the first charge from infinity since there is no electric field to fight against (as the

system is empty i.e. charge free).

Hence, W1 = 0 J

Now bring in another point charge Q2 from infinity to position r2 in the system. In this case we

have to do work against the electric field generated by the first charge Q1.

Hence, W2 = Q2 V21

where V21 is the electrostatic potential at point r2 due to Q1.


- Work done W2 is also given as:

- Now bring in another point charge Q3 from infinity to position r3 in the system. In this case

we have to do work against the electric field generated by Q1 and Q2.

Hence, W3 = Q3 V31 + Q3 V32 = Q3 ( V31 + V32 )

where V31 and V32 are electrostatic potential at point r3 due to Q1 and Q2 respectively.

- The work done is simply the sum of the work done against the electric field generated by

point charge Q1 and Q2 taken in isolation:

- Thus the total work done in assembling the three charges is given as:

WE = W1 + W2 + W3

= 0 + Q2 V21 + Q3 ( V31 + V32 )

- Also total work done ( WE ) is given as:


- If the charges were positioned in reverse order, then the total work done in assembling

them is given as:

WE = W3 + W2+ W1

= 0 + Q2V23 + Q3( V12+ V13)

where V23 is the electrostatic potential at point r2 due to Q3 and V12 and V13 are electrostatic

potential at point r1 due to Q2 and Q3 respectively.

- Adding the above two equations we have,

2WE = Q1 ( V12 + V13) + Q2 ( V21 + V23) + Q3 ( V31 + V32)

= Q1 V1 + Q2 V2 + Q3 V3

Hence

WE =1 / 2 [Q1V1 + Q2V2 + Q3V3]

where V1, V2 and V3 are total potentials at position r1, r2 and r3 respectively.

- The result can be generalized for N point charges as:


- The above equation has three interpretation:

a) This equation represents the potential energy of the system.

b) This is the work done in bringing the static charges from infinity and assembling

them in the required system.

c) This is the kinetic energy which would be released if the system gets dissolved i.e.

the charges returns back to infinity.

- In place of point charge, if the system has continuous charge distribution ( line, surface or

volume charge), then the total work done in assembling them is given as:

- Since ρv = ∇ . D and E = - ∇ V,

For remaining derivation refer ur class note


Substituting the values in the above equation, work done in assembling a volume charge

distribution in terms of electric field and flux density is given as:

The above equation tells us that the potential energy of a continuous charge distribution is

stored in an electric field.

The electrostatic energy density wE is defined as:

Q.1 Point charges Q1 = 1 nC, Q2 = -2 nC and Q3 = 3 nC and are positioned one at a time

and in that order at (0, 0, 0), (1, 0, 0) and (0, 0, -1) respectively. Calculate the energy in the

system after each charge is positioned?

Ans:

Initially the system is assumed to be charge free.

The energy required to bring Q1 into the system is 0 J.

After Q1: Energy in the system = 0 J

The energy required to bring Q2 into the system is


Hence, After Q2: Energy in the system = -18 nJ

The energy required to bring Q3 into the system is:

W3 = Q3 ( V31 + V32 ) + Q2 V21


Q2. Determine the work necessary to transfer charges Q1 = 1 mC and Q2 = -2 mC from

infinity to points (-2, 6, 1) and (3, -4, 0) respectively.

Ans:

No work is done in transferring the first charge Q1. However work done to transfer the point

charge Q2 is given as:

Q3. A point charge Q is placed at the origin. Calculate the energy stored in region r > a?

Ans:

Work done in assembling a volume charge distribution in terms of electric field and flux

density is given as:

Electric field intensity due to a point charge Q placed at origin is given as:


Hence energy stored in a region r > a is given as:

emtl unit-1

Documents