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EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

ENGINEERING ELECTROMAGNETICS [17EC36] B.E., III Semester, Electronics & Communication Engineering/ Telecommunication Engineering (VTU) [As per Choice Based Credit System (CBCS) scheme]

Module-2

A

I

Gauss’s law and Divergence

Gauss’s law, Divergence, Maxwell’s First equation (Electrostatics), Vector operator

K

Divergence theorem

) and

BA LE

Energy, Potential and Conductors

Energy expended in moving a point charge in an electric field, The line integral, Definition of potential difference and potential, The potential field of point charge, Current and Current density, Continuity of current.

Text Books:

1. W.H. Hayt and J.A. Buck, “Engineering Electromagnetics”, 7th Edition, Tata

IT EJ

McGraw-Hill, 2009, ISBN-978-0-07-061223-5.

Reference Books:

1. Ian John Krauss and Daniel A Fleisch, “ Electromagnetics with applications”, Mc Graw Hill.

V

2. N. Narayana Rao, “Fundamentals of Electromagnetics for Engineering”, Pearson.

RA

PREPARED BY:

RAVITEJA BALEKAI Asst Professor ECE Dept, GMIT Davangere 577006 Cell: +919739223504 Mail: [email protected] , [email protected] Website: https://ravitejb.wixsite.com/ravitej

DEPT. OF E&CE, GMIT

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Page No - 1

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Gauss’s Law Statement: “The electric flux passing through any closed surface is equal to the total charge enclosed by that surface”. Mathematical representation of Gauss’s law Consider any object of irregular shape as shown in figure. The total charge enclosed by the irregular

I

closed surface is Q coulombs. It may be in any form of distribution. Consider a small differential surface

A

dS at point P. The flux density at point P is D and its direction is such that it makes an angle θ with the normal direction at point P. The flux d passing through the surface dS is the product of the component of

K

D in the direction normal to the dS and d

LE

Mathematically this can be represented as d = Dn dS

BA

where Dn = Component of D in the direction of normal to the surface dS.

Closed irregular surface

IT EJ

dS

Q

Fig.1 Flux through irregular shaped closed surface

RA

V

ɵ

Dn = d =

cos θ cos θ dS

From the definition of dot product, =

cos θAB d =

DEPT. OF E&CE, GMIT

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Page No - 2

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

This is the flux passing through incremental surface area dS. Hence the total flux passing through the entire closed surface is expressed as =

=

Now irrespective of the shape of the surface and charge distribution, total flux passing through the surface is the total charge enclosed by the surface.

I

=Q

Problems

K

1. Given the electric flux density, D = 0.3r2 ar nC/m2 in free space

A

=

LE

a) Find E at point P(r =2 , θ=25° , ϕ=90°). b) Find the total charge within the sphere r=3.

c) Find the total electric flux leaving the sphere r=4.

D = E εo

a)

= =

r=3

Q=D

= 305.40

C

=

V

c)

ar =135.5 ar V/m

IT EJ

E=

b)

BA

D = 0.3r2 ar nC/m2

r2 sinθ dθ dφ

RA

=

= 965.22 nC

DEPT. OF E&CE, GMIT

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Page No - 3

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Point Charge

ϕ

r =a

Ds

Q

K

A

I

ɵ

LE

Fig.2 Application of Gauss’s law to the field of a point charge Q on spherical closed surface of radius ‘a’.

magnitude at every point on it.

BA

The electric flux density D is everywhere normal to the spherical surface and has a constant

The electric field intensity of the point charge has been found to be

Since D= E

- - - - - - - - - (1)

IT EJ

E=

we have D =

- - - - - - - - - (2)

At the surface of the sphere, Ds=

V

The differential element of area on a spherical surface is in spherical coordinates dsr = r dθ r sin θ dφ ar = a dθ a sin θ dφ ar = a2 sin θ dθ dφ ar

RA

Ds . dS =

.

a2 sin θ dθ dφ ar

(

Leading to the closed surface integral sinθ dθ dφ = Q

=

DEPT. OF E&CE, GMIT

- - - - - - - - - (3)

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Page No - 4

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

This proves the Gauss’s law that Q coulombs of flux crosses the surface if Q coulombs of charge is enclosed by that surface.

Infinite Line Charge z

L

A

I

r

LE

-z

BA

x

K

dS

Fig. 3 Infinite line charge

Consider an infinite line charge of density

C/m lying along z-axis from - ꚙ to ꚙ. Consider the Gaussian

surface as the right circular cylinder with z-axis as its axis and radius ‘r’ as shown in the figure. Length of

IT EJ

the cylinder is ‘L’ mtr. The flux density at any point on the surface is directed radially outwards i.e in the direction according to cylindrical coordinate system. As the line charge is along z-axis, there cannot be any component of

RA

Now,

V

radial component.

Q=

Q=

Q= +

+

=

. r dϕ dz

has only

- - - - - - - - - (1)

= =

in z direction . So

- - - - - - - - - (2) =0

=

. r dϕ dz

Q = 2 r Dr L - - - - - - - - - (3)

DEPT. OF E&CE, GMIT

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Page No - 5

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Dr =

But

= Dr =

(Due to infinite line charge)

E=

Coaxial Cable b

BA

LE

a

K

A

I

V/m - - - - - - - - - (4)

IT EJ

L

Fig. 4 Coaxial cable

Consider a coaxial cable as shown in figure. The radius of inner conductor is ‘a’ while the radius of the outer conductor is ‘b’. The length of the cable is ‘L’. The charge distribution on the outer surface of the

can be expressed interms of

RA

Hence

C/m2. The total surface area of the inner conductor is 2 aL.

V

inner conductor is having density

=

= =2 a

Thus line charge density of inner conductor is

- - - - - - - - - (1) C/m.

Consider the right circular cylinder of length L as the Gaussian surface. Due to the symmetry,

has only

radial component.

DEPT. OF E&CE, GMIT

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Page No - 6

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Q = 2 r Dr L (From infinite Line charge) where a < r < b - - - - - - - - - (2) Total charge on the inner conductor is to be obtained by evaluating the surface integral of the surface charge distribution. Q=

r dϕ dz

dS =

. a dϕ dz = 2 a L

- - - - - - - - - (3)

I

Q=

A

Equating (2) and (3)

K

2 r Dr L = 2 a L

=

BA

From eqn (1)

- - - - - - - - - (4)

LE

Dr =

D=

(a < r < b ) V/m - - - - - - - - - (5)

IT EJ

E=

This is same as obtained for infinite line charge. Every flux line starting from the positive charge on the inner cylinder must terminate on the negative charge on the inner surface of the outer cylinder. Hence the

RA

But

V

total charge on the inner surface of the outer cylinder is

DEPT. OF E&CE, GMIT

Qouter cylinder = - 2 a L Qouter cylinder = 2 b L

2 bL =-

=-2 aL - - - - - - - - - (6)

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Page No - 7

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Gauss’s law applied to Differential volume element P

K

Fig. 5 Differential volume element

A

I

D = Do = Dxo ax + Dyo ay +Dzo az

Consider a closed Gaussian differential surface in the form of rectangular box, which is a

LE

differential volume element. The sides of this element are ∆x ,

and

. Let us consider a point P

located in a rectangular coordinate system as shown in figure. The value of D at the point P may be

BA

expressed in rectangular components,

Do = Dxo ax + Dyo ay +Dzo az - - - - - - - - - (1)

The components of Dxo ,Dyo and Dzo vary with distance in the respective directions.

IT EJ

According to Gauss’s law,

- - - - - - - - - (2)

={

- - - - - - - - - (3)

V

Q=

is varying with distance for small

Consider the front surface of the differential element. Though

RA

surface like front surface it can be assumed constant. And

=

=

Front

= Dx Front

.(

) - - - - - - - - - (4) .(

) - - - - - - - - - (5)

Dx Front = Dx0 + [ Rate of change of Dx with x ] x [Distance of surface from P ]

DEPT. OF E&CE, GMIT

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Page No - 8

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

Dx Front = Dx0 +

ASST. PROFESSOR

- - - - - - - - - (6)

]

The point P is at the centre so distance of surface in x direction from P is

]

Dx0 +

=

- - - - - - - - - (7)

Back

.( .(

- - - - - - - - - (9)

K

= Dx Back

- - - - - - - - - (8)

A

=

I

Consider the integral over back surface

positive

Back

for expressing

LE

[ The flux is entering from back side and leaving from front in positive x direction hence is used . While the surface is considered from point P is in negative x direction hence ]

is used

BA

Dx Back = Dx0 - [ Rate of change of Dx with x ] x [Distance of surface from P ] Dx Back = Dx0 -

- Dx0 +

IT EJ

=

- - - - - - - - - (10)

]

- - - - - - - - - (11)

Combining (7) and (11)

+

=

- - - - - - - - - (12)

=

- - - - - - - - - (13)

RA

V

Similarly we can write,

+

+

=

- - - - - - - - - (14)

From eqn (3)

=

DEPT. OF E&CE, GMIT

- - - - - - - - - (15)

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Page No - 9

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI =

ASST. PROFESSOR - - - - - - - - - (16)

Problems 1. In free space, let D = 8xyz4ax+4x2z4ay+16x2yz3az pC/m2. (a) Find the total electric flux passing through the rectangular surface z = 2, 0 < x < 2, 1 < y < 3, in the a z direction. (b) Find E at located at P (2,−1, 3) and having a volume of 10−12 m3.

Q=

Along z direction Q=

Q=

LE

a)

K

D = 8xyz4ax+4x2z4ay+16x2yz3az pC/m2

C

BA

Q = 1365 x

=

b) E =

A

I

P (2,−1, 3). (c) Find an approximate value for the total charge contained in an incremental sphere

IT EJ

E at P (2, −1, 3) E=

Q=

V

c)

RA

Q at P (2, −1, 3)

=

= 8yz

=

=

Q=

DEPT. OF E&CE, GMIT

= (-146.4 ax+146.4ay-195.2az) v/m

=

= 10−12 m3

= - 648 x 10−12

=

=0

= -1728 x 10−12

=

ravitejb.wixsite.com/ravitej

= - 2376

C

Page No - 10

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Divergence Applying Gauss’s law to the differential volume element, we have obtained the relation, ΔV

Q=

- - - - - - - - - (1)

This is the charge enclosed in the volume ΔV. by Gauss’s law

- - - - - - - - - (2)

I

But

A

To apply Gauss’s law, we have assumed a differential volume element as the Gaussian surface, over which

K

D is constant. Hence equations (1) and (2) can be equated in limiting case as ΔV →0. ΔV

LE

=

= is any vector then,

- - - - - - - - - (3)

BA

This in general if

=

=

is called a divergence and is given by

IT EJ

This mathematical operation on div

=

=

- - - - - - - - - (4)

= div

=

RA

.

V

Divergence in different coordinate systems

.

= div

=

.

= div

=

DEPT. OF E&CE, GMIT

(Cartesian) (Cylindrical)

θ θ

θ

θ

θ

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(Spherical)

Page No - 11

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Problems 1. In each of the following parts, find a numerical value for div D at the point specified a) D= (2xyz-y2)ax + (x2z-2xy)ay+x2yaz c/m2 at P(2, 3, -1). b) D=2rz2sin2Φ ar + rz2 sin2ΦaΦ +2r2zsin2Φaz c/m2 at P(2, 1100, -1). c) D=2r sinθ cosΦ ar + r cosθ cosΦ aθ - rsinΦaΦ c/m2 at P(1, 300, 500).

= 2yz

=

div

A = -2x

=

=0

LE

=

=

K

div

I

a) D= (2xyz-y2) ax + (x2z-2xy) ay+x2y az c/m2

= 2yz-2x

=

At P(2,3,-1)

BA

= 2 (3) (-1) – 2 (2) = -10 C/m3

b) D=2rz2sin2Φ ar + rz2 sin2Φ aΦ +2r2zsin2Φ az c/m2

=

=

=

RA

=

= 4 z2 sin2Φ

V

=

=

IT EJ

div

div

=

= 4 z2 sin2Φ +

+

At P (2,1100,-1)

= 3.53 - 1.532 + 7.064 = 9.062 C/m3

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Page No - 12

RAVITEJ BALEKAI

c) D=2r sinθ cosΦ ar + r cosθ cosΦ aθ - rsinΦ aΦ c/m div

=

θ θ

=

θ

θ

θ

=6

=

θ

θ θ

=

=

θ

div

θ

I

=

θ

θ

=

θ θ

+

=6

θ

θ

-

θ

K

θ

θ

θ

θ

At P (1,300,500)

LE

θ θ

ASST. PROFESSOR

2

A

EE 17EC36 - MODULE - 2

Maxwell’s First Equation

div

is given by,

IT EJ

The divergence of electric flux density

BA

= 1.93 + 0.643 – 1.29 = 1.283 C/m3

=

- - - - - - - - - (1)

According to Gauss’s law, it is known that =

- - - - - - - - -(2)

RA

V

Expressing Gauss’s law per unit volume

=

Taking lim ΔV →0 i.e volume shrinks to zero,

But

= =

- - - - - - - - - (3) at that point

- - - - - - - - - (4)

The equation (4) gives the volume charge density at the point where divergence is obtained.

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Page No - 13

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Equating (1) and (4), div - - - - - - - (5)

.D=

This is volume charge density around a point. The equation (5) is called Maxwell’s first equation applied to electrostatics. This is also called the Gauss’s law in point form or Gauss’s law in differential form.

I

Statement

A

“The divergence of electric flux density in a medium at a point (differential volume shrinking to zero), is

LE

K

equal to the volume charge density (charge per unit volume) at the same point ”.

Divergence Theorem

BA

From the Gauss’s law we can write, Q=

- - - - - - (1)

While the charge enclosed in a volume is given by,

IT EJ

Q=

- - - - - - (2)

But according to Gauss’s law in the point form,

.D=

V

Q=

- - - - - - (3)

RA

Equating (1) and (3),

=

- - - - - - (4)

The equation (4) is called divergence theorem. Statement

“The integral of the normal component of any vector field over a closed surface is equal to the integral of the divergence of this vector field throughout the volume enclosed by that closed surface ”.

DEPT. OF E&CE, GMIT

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Page No - 14

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Problems C/m2 . Find

1. Given that the field D = a) Volume charge density

b) The total electric flux leaving the surface of the spherical volume

of radius 2 m. a)

.D=

θ

=

Φ

θ

=

Ψ=

b)

= θ

Φ

θ

Φ

C/m3

inθ dr dθ dϕ

BA

=

θ

A

θ

θ θ

K

=

LE

.D

I

=0

Ψ=0C

2. Evaluate both sides of the divergence theorem for the field D = 2xyax + x2ay C/m2 and the

IT EJ

rectangular parellelepiped formed by the planes x = 0 and 1, y = 0 and 2, and z = 0 and 3. The divergence theorem states that

=

V

Evaluating the surface integral first, we note that D is parallel to the surfaces at z = 0 and z = 3, so D . dS = 0 there. For the remaining four surfaces,we have

RA

=

+

=

+

x=0

. (- dy dz ax ) +

y=0

. (- dx dz ay ) + x=0

x=1

y=2

. (- dy dz ax ) +

y=0

. (dy dz ax)

. (- dx dz ay ) +

. ( dx dz ay ) x=1

. (dy dz ax)

y=2

. ( dx dz ay )

= 12

DEPT. OF E&CE, GMIT

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Page No - 15

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

=

=

ASST. PROFESSOR

+

=

= 2y

dx dy dz = 12

3. Given the field D = 6r sin

aϕ C/m2, evaluate both sides of the divergence

ar + 1.5r cos

theorem for the region bounded by r = 2, ϕ = 0, ϕ = π, z = 0, and z = 5.

I

The divergence theorem states that

+

. (-r dϕ dz ar ) +

ϕ=0

. (- dr dz aϕ ) +

=

r=2

. (r dϕ dz ar )

ϕ=

. ( dr dz aϕ )

r=0 (-r

dϕ dz ar )

r=2 (r

dϕ dz ar )

BA

+

K

r=0

LE

=

A

=

+ + = 225

=

RA

V

=

IT EJ

ϕ=

= 12 sin

+

ϕ=0 (-

dr dz aϕ )

( dr dz aϕ )

+0

- 0.75 sin

=

dr dϕ dz

= 225

DEPT. OF E&CE, GMIT

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Page No - 16

EE 17EC36 - MODULE - 2 -r

RAVITEJ BALEKAI

ASST. PROFESSOR

2

4. Given D=30e ar - 2zaz c/m in cylindrical coordinates. Evaluate both sides of the divergence theorem for volume enclosed by r=2, z=0, z=5. The divergence theorem states that =

+

+

z=5

=

. (r z=0

dϕ dz ar ) = 255.1

dr d az ) = -40 π . (r

dr d az ) = 0

+

=

+

=-

IT EJ

=

+

= 129.43

BA

+ =

A

=

. (r

K

r=2

LE

=

I

Consider dS normal to ar direction which is for the side surface.

+

–

–2

dr dϕ dz

RA

V

= 129.43

DEPT. OF E&CE, GMIT

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Page No - 17

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Energy, Potential and Conductor Work Done: The electric field intensity is defined as the force on a unit test charge at that point at which we want to find the value of E. Consider an electric field due to a positive charge Q. If a unit test positive charge Qt is placed at any point in this field, it experiences a repulsive force and tends to move in the direction of the force. But if a positive test charge Qt is to be moved towards the positive base charge Q

A

I

then it is required to be moved against the electric field of the charge Q. i.e against the repulsive force exerted by charge Q on the test charge Qt. While doing so, an external source has to do work to move the

K

test charge Qt against the electric field. This movement of charge requires to expand the energy. This work

LE

done becomes the potential energy of the test charge Qt, at the point at which it is moved.

Consider a positive charge Q and its electric field E. If a positive test charge Qt is placed in this field, it will move due to the force of repulsion. Let the movement of the charge Qt is dl. The direction in which the movement has taken place is denoted by unit vector

BA

, in the direction of dl. This is shown in

figure.

+ Qt

IT EJ

dL

+Q

V

Fig. 6

RA

According to Coulomb’s law the force exerted by the field E is given by, F = Qt E

But the component of this force exerted by the field in the direction of dl, is responsible to move the charge Qt , through the distance dl. w.k.t the component of a vector in the direction of the unit vector is the dot product of the vector with that unit vector. Thus the component of F in the direction of unit vector

given by, Fl = F.

DEPT. OF E&CE, GMIT

= Qt E .

N

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Page No - 18

is

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

This is the force responsible to move the charge Qt through the distance dl, in the direction of the field. To keep the charge in equilibrium, it is necessary to apply the force which is equal and opposite to the force exerted by the field in the direction dl. Fapplied = - Fl = - Qt E .

N

In this case, the work is said to be done.

A

I

Thus there is expenditure of energy which is given by the product of force and the distance.

through a distance dl, against the direction of field E is given by,

dW = - Qt

.

dl

LE

dW = Fapplied x dl = - Qt E .

K

Hence mathematically the differential work done by an external source in moving the charge Q t

Joules

BA

Thus if a charge Q is moved from initial position to the final position, against the direction of electric field E then the total work done is obtained by integrating the differential work done over the distance from initial position to the final position.

IT EJ

W=

W=

Joules

The Line Integral

V

Consider that the charge is moved from initial position B to the final position A, against the electric field E

RA

then the work done is given by, W=

This is called the line integral, where

gives the components of E along the direction

.

Mathematical procedure involved in such a line integral is, 1. Choose any arbitrary path B to A. 2. Break up the path into number of very small segments, which are called differential lengths. 3. Find the component of E along each segments.

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Page No - 19

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

4. Adding all such components and multiplying by charge, the required work done can be obtained. Thus line integral is basically a summation and accurate result is obtained when the number of segments becomes infinite. Consider an uniform electric field E. The charge is moved from B to A along the path shown in figure. The path B to A is divided into number of small segments. The line integral from B to A can be expressed as

+

+ ….._

+

]

A

W=-Q[

E2

E3 dL2

+ …..+

+

dL3

]

E4

dL4

dL1

B

+

A E5

dL5

IT EJ

E1

=

BA

W=-QE.[

=

LE

=

K

But the electric field is uniform and is equal in all directions, =

I

the summation of dot products.

V

Fig. 7

RA

The sum of all such vectors is the vector

joining initial point to final point. W=-QE.

Key Point: The work done depends on Q, This is true for nonuniform electric field

and does not depend on the path joining B to A.

as well.

The work done in moving a charge from one location B to another A, in a static, uniform or nonuniform electric field E is independent of the path selected.

DEPT. OF E&CE, GMIT

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Page No - 20

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Differential length vector Cartesian = dx ax + dy ay + dz az Cylindrical

I

= dr ar + r dφ a φ + dz az

A

Spherical

K

= dr ar + r dθ aθ + r sinθ dφ aφ

LE

Problems

1. Find the work done in moving a charge of 2 C from (2, 0, 0) m to (0, 2, 0) m along the straight

BA

line path joining two points, if the electric field is E= 12xax-4yay v/m. W=

=

IT EJ

=

= 64 J

=

(8xyzax + 4x2zay - 4x2yaz) v/m , find the differential amount of

2. Given the electric field E=

work done in moving a 6nc charge a distance of 2μm, starting at P(2, -2, 3) and proceeding in

RA

a)

V

the direction aL a) - ax+ ay+ az b)

ax -

dW = - Qt E .

= - 6 x 10-9

ay - az c)

ax +

ay

dl

(8xyzax + 4x2zay - 4x2yaz) . ( - ax+ ay + az ) (2 x 10-6)

Use Dot product operation = - 12 x 10-15 ( -

+

-

az )

At P (1, -2, 3) dW = -149.34 x 10-15 J

DEPT. OF E&CE, GMIT

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Page No - 21

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

b)

dW = - Qt E .

= - 6 x 10-9

ASST. PROFESSOR dl

(8xyzax + 4x2zay - 4x2yaz) . (

ax - ay - az ) (2 x 10-6)

Use Dot product operation At P (1, -2, 3)

= - 6 x 10-9

dl

A

dW = - Qt E .

(8xyzax + 4x2zay - 4x2yaz) . ( ax+ ay) (2 x 10-6)

K

c)

I

dW = 149.34 x 10-15 J

= - 12 x 10-15 (

+

LE

Use Dot product operation

)

BA

At P (1, -2, 3)

dW = 0

3. Calculate the work done in moving a 4 C charge from B (1, 0 ,0) to A (0, 2, 0) along the path y = 2 – 2x, z = 0 in the field E a) 5 ax v/m b) 5 x ax v/m c) 5 x ax + 5 y ay v/m

IT EJ

a)

W=

W=

= -20

RA

V

b)

= 20 J

W= W= = -20

= 10 J

c)

W= W= = -4 [

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= - 30 J

Page No - 22

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Potential difference The workdone in moving a point charge Q from point B to A in the electric field E is given by, W=-Q

- - - - - - (1)

If the charge Q is selected as unit test charge then from the above equation we get the workdone in moving unit charge from B to A in the field E.

A

I

Thus “Workdone per unit charge in moving unit charge from B to A in the field E is called potential difference between the points B to A”. Potential difference = V = -

K

- - - - - - (2)

LE

NOTE: If B is the initial point and A is the final point then the potential difference is denoted as VAB which indicates the potential difference between the points A and B and unit charge is moved from B to A. - - - - - - (3)

BA

VAB = -

Potential due to point charge

Consider a point charge, located at the origin of a spherical coordinate system, producing E radially

IT EJ

in all the directions as shown in the figure.

rA

RA

V

Q

A

B

rB

Fig. 8 Potential due to point charge Q

Assuming free space, the field E due to a point charge Q at a point having radial distance r from

origin is given by E=

DEPT. OF E&CE, GMIT

- - - - - - (1)

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Page No - 23

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Consider a unit charge which is placed at a point B which is at a radial distance of rB from the origin. It is moved against the direction of E from point B to point A. The point A is at a radial distance of rA from the origin. The differential length in spherical system is, dL = dr ar + r dθ aθ + r sinθ dφ aφ - - - - - - (2)

A

I

Hence the potential difference VAB between points A and B is given by,

VAB = -

K

VAB = -

(dr ar + r dθ aθ + r sinθ dφ aφ )

LE

=-

=

Concept of Absolute Potential

volts

BA

VAB =

IT EJ

Instead of potential difference, it is more convenient to express absolute potentials of various points in the field. Such absolute potentials are measured with respect to a specified reference position. Such a reference position is assumed to be at zero potential.

Most widely used reference which is used to develop the concept of absolute potential is infinity. The

V

potential at infinity is treated to be zero and all the potentials at various points in the field are defined with reference to infinity.

RA

Consider potential difference VAB due to movement of unit charge from B to A in a field of a point

charge Q. It is given by

VAB =

Now let the charge is moved from infinity to the point A i.e rB = VA =

DEPT. OF E&CE, GMIT

=

, hence - - - - - - (1)

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Page No - 24

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Equation (1) is called potential of point A denoted as VA. This is also called Absolute potential of point A. Similarly absolute potential of point B can be defined as VB =

- - - - - - (2)

This is the workdone in moving unit charge from infinity at point B.

I

Hence the potential difference can be expressed as the difference between the absolute potentials of the two

A

points.

K

VAB = VA - VB - - - - - - (3)

LE

Thus absolute potential can be defined as,

“The workdone in moving a unit test charge from the infinity (or reference point at which potential is zero) to the point, against the direction of the field”.

BA

Thus, the absolute potential at any point which is at a distance r from the origin of a spherical system, where point charge Q is located, is given by

- - - - - - (4)

IT EJ

V=

Potential due to several Point Charges

Consider the various point charges Q1, Q2, Q3 …… Qn located at the distances r1, r2, r3 …… rn from the origin as shown in the figure. The potential due to all these point charges, at point P is to be determined.

RA

V

Use superposition principle.

Q1

Q2

Q3

r1 r2

P

r3 rn

Qn Fig. 9 Potential due to several point charges

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EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Consider the point charge Q1, the potential VP1 due to Q1 is given by, VP1 =

I

The potential VP2 due to Q2 is given by,

A

VP2 =

K

The potential VPn due to Qn is given by,

LE

VPn =

individual point charges,

BA

As the potential is scalar, the net potential at point P is the algebraic sum of the potentials at P due to

VP = VP1 + VP2 + …….. VPn

VP =

Problems

+ …….

+

IT EJ

=

volts

RA

V

1. A point charge Q= 0.4nC is located at the origin. Obtain the absolute potential of A (2, 2, 3). A (2, 2, 3) V= =

Q= 0.4 nC(0, 0, 0) =

V=

DEPT. OF E&CE, GMIT

= 0.8719 volts

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Page No - 26

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

2. If Q = 0.4 nC is located at (2, 3, 3) then obtain the absolute potential of point A (2, 2, 3). A (2, 2, 3) VA =

Q=0.4 nC (2, 3, 3)

=

=1

=3.595 volts

A

I

VA =

points A and B.

VB = rB = Distance between point B and Q

A (2, 2, 3)

Q=0.4 nC(2, 3, 3) B (-2, 3, 3)

=4

VB =

IT EJ

=

BA

LE

VAB = VA - VB Where VA and VB are the absolute potentials of A and B Now VA =3.595 volts

K

3. If the point B is at (-2, 3, 3) in the above example. Obtain the potential difference between the

=

= 0.8987 volts

VAB = VA - VB = 2.6962 volts

V

4. If three charges 3μC, -4μC and 5μC are located at (0,0,0), (2,-1,3) and (0,4,-2) respectively. Find

RA

the potential at (1,0,1) assuming zero potential at infinity.

Q1 (0,0,0)

Q2 (2,-1,3)

Q3 (0,4,-2)

r1 r2

P(1,0,1)

r3

VP = V1 + V2 + V3

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Page No - 27

EE 17EC36 - MODULE - 2

=

RAVITEJ BALEKAI

+

+

=

=

=

+

=

ASST. PROFESSOR

=

= 13.2 x 103 V

+

5. An electric field is expressed in rectangular coordinates by E = 6x2 ax + 6y ay + 4 az V/m. c) VN if V=2 at P (1,2,-4)

2

=-

=-

-

-

VMQ = -

M (2,6,-1)

=-

-

-

c) VN if V=2 at P (1,2,-4) VN = VNQ = -

(To calculate V at N , consider reference at point Q because V=0 at Q)

=-

-

-

= 19 V

V

=-

= -120 V

IT EJ

=-

= -139 V

BA

b) VM if V=0 at Q (4,-2,-35)

LE

VMN = -

M (2,6,-1) and N (-3,-3,2)

K

a) E = 6x ax + 6y ay + 4 az

A

b) VM if V=0 at Q (4,-2,-35)

I

Find a) VMN if points M and N are specified by M (2,6,-1) and N (-3,-3,2)

RA

Potential calculation when reference other than infinity The expressions derived uptill now are under the assumption that the reference position of zero

potential is at infinity.

If any other point than infinity is selected as the reference than the potential at a point A due to point charge Q at the origin becomes, VA =

+C

Where C = constant to be determined at chosen reference point where v=0.

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Page No - 28

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Problem 1. A point charge of 6nC is located at the origin in free space, find potential of point P if P is located at (0.2, -0.4, 0.4) and a) V= 0 at infinity

b) V=0 at (1, 0, 0)

c) V= 20V at (-0.5, 1, -1)

a) The reference is at infinity

=

Q

b)

LE

K

= 89.87 V

A

= 0.6

VP =

RP

V=0 at (1, 0, 0) , Thus the reference is not at infinity. In such case potential at P is

BA

P(0.2,-0.4,0.4)

VP =

Q (0,0,0)

RP

RR

R(1,0,0)

IT EJ

Now VR at (1,0,0) is zero

VR =

=

=1

+C

C = -53.92

V

0=

RA

VP =

= 89.87 – 53.92 = 35.95 V

DEPT. OF E&CE, GMIT

P

I

VP =

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Page No - 29

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

c) Now V = 20V at (-0.5, 1, -1) Let this is M (-0.5, 1,-1). The reference is not given as infinity.

VM =

Rp =

Q (0,0,0)

= 1.5

+C

C = -15.95

M(-0.5,1,-1)

K

A

= 89.87 – 15.95 = 73.92V

VP =

RM

I

20 =

P(0.2,-0.4,0.4)

LE

Potential Gradient

Consider an electric field E due to a positive charge placed at the origin of a sphere =

BA

V=-

The potential decrease as distance of point from the charge increases. It is known that the line integral of E between the two points gives a potential difference between the two points. For an elementary length ΔL

IT EJ

we can write

VAB = ΔV = -

The rate of change of potential with respect to the distance is called the potential gradient. = Potential gradient

V

=

and V

RA

Relation between Consider

due to a particular charge distribution in space. The electric field

from point to point in space. Consider a vector incremental direction of

and potential V is changing

making an angle θ with respect to the

, as shown in figure.

To find incremental potential we use, ΔV = -

DEPT. OF E&CE, GMIT

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Page No - 30

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI = ΔL

=E Where

ASST. PROFESSOR

= unit vector in the direction of ΔL ΔL θ

K

A

I

P

ΔV = -

LE

Fig. 10 Incremental length at an angle θ

cos θ (using Dot product operation) = Potential gradient

BA

=

= -

cos θ

IT EJ

is maximum only when cos θ = -1 (θ = 180°) . This indicates that to .

must be in the direction opposite

The maximum value of rate of change of potential with distance (dV / dL) is called gradient of V.

V

= - (grad V)

RA

=-

V

The gradient V in different coordinate systems V=

(Cartesian)

V=

(Cylindrical)

V=

θ

θ

DEPT. OF E&CE, GMIT

(Spherical)

θ

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Page No - 31

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Problem r cos

1. Given the potential field in cylindrical coordinates V =

=60° , z= 2m, find values at P for a) V e) aN

f)

V, and point P at r=3m ,

c) Magnitude of E

b) E

in free space

r cos

a) V =

at P ( r=3m , ϕ =60° , z= 2m)

A

V

cos

-

-

At P ( r=3m , ϕ =60° , z= 2m) = - 10 =

+ 24

V/m

| max = E = 31.23 V/m

V

e) E = -

}

= 31.23 V/m

=

d)

r z cos

IT EJ

c)

+ 17.3

r cos

BA

=-{

r cos

LE

r cos

V=

K

=-

I

V = 30 V b)

d)

=

= 0.32

- 0.55

-0.77

RA

=

f)

=?

D=E

= 8.854 x 10-12 {

DEPT. OF E&CE, GMIT

cos

+

ravitejb.wixsite.com/ravitej

+

r z cos

}

Page No - 32

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

D=

.

= div

cos

=

+

ASST. PROFESSOR

+

r z cos

=

K

A

r z cos

I

=

LE

=

=

r cos

BA

+

At P ( r=3m , ϕ =60° , z= 2m)

IT EJ

- 234 pC/m3

RA

V

=

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Page No - 33

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Current and Current Density The current is defined as the rate of flow of charge and it is measured in amperes. I= The current is considered to be the motion of the positive charges. The conventional current is due

I

to the flow of electrons, which are negatively charged. Hence the direction of conventional current is

A

assumed to be opposite to the direction of flow of the electrons.

The current which exists in the conductors, due to the drifting of electrons, under the influence of

K

the applied voltage is called drift current. While in dielectrics, there can be flow of charges, under the

LE

influence of electric field intensity. Such a current is called the displacement current or convection current. The analysis of such currents, in the field theory is based on defining a current density at a point in

BA

the field. The current density is a vector quantity associated with the current and denoted as J. The current density is defined as the current passing through the unit surface area, when the surface is held normal to the direction of the current. The current density is measured in amperes per square meters (A/ m2).

IT EJ

Relation between I and J

Consider a surface S and I is the current passing through the surface. The direction of current is

RA

V

normal to the surface S and hence direction of J is also normal to the surface S.

J dS

Fig. 11 J and dS are normal

I=

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Page No - 34

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

Continuity Equation The continuity equation of the current is based on the principle of conservation of charge. The principle states that, The charges can neither be created nor be destroyed. Consider a closed surface S with a current density J, then the total current I crossing the surface S is given by, - - - - - - - - - (1)

I

I=

A

The current flows outwards from the closed surface. It has been mentioned earlier that the current means the flow of positive charges. Hence the current I is constituted due to outward flow of positive

K

charges from closed surface S. According to principle of conservation of charge, there must be decrease of

LE

an equal amount of positive charge inside the closed surface. Hence the outward rate of flow of positive charge gets balanced by the rate of decrease of charge inside the closed surface.

BA

Let Qi = Charge within the closed surface

= Rate of decrease of charge inside the closed surface. The negative sign indicates decrease in charge.

IT EJ

Due to principle of conservation of charge, this rate of decrease is same as rate of outward flow of charge, which is a current.

I=

=

- - - - - - - - - (2)

V

This is the integral form of continuity equation of the current.

RA

Using the divergence theorem, convert the surface integral in integral form to the volume integral. = =

- - - - - - - - - (3) - - - - - - - - - (4)

But Qi =

=

- - - - - - - - - (5)

For a constant surface, the derivative becomes the partial derivative.

DEPT. OF E&CE, GMIT

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Page No - 35

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI =

ASST. PROFESSOR

- - - - - - - - - (6)

If the relation is true for any volume, it must be true even for incremental volume ΔV. ==-

- - - - - - - - - (7)

A

I

This is the point form or differential form of the continuity equation of the current.

1. Given the vector current density J = 10

– 4r

z

ϕ

mA/m2

LE

(a) Find the current density at P (r = 3, ϕ = 30◦, z = 2)

K

Problem

(b) Determine the total current flowing outward through the circular band r = 3, 0 < ϕ < 2π, 2 < z < 2.8. J = (10 At P (r = 3, ϕ = 30◦, z = 2)

b)

ϕ

)x

–9

A/ m2

)x

IT EJ

J = (180

– 4r

z

BA

a)

I=

=

RA

V

.J=

=

–

= . J = ( 30 r z + 8 sin ϕ cos ϕ ) x

I= ϕ

ϕ

ϕ

dr dϕ dz

I = 3.245 A

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Page No - 36

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

ASSIGNMENT

TOPIC: GAUSS'S LAW AND DIVERGENCE 1. State and prove and Gauss's law. 2. Show that

s outer

= (-a/b)

s inner.

b) a uniform line charge

L=20mc/m

on the x-axis.

A

a) a point charge Q= 55mc at Q(-2,3,-6).

I

3. Calculate D in rectangular coordinates at point P(2,-3,6) produced by

K

c) a uniform surface charge density s=120μC /m2 on the plane z= -5m

4. Given the electric flux density, D= 0.3r2ar nc/m2 in free space. a) Find E at point P(2,250,900)

LE

b) Find the total charge within the sphere r=3.

c) Find the total electric flux leaving the sphere r=4.

BA

5. In each of the following parts, find a numerical value for div D at the point specified a) D= (2xyz-y2)ax + (x2z-2xy)ay+x2yaz c/m2 at P(2,3,-1).

b) D=2rz2sin2Φ ar + rz2 sin2ΦaΦ +2r2zsin2Φaz c/m2 at P(2,1100,-1). c) D=2r sinθ cosΦ ar + r cosθ cosΦ aθ - rsinΦaΦ c/m2 at P(1,300,500).

IT EJ

6. State Gauss's law in point form.

7. Determine an expression for the volume charge density associated with each D field following a) D= 5sinθ aθ + 5sinΦ aΦ c/m2 at P(0.5, /4, /4). b) D= (4xy/z) ax + (2x2/z) ay +- (2x2y/z2) az.

V

c) D= z sinΦar + cosθ sinΦ aθ + rsinΦaz .

8. State and prove divergence theorem.

RA

9. Given D=5r ar c/m2, prove divergence theorem for a shell region enclosed by spherical surface at r=a and r=b (b>a) and centered at the origin.

10. Given D=30e-rar - 2zaz c/m2 in cylindrical coordinates. Evaluate both sides of the divergence theorem for volume enclosed by r=2, z=0, z=5.

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Page No - 37

EE 17EC36 - MODULE - 2

RAVITEJ BALEKAI

ASST. PROFESSOR

TOPIC: ENERGY, POTENTIAL AND CONDUCTOR 1. Find the work done in moving a charge of 2 C from (2,0,0)m to (0,2,0) m along the straight line path joining two points, if the electric field is E= 12xax-4yay v/m. 2. Given the electric field E= (6xyzax + 2x2zay-4x2yaz)v/m , find the differential amount of work done in moving a 3nc charge a distance of 5μm,starting at P(1,-2,3) and proceeding in the direction aL a) -6/7 ax+2/7 ay+5/7az b) 6/7 ax- 4/7ay-5/7 az y/4) in cartesian at P(0,0,1) .

I

b) V= 100rcosΦ /(z2+1) at Q(4,500,3).

a) V= Eoe-xsin(

c) V= 60sinθ/r2 at R(3,600,250)

A

3. Find V, E and D in free space

5. Derive the potential due to several point charges.

K

4. Derive the potential due to point charge.

LE

6. A point charge of 6nc is located at origin in free space, find potential of point P if P is located at (0.2,-0.4,0.4) and a) V=0 at infinity b) V=0 at (1,0,0) c) V=20V at (-0.5,1,-1).

BA

7. Derive an expression for J, v and velocity of the charge element. 8. With usual notations derive the expressions ᴧ.J= -∂ v /∂ t. 9. Find the current

a) J= 2x2ax 2xy3ay + 2xyaz A/m2 in outward direction of a cube of 1m, with one corner at the

IT EJ

origin and edges parallel to the coordinate axes. b) J= 10e-100 r

A/m2

c) J= 1

A/m2 r = 3m, 0 < θ < /6, 0 < Φ < 2 .

RA

V

0.01≤ r ≤ 0.02 ,0 ≤ z ≤1m.

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Page No - 38