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					Course No.             TEE-353
Course Title           Electromagnetic Theory
Credit break-up        2(2-1-0)
Pre-requisite          Engineering Mathematics-II (BPM-132)

Catalogue Description:

         Vector approach to static electric and magnetic fields, Gauss Theorem, divergence, scalar and vector
potential functions, Laplace and Poisson's equations, Curl and Stroke's theorem, energy storage, mapping of
fields, forces on conductors in magnetic fields, capacitances and inductances, theory of images, inversion.
Maxwell’s equation, Transmission Lines.

Syllabus:
        Review: Scalar and vector fields; Vector representation of surfaces. physical representation of
gradient, divergence and curl, Gauss's law, Stoke's theorem; Helmholtz theorem; different coordinate
systems. The static electric field: The force between point charges and Coulomb's law; Electric Field
Intensity; The electric field of several point charges and the principle of superposition of fields, The electric
scalar potential; Use of Gauss's flux theorem, Solution of Laplace's and Poission's equation in one
dimension; Method of images applied to plane boundaries; Electric dipoles; Polarizability; Electric flux
density; Boundary conditions; Capacitance; Electrostatic shielding, Electrostatic energy. The static
magnetic field: Ampere's force law; Magnetic flux density; Vector potential; ampere's circuital law;
Magnetic dipoles; Magnetic field intensity; Polarization currents; Boundary conditions, Scalar potential;
Faraday's law; Motional emf; Inductance. Time Varying fields: Continuity equation: Displacement current;
Maxwell's equations; Boundary conditions; Wave equation and its solution in different media; Phaser
notation, Polarization; Reflection and refraction of plane waves at plane boundaries. Transmission Lines:
Coaxial, Two-wire and Infinite-plane transmission lines. The infinite uniform transmission lines,
Comparison of circuit and field quantities, Characteristics-impedance determinations, The terminated
uniform transmission line, Transmission line charts.

Books:
William H. Hayt, “Engineering Electromagnetics”, Tata McGraw-Hill
Joseph A. Edminister, “Schaum’s Outline of Theory and Problems of Electromagnetics”, McGraw-Hill
Evaluation:
           First Prefinal Exam:      25
           Second Prefinal Exam: 25
           Final Exam:               50
                                            Vector Analysis
Cartesian Coordinate System:               Point P is represented as (x,y,z).
Circular Cylindrical Coordinate System:    Point P is represented as (,,z).
Spherical Coordinate System:               Point P is represented as (r,,).

x: distance from the origin to the intersection of a perpendicular dropped from the point P to the x axis
y: distance from the origin to the intersection of a perpendicular dropped from the point P to the y axis
z: distance from the origin to the intersection of a perpendicular dropped from the point P to the z axis
distance from the origin to the intersection of a perpendicular dropped from the point P to the xy plane (or
        z = 0 plane) or distance of the point P from the z axis in a plane normal to the z axis
angle between x-axis and a line between the origin to the intersection of a perpendicular dropped from the
        point P to the xy plane or z = 0 plane
r: distance of the point P from the origin
: angle between z-axis and a line between the origin to point P

P: common intersection of three surfaces:
      Cartesian Coordinate System:                    x = constant, y = constant, and z = constant
     Circular Cylindrical Coordinate System:          = constant,  = constant, and z = constant
                                                                                                               1
       Spherical Coordinate System:                   r = constant,  = constant, and  = constant
x = constant:        infinite plane                          -  x  
y = constant:        infinite plane                          -  y  
z = constant:        infinite plane                          -  z  
 = constant:        circular cylinder                       0
 = constant:        half plane                              0   2
r=constant:         sphere (center at the origin)            0r
 = constant:        cone (vertex at the origin)             0   




                                                                               x2  y2
                                                                             r  x2  y2  z2
                                                                                        z
                                                                               cos1  
                                                                                        r
                                                                                         y
                                                                               tan 1  
                                                                                         x
                                                                             x   cos
                                                                             y   sin 
                                                                             z  r cos
                                                                               r sin 




                                               x = distance from the yz plane
                                               y = distance from the xz plane
                                               z = distance from the xy plane
                                                = distance from the z axis
                                               r = distance from the origin
                                                = angle between z-axis & the line from origin to the point
                                                = angle between xz plane and a plane containing z axis and
                                               the point




Position Vector:
       Position vector of a point P: Vector from the origin to the point P
                                      
        P(x, y, z)  rP  xa x  ya y  za z


Vector from P to Q,
      
RPQ  rQ  rP                       (Use this formula with caution in cylindrical and spherical coordinates.)
      Unit Vectors have fixed directions (independent of the location of P) only in the Cartesian system.
                       ˆ ˆ ˆ ˆ ˆ ˆ                  ˆ
       All Unit vectors a , a , a , a , a , a , and a are normal to the corresponding surfaces and point
                         x y z   r                  
                                                                                                                2
       towards increasing coordinate value.
      All these systems are right handed.
        ax  a y  az ;
        ˆ ˆ        ˆ          a   a  a z ;
                              ˆ     ˆ    ˆ            ar  a  a
                                                      ˆ ˆ       ˆ
ˆ
A  Ax a x  Ay a y  Az a z  A a   A a  Az a z  Ar ar  A a  A a
       ˆ        ˆ        ˆ        ˆ        ˆ       ˆ        ˆ       ˆ       ˆ
                                             
                                             A
Unit Vector along the direction of A,   aA  
                                        ˆ
                                             A

                                                                       
                                                                       R PQ
Unit vector in the direction from Point P towards Point Q,    ˆ
                                                              a PQ    
                                                                       R PQ
      
where RPQ = RPQ = distance between points P and Q
(Use this formula with caution in cylindrical and spherical coordinates.)
Dot Product: Scalar, Magnitude of one vector × Projection of another vector in the direction of the first
  
 A  B  AB cos  AB
Cross Product: Vector
  
 A  B  AB sin  AB a N
                     ˆ
                                                              
 ˆ
a N : a unit vector normal to the plane determined by A and B when they are drawn from a common point.
                                                                                                      
The normal selected is in the direction of advance of a right-handed screw when A is turned towards B .
a x  a   cos
ˆ ˆ                                 a x  a r  sin  cos
                                    ˆ ˆ                                       a   a r  sin 
                                                                              ˆ ˆ
a x  a   sin 
ˆ ˆ                                 a x  a  cos cos
                                    ˆ ˆ                                       a   a  cos
                                                                              ˆ ˆ
ax  az  0
ˆ ˆ                                 a x  a   sin 
                                    ˆ ˆ                                       a   a  0
                                                                              ˆ ˆ


a y  a   sin 
ˆ ˆ                                 a y  a r  sin  sin 
                                    ˆ ˆ                                       a  a r  0
                                                                              ˆ ˆ
a y  a  cos
ˆ ˆ                                 a y  a  cos sin 
                                    ˆ ˆ                                       a  a  0
                                                                              ˆ ˆ
ay  az  0
ˆ ˆ                                 a y  a  cos
                                    ˆ ˆ                                       a  a  1
                                                                              ˆ ˆ

az  a  0
ˆ ˆ                                 a z  a r  cos
                                    ˆ ˆ                                       a z  a r  cos
                                                                              ˆ ˆ
a z  a  0
ˆ ˆ                                 a z  a   sin 
                                    ˆ ˆ                                       a z  a   sin 
                                                                              ˆ ˆ
az  az  1
ˆ ˆ                                 a z  a  0
                                    ˆ ˆ                                       a z  a  0
                                                                              ˆ ˆ


Scalar Field: Scalar function of a position vector
Vector Field: Vector function of a position vector

Differential Volume, Surface, and Line Elements:




                                                                                                            3
                                       Line Elements:
                                          ˆ
                                       dxa x
                                         ˆ
                                       dya y
                                         ˆ
                                       dza z
                                       Surface Elements:
                                             ˆ
                                       dydza x
                                           ˆ
                                       dzdxa y
                                            ˆ
                                       dxdya z
                                       Volume Element:
                                       dv  dxdydz

                                       Line Elements:
                                       da 
                                          ˆ
                                       da
                                          ˆ
                                         ˆ
                                       dza z
                                       Surface Elements:
                                       ddza 
                                             ˆ
                                       dzda
                                           ˆ
                                       dda z
                                             ˆ
                                       Volume Element:
                                       dv  dddz




                                       Line Elements:
                                         ˆ
                                       dra r
                                       rda
                                          ˆ
                                       r sin da
                                                 ˆ
                                       Surface Elements:
                                       r 2 sin dda r
                                                    ˆ
                                       r sin drda
                                                  ˆ
                                       rdrddaˆ
                                       Volume Element:
                                       dv  r 2 sin drdd



                 Coulomb’s Law and Electric Field Intensity

Coulomb’s Law:
                                                 1 Q1Q2
                                          F2               ˆ
                                                           a12
                                                4  R12 2
                                           
                                          F2 = Force on Q2 due to Q1
                                           
                                          R12 =Directed line segment from Q1 to Q2
                                                 
                                          R12= | R12 |
                                                
                                          ˆ
                                          a12 = R12 / R12

                                                                                     4
                                         
R12 = R21; F1 = F2; a12   a 21 ; F1   F2
                    ˆ       ˆ
For free space,
=0=8.854×10-12 F/m

Electric Field Intensity:
     1 Q
E            ˆ
              aR
    4  R 2

                                                                     
R =Directed line segment from the location of Q to the point at which E is desired.
     
R=| R |
       
ˆ
a R = R /R




Charge Distributions: Volume charge, Surface (Sheet) charge, Line charge, and Point charge

Volume charge density:
      lim  Q 
v             
     v  0  v 
Q    v dv
     V
    v aR
        ˆ
E          dv
  V 4  R
           2




Surface charge density:
       lim  Q 
S             
     S  0  S 
Q    S dS
     S
     S aR
         ˆ
E           dS
  S 4  R
            2




Line charge density:

         lim   Q 
l               
       l  0  l 
Q    l dl
     L
    l aR
        ˆ
E          dl
  L 4  R
           2




                            Streamlines of Fields:

                            The equation of a streamline is obtained by solving the differential equation
                            E y dy
                                
                            E x dx

                                                                                                            5
                           Electric Flux Density, Gauss’s Law, and Divergence
                            
Electric flux density, D =  E
  1 Q                aˆ        a
                                  ˆ        aˆ
D           a R   v R dv   S R dS   l R dl
             ˆ
     4 R 2        V 4R
                          2
                              S 4R
                                    2
                                         L 4R
                                               2

                      
Electric flux,  =  D  dS
                   S
Gauss’s Law: The electric flux passing through any closed surface is equal to the total charge enclosed
                       
by that surface. Q =  D  dS .
                       S
The solution is easy if we are able to choose a closed surface which satisfies two conditions:
    
1. D is everywhere either normal or tangential.
2. Wherever it is normal, D = constant.

First step is to investigate the symmetry of the field by asking
1. With which coordinate does the field vary?
2. Which components of the field are present?

D due to a Uniform Line Charge:

                                                                   Dz = DΦ = 0
                                                                     
                                                                    D : tangential at surface 1 & 3
                                                                      : normal at surface 2
                                                                      
                                                                   | D |=Dρ = constant at surface 2
                                                                   Applying Gauss’ Law,
                                                                                                       
                                                                   Q   D  dS   D  dS   D  dS   D  dS
                                                                        1 2  3       1        2       3

                                                                    0  D   2 L
                                                                                   Q       Q/L   
                                                                    D                       l
                                                                               2 L       2  2
                                                                      
                                                                   D  l aˆ
                                                                      2
                                                                       l
                                                                   E        ˆ
                                                                             a
                                                                      2 

D due to a Uniform Surface Charge:
                                                              Dρ = DΦ = 0
                                                                
                                                               D : tangential at surface 2
                                                                 : normal at surfaces 1 & 3
                                                                 
                                                              | D |= Dz = constantat surfaces 1 & 3
                                                              Applying Gauss’ Law,




                                                                                                              6
                                                                                                          
                                                                      Q   D  dS   D  dS   D  dS   D  dS
                                                                                1 2  3               1               2               3

                                                                             ˆ          ˆ  
                                                                       D z a z    a z  0   D z a z     2 a z
                                                                                                   2
                                                                                                         ˆ             ˆ                 
                                                                       D z   2  0  D z   2
                                                                                           Q               S
                                                                       Dz                            
                                                                                       2     2
                                                                                                           2
                                                                        
                                                                       D  S az
                                                                             ˆ
                                                                           2
                                                                        
                                                                       E  S az
                                                                             ˆ
                                                                          2

Divergence:
                   
    
div A 
           lim    A  dS
       v  0 v
Divergence in Cartesian Coordinates:
                                                             A  dS   A  dS   A  dS   A  dS   A  dS
                                                             S          front              back                 left       right

                                                                                             A  dS   A  dS
                                                                                                                top        bottom

                                                                      x Ax                 x Ax 
                                                              Axo          yz   Axo           yz
                                                                       2 x                   2 x 
                                                                         y A y                y A y 
                                                                 A yo 
                                                                 
                                                                                  xz   A yo          xz
                                                                         2 y   
                                                                                          
                                                                                                  2 y   
                                                                         z Az                 z Az 
                                                                  Azo          xy   Azo           xy
                                                                          2 z                   2 z 
                                                              A    A y A y 
                                                             x 
                                                              x             xyz
                                                                     y    y 
                                                                               
                                                              A    A y A y 
                                                             x 
                                                              x             v
                                                                     y    y 
                                                                               
                                                                    A     A y A y 
                                                             div A   x 
                                                                       x            
                                                                             y   y 

Divergence in Cylindrical Coordinates:
                           A
              A   1   Az
      1 
div A 
                              z
Divergence in Spherical Coordinates:
               r Ar   r sin   sin A   r sin  A
      1  2                1                        1
div A  2
         r r                                               
Divergence Theorem:
                
S
  A  dS   div Adv
            V
Gauss’s Law in Point Form:
                              dQ
Q   D  dS   div Ddv  div D      v
     S          V                  dv

                                                  Energy and Potential
                                                                                                                                           7
Work Done in Moving a Point Charge:




dW= Fa  dl =  F  dl =  QE  dl
         final            final
W  Q  E  dl  Q  E L dl
            init                init




W  Q( E L1 L1  E L 2 L2  E L 3 L3  .......)

Potential of a Point and Potential Difference between Two Points:
Potential of point A with respect to point B: Work done in moving a unit positive charge from B to A
            A
V AB    E  dl
            B
VAB = VAC - VBC = VA - VB (If reference point C is at zero potential e.g. infinity)
VA =Potential of point A
VB =Potential of point B

Potential Field of a Point Charge:

    1 Q              1 Q
E            aR 
               ˆ               ˆ
                               ar
    4  R  2
                    4  r 2
                  1 Q 
                            a r   dra r  rda  r sin da   
                                                                        1 Q
dV   E  dl             ˆ         ˆ        ˆ             ˆ                 dr
                  4  r                                             4  r 2
                          2


        A           rA B     1 Q          Q 1     1   
V AB   dV                      dr        
                                                r       
                                                         
                            4  r       4   A rB
                                   2
        B          rB
                                                         
        Q  1 1      Q            Q
VA           
            r     4  r  V  4  r
       4   A         A



Potential of a Charge Distribution:


                                                                                                       8
                v dv       dS        l dl
      V              S       
            V 4  R    S 4  R   L 4  R

      Conservative Property of the Electrostatic Field:
      Closed line integral: zero
       E  dl  0
          L




 E  dl          E  dl     E  dl   VBA  V AB  (VB  V A )  (V A  VB )  0
                 B                 A
                             B          
 I  II        A           I             II

Potential Gradient:

dV = -E dl cos
 dV
     = -E cos
 dl
               dV
Maximum of          is E, and it occurs when  = 180o, i.e., field is opposite
                dl
to the direction in which V is increasing most rapidly.

   dV                        dV
E                  aN  
                     ˆ           a N   grad V
                                 ˆ
    dl        m ax            dN
              dV                 dV
grad V                           ˆ
                                    aN
              dl     m ax        dN
 
 E   grad V
If dl is directed along an equipotential surface, dV = -E dl cos90o = 0

Gradient in Cartesian Coordinates:
       
dV   E  dl   grad V   dl  grad V  dl
dl  dxa x  dya y  dza z
       ˆ       ˆ       ˆ
      V        V         V
dV        dx     dy         dz
      x        y         z
V        V       V
    dx       dy        dz  grad V  dxa x  dya y  dza z    grad V  x dx   grad V  y dy   grad V  z dz
                                             ˆ       ˆ         ˆ
 x        y      z

grad V  x  V
              x
              V
grad V  y 
              y

grad V  z  V
              z
                                                                 V      V         V
grad V   grad V  x a x   grad V  y a y  grad V  z a z 
                      ˆ                  ˆ                 ˆ        ax 
                                                                    ˆ         ay 
                                                                              ˆ          ˆ
                                                                                        az
                                                                 x      y         z


Gradient in Cylindrical Coordinates:

         V       1 V      V
grad V      a 
             ˆ         a 
                       ˆ       ˆ
                               az
                       z
Gradient in Spherical Coordinates:
                                                                                                                        9
        V       1 V         1 V
gradV      ar 
            ˆ         a 
                      ˆ               ˆ
                                      a
        r       r       r sin  
The Dipole:




        Q  1     1      Q R2  R1        Q d cos
V            
             R     
                     4  R R  4  r 2
      4   1 R2              1 2
                Qd cos       Qd sin 
E   grad V             ar 
                          ˆ             ˆ
                                        a
                 2  r 3
                               2  r 3
       
p  Qd

d  a r  d cos
    ˆ
        
        p  ar
            ˆ
V
      2  r 2

Energy Density in the Electrostatic Field:

Consider a system of n point charges Q1, Q2,… Qn.
Vij = potential at the location of Qi due to the presence of Qj.
                                        n
Vi = potential at the location of Qi=  Vij
                                       j 1
                                       j i

Wi = work to position Qi, charges are being positioned in a sequence Q1, Q2,… Qn.
Wi’= work to position Qi, charges are being positioned in a sequence Qn, Qn-1,… Q1.

             i 1
Wi  Qi  Vij
             j 1
              in
Wi '  Q i   V
             j  i 1
                        ij


                                n   i 1         in         1 n
       1 n                                                             n
                                                                             1 n
WE          Wi  Wi '  1   Qi  Vij  Qi          Vij    Qi  Vij   QiVi
                                                    2 i 1 j 1
       2 i 1               2 i 1  j 1
                                                 j  i 1                  2 i 1
                                                                      j i

       1
       2 V
WE          vVdv




                                                                                       10
                                                Tutorial Sheet-1

1. Transform (i) point P(1,2,3) from Cartesian to cylindrical coordinates, (ii) point P(1,30o,3) from
   cylindrical to Cartesian coordinates, (iii) point P(1,2,3) from Cartesian to spherical coordinates, (iv)
   point P(1,30o,3) from cylindrical to spherical coordinates, (v) point P(1,2,3) from Cartesian to spherical
   coordinates, (vi) point P(1,30o,45o) from spherical to Cartesian coordinates, and (vii) point P(1,30o,45o)
   from spherical to cylindrical coordinates.
                                                    
                                                                           ˆ ˆ                ˆ
2. Given points A(-1,-3,-4) and B(2,2,2), find: rA , rB , R AB , rA , rB , a A , a B , R AB & a AB .
                                                                                                       
3. Given points A(2,5,-1), B(3,-2,4) and C(-2,3,1), find: RAB  RAC ; the angle between R AB and R AC ; the
                                                                                                
   length of the projection of R AB on R AC ; the vector projection of R AB on R AC , R AB  R AC ; the area of the
   triangle defined by the points A, B, and C; and a unit vector perpendicular to the plane in which the
   triangle is located.
4. Given points A(x = 2, y = 3, z = -1) and B(ρ = 4,Φ = -50o, z = 2), find the distances from A to the origin,
   B to the origin, and A to B. Also find a unit vector in cylindrical coordinates at point A directed toward
   point B.
5. Given points A(x = 2, y = 3, z = -1) and B(r = 4, θ = 25o, Φ = 120o), find the spherical coordinates of A,
   the Cartesian coordinates of B, and the distance from A to B.
                                                                                                        ˆ
6. Transform each of the following vectors to cylindrical coordinates at the point specified: 5a x at P(ρ =
   4,Φ = 120 , z = 2), 5a x at Q(x = 3, y = 4, z = -1), and 4a x  2a y  4a z at A(x = 2, y = 3, z = 5).
             o
                        ˆ                                    ˆ      ˆ      ˆ
                      
7. Express the field F  2xyza x  5x  y  z a z in cylindrical coordinates. Find F at P(ρ = 2,Φ = 60o, z =
                                ˆ                 ˆ
   3).
                                                                                                    ˆ
8. Transform each of the following vectors to spherical coordinates at the point specified: 5a x at P(r = 4, θ
   = 25 , Φ = 120 ), 5a x at Q(x = 2, y = 3, z = -1), and 4a x  2a y  4a z at A(x = -2, y = -3, z = 4).
       o         o
                      ˆ                                    ˆ      ˆ      ˆ

9. A closed surface is defined in spherical coordinates by 3<r<5, 0.1π< θ<0.3π, 1.2π< Φ <1.6π. Find the
   volume enclosed, the total surface area and the distance from A(r = 3, θ = 0.1π , Φ = 1.2π) to B(r = 53, θ
   = 0.3π , Φ = 1.6π). [Ans: 14.91, 36.8, 3.86].
10. Find the total volume defined by 4< ρ<6, 30o< Φ<60o, 2<z<5. What is the length of the longest straight
    line that lies entirely within the volume? Find the total area of the surface.


                                              Tutorial Sheet-2


1. A 2 mC positive charge Q1 is located in vacuum at (3,-2,-4), and a 5 C negative charge Q2 is at (1,-4,2).
   Find the vector force on the negative charge. What is the magnitude of the force on the positive charge?
                ˆ          ˆ         ˆ
   [Ans: 0.616 a x +0.616 a y -1.848 a z N, 2.04 N].
              
2. Calculate E at A (3,-4,2) in free space caused by(a) a charge Q1 = 2 C at B(0,0,0); (b) a charge Q2 = 3
   C at B(-1,2,3); (c) a charge Q1 = 2 C at B(0,0,0) and a charge Q2 = 3 C at B(-1,2,3). [Ans: 345 a x -
                                                                                                      ˆ
       ˆ         ˆ             ˆ       ˆ        ˆ         ˆ        ˆ          ˆ
   460 a y +230 a z V/m; 280 a x -419 a y -69.9 a z ; 625 a x -880 a y +160.3 a z ].

3. Find the total charge inside each of the volumes indicated: (a) v = 10z2e-0.1xsiny, -1  x  2, 0  y  1,
   3  z  3.6; (b) v = 4xyz2, 0    2, 0    /2, 0  z  3; (c) v = 3 cos2 cos2/[2r2(r2+1)],
   universe. [Ans: 119.5 C, 72 C, 15.5 C].
4. Charge is distributed uniformly along an infinite straight line with constant density l. Develop the
                   
   expression for E at the general point P.

                                                                                                                11
                                 
5. Develop an expression for E due to charge uniformly distributed over an infinite plane with density S.
                                                                                                   
6. An infinitely long, uniform line charge is located at y = 3, z = 5. If l = 30 nC/m, find E at: (a) the
                                                     ˆ         ˆ             ˆ         ˆ             ˆ
   origin; (b) A (0.6.1); (c) B (5.6.1). [Ans: -47.6 a y -79.3 a z V/m; 64.7 a y -86.3 a z V/m; 64.7 a y -
        ˆ
   86.3 a z V/m].
7. Four infinite uniform sheets of charge are located as follows: 20 pC/m2 at y = 7, -8 pC/m2 at y = 3, 20
                                                                       
   pC/m2 at y = 7, 6 pC/m2 at y = -1, and -18 pC/m2 at y = -4. Find E at the point: (a) A(2,6,-4); (b)
                                                                 ˆ               ˆ              ˆ
   B(0,0,0); (c) C(-1,-1.1,5); (d) D(106, 106, 106). [Ans: -2.26 a y V/m; -1.355 a y V/m; -2.03 a y V/m; 0].
8. Obtain the equation of the streamline that passes through the point P(-2,7,10) in the field =: (a) 2(y-1)
   ˆ       ˆ            ˆ            ˆ
   a x +2x a y ; (b) ey a x +(x+1)ey a y . [Ans: (y-1)2-x2=32; 2y-(x+1)2=13].

9. A uniform line charge of l = (2×10-8/6) C/m lies along the x axis and a uniform sheet of charge is
                                                                           
    located at y = 5 m. Along the line y = 3 m, z = 3 m the electric field E has only a z-component. What is
    S for the field? [Ans: 125 pC/m2].
                                                                                        
10. The circular region,  < a, z=0, carries a uniform surface charge density S. Find E at P(0,0,h), h>0.


                                               Tutorial Sheet 3


1. A 25 C point charge is located at the origin. Calculate the electric flux passing through: (a) that portion
   of the sphere r = 20 cm bounded by  = 0 and ,  = 0 and /2; (b) the closed surface  = 0.8 m, z = 0.5
   m; (c) the plane z = 4 m. [Ans: 6.25 C, 25 C, 12.5 C]
           
2. Find D in Cartesian coordinates at P(6, 8, -10) caused by: (a) a point charge of 30 mC at the origin; (b)
   a uniform line charge l = 40 C/m on the z axis; (c) a uniform surface charge density s = 57.2 C/m2
   on the plane x = 9. [Ans: 5.06 a x +6.75 a y -8.44 a z C/m2 ; 0.382 a x +0.509 a y C/m2 ; -28.6 a x C/m2 ]
                                     ˆ       ˆ        ˆ                 ˆ          ˆ                 ˆ
         
              ˆ
3. Let D =r a r /3 nC/m2 in free space. Find: (a) E at r = 0.2 m; (b) total charge within the sphere r = 0.2 m;
   (c) total electric flux leaving the sphere r = 0.3 m. [Ans: 7.53 V/m; 33.5 pC; 113.1 pC]
4. Find the total electric flux leaving the spherical surface r = 2.5 m given the charge configuration: (a) Q =
   2  x nC on the x axis at x = 0, 1, 2, …. M; (b) a line charge l = 1/(z2+1) nC/m on the z axis; (c) a
        2



   surface charge s = 1/(x2+y2+4) nC/m2 on the z = 0 plane. [Ans: 2.125 nC, 2.38 nC, 2.96 nC]
5. Surface charge densities of 200, -50, and sx C/m2 are located at r = 3, 5, and 7 cm, respectively. Find
    
   D at r =: (a) 2 cm; (b) 4 cm; (c) 6 cm. (d) Find sx if = 0 at r = 7.32 cm. [Ans: 0, 112.5 a r C/m2, -
                                                                                                 ˆ
   11.22 C/m ]2

        
                ˆ          ˆ          ˆ
6. Let D =y2z3 a x +2xyz3 a y +3xy2z2 a z pC/m2 in free space. Find the total electric flux passing through the
   surface x = 3, 0  y  2, 0  z  1 in a direction away from the origin. Find E at P(3,2,1). Find the total
   charge contained in an incremental sphere having a radius of 2 m centered at P(3,2,1). [Ans: 0.667 pC,
   4.31 V/m, 2.61 × 10-27 C]
                                                                                                 
7. Find the volume charge density that is associated with each of the following fields: (a) D =
                                     
   xy2 a x +yx2 a y +z a z C/m2; (b) D = z2sin2 a  +z2sincos a  +2zsin2 a z C/m2; (c) a r C/m2. [Ans:
       ˆ        ˆ      ˆ                           ˆ               ˆ             ˆ             ˆ
   x2+y2+1 C/m3; z2+2zsin2 C/m3; 2/r C/m3]
                            
                                          ˆ            ˆ
8. Given the flux density D =(2cos/r3) a r +(sin/r3) a C/m2, evaluate both sides of the divergence
   theorem for the region defined by 1 < r < 2, 0 <  < /2, 0 <  < /2. [Ans: 0; 0]
                                       
9. Let D =x a x and find the value of  DdS over the surface of the sphere r = 1. [Ans: 4.19]
            ˆ
                                       S


                                                                                                               12
                                                      
10. Given that D =0z a z in the region -1  z  1 and D =(0z/|z|) a z elsewhere, find the charge density. [Ans:
                        ˆ                                           ˆ
    0 for -1  z  1 and 0 elsewhere]


                                               Tutorial Sheet-4


1. An electric field is given as E = 6y2zax + 12xyzay + 6xy2az V/m. An incremental path is represented by
   ∆L = -3ax + 5ay -2az m. Find the work done in moving a 2C charge along this path if the location of
   the path is at: (a) A(0,2,5); (b) B(1,1,1); (c) C(-0.7,-2,-0.3). Ans: 720pJ; -60pJ; -60pJ.
2. Find the work done in moving a 5C charge from the origin to P(2,-1,4) through the field E = 2xyzax +
   x2zay + x2yaz V/m via the pat: (a) straight line segments: (0,0,0) to (2,0,0) to (2,-1,0) to (2,-1,4); (b)
   straight line: x = -2y, z = 2x; (c) curve: x = -2y3, z = 4y2. Ans: 80J; 80J; 80J.
3. A time-varying E field need not be conservative. Let E = xay V/m at a certain instant of time, and
   calculate the work required to move a 3C point charge from (1,3,5) to (2,0,3) along the straight-line
   segments joining: (a) (1,3,5) to (2,3,5) to (2,0,5) to (2,0,3); (b) (1,3,5) to (1,3,3) to (1,0,3) to (2,0,3).
   Ans: 18J; 9J.
4. Let E = (-6y/x2)ax + (6/x) ay + 5 az V/m and calculate: (a) VPQ given P(-7,2,1) and Q(4,1,2); (b) VP if V
   = 0 at Q; (c) VP if V = 0 at (2,0,-1). Ans: 8.21V; 8.21V; -8.29V.
5. A point charge of 6nC is located at the origin in free space. Find VP if point P is located at P(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). Ans: 89.9V; 36.0V;
   73.9V.
6. Assume a zero reference at infinity, and find the potential at P(0,0,10) that is caused by this charge
   configuration in free space: (a) 20nC at the origin; (b) 10 nC/m along the line x = 0, z = 0, -1 < y < 1; (c)
   10 nC/m along the line x = 0, y = 0, -1 < z < 1. Ans: 17.98V; 17.95V; 18.04V.
7. If V = (60 sin )/r2 V in free space and point P is located at r = 3 m,  = 60o,  = 25o, find: (a) VP; (b) EP;
   (c) dV/dN at P; (d) aN at P; (e) v at P. Ans: 5.77V; 3.85ar – 1.111a V/m; 4.01 V/m; - 0.961ar +
   0.277a; - 7.57pC/m3.
8. A dipole of moment p = - 4ax + 5ay + 3az nC-m is located at D(1,2,-1) in free space. Find V at: (a)
   PA(0,0,0); (b) PB(1,2,0); (c) PC(1,2,-2); (d) PD(2,6,1). Ans: -1.835 V; 27.0 V; -27.0 V; 2.02 V.
9. Point charges of +3C and -3C are located at (0,0,1 mm) and (0,0,-1 mm), respectively, in free space.
   (a) Find p. (b) Find E in spherical components at P(r = 2,  = 40o,  = 50o). (c) Find E in spherical
   components at (1,2,1.5). Ans: 6az nC-m; 10.33ar + 4.33a V/m; 3.08ar + 2.29 a V/m.
10. Find the energy stored in free space for the region, 0 < r <a, 0 <  < , 0 < z < 2, given the potential field
    V = : (a) V0r/a; (b) V0r/a cos 2. Ans: 1.5710V02; 1.3740V02.


                                               Tutorial Sheet-5
1. The vector current density is given as J = (4/r2) cos  ar + 20e-2r sin  a - r sin  cos am(a) Find
   J at r = 3,  = 0, (b) Find the total current passing through the spherical cap r = 3, 0 <  < 20o, 0 <
    < 2, in the ar direction. Ans: 0.444 arm1.470 A.
2. Assume that an electron beam carries a total current of -500A in the az direction, and has a current
   density Jz that is not a function of  and  in the region 0  m. If the electron velocities are
   given by vz = 8  107 z m/s, calculate v at  = 0 and z =: (a) 1 mm; (b) 2 cm; (c) 1 m. Ans: - 0.1989
   C/m3; -9.95 mC/m3; -198.9 C/m3.
3. Find the magnitude of the electric field intensity in a sample of silver having  = 6.17  107 mho/m and
   e = 0.0056 m2/V-s if: (a) the drift velocity is 1 mm/s; (b) the current density is 107 A/m2; (c) the
   sample is a cube, 3 mm on a side, carrying a total current of 80 A; (d) ) the sample is a cube, 3 mm on a

                                                                                                               13
   side, having a potential difference of 0.5 mV between opposite faces. Ans: 0.1786 V/m; 0.1621 V/m;
   0.1441 V/m; 0.1667 V/m.
4. An aluminum conductor is 1000 ft long and has a circular cross-section with a diameter of 0.8 in. If
   there is a dc voltage of 1.2 V between the ends, find: (a) the current density; (b) the current; (c) the
   power dissipated. Ans: 1.504  105 A/m2; 48.8 A; 58.5 W.
5. A potential field is given as V = 100e-5x sin 3y cos 4z V. Let point P(0.1,/12,/24) be located at a
   conductor-free space boundary. At point P, find the magnitude of: (a) V; (b) E; (c) EN; (d) Et; (e) s.
   Ans: 37.1 V; 233 V/m; 0; 2.06 nC/m2.
6. A point charge of 25 nC is located in free space at P(2,-3,5), and a perfectly conducting plane is at z = 2.
   Find: (a) V at (3,2,4); (b) E at (3,2,4); (c) s at(3,2,2). Ans: 11.78 V; 0.985ax + 4.92 ay – 4.69az V/m; -
   57.6 pC/m2.
7. A certain homogeneous slab of lossless dielectric material is characterized by an electric susceptibility
   of 0.12 and carries a uniform electric flux density within it of 1.6 nC/m2. Find: (a) E; (b) P; (c) the
   average dipole moment if there are 2  1019 dipoles per cubic meter; (d) the voltage between two
   equipotentials 1 in apart. Ans: 161.3 V/m; 171.4 pC/m2; 8.57  10-30 C; 4.10 V.
8. The region y < 0 contains a dielectric material for which R1 = 2.5, while the region y > 0 is
   characterized by R2 = 4. Let E1 = -30ax + 50ay + 70az V/m, and find: (a) EN1; (b) Et1; (c) Et1; (d) E1; (e)
   1; (f) DN2; (g) Dt2; (h) D2; (i) P2; (j) 2. Ans: 50 V/m; -30ax + 70az V/m; 76.2 V/m; 91.1 V/m; 56.7o;
   1.107 nC/m2; 2.7 nC/m2; -1.062ax + 1.107ay + 2.48az nC/m2; -0.797ax + 0.830ay + 1.859az nC/m2;
   67.7o.
9. Find the relative permittivity of the dielectric material used in a parallel plate capacitor if: (a) C = 40 nF,
   d = 0.1 mm, and S = 0.15 m2; (b) d = 0.2 mm, E = 500 kV/m, and S = 10 C/m2; (c) D = 50 C/m2 and
   the energy density is 20 J/m3. Ans: 3.01; 2.26; 7.06.
10. Find the capacitance of: (a) 20 cm of 58C/U coaxial cable having an inner conductor 0.0295 in in
    diameter, an outer conductor having an inside diameter of 0.116 in, and a polyethylene dielectric; (b) a
    conducting sphere 1 cm in diameter, covered with a layer of polyethylene 1 cm thick, in free space; (c) a
    conducting sphere 1 cm in diameter, covered with a layer of polyethylene 1 cm thick, and surrounded by
    a concentric conducting sphere 1.5 cm in radius. Ans: 18.37 pF; 0.885 pF; 1.886 pF.


                                               Tutorial Sheet-9


1. Given the magnetic flux density, B = 6 cos 106t sin 0.01x az mT, find: (a) the magnetic flux passing
   through the surface z = 0, 0 < x < 20 m, 0 < y < 3 m, at t = 1 s; (b) the value of the closed line integral
   of E around the perimeter of the surface specified above, at t = 1 s. Ans: 19.39 mWb; 30.2 kV.
2. Find the amplitude of the displacement current density: (a) in the air near a car antenna, where the field
   strength of an FM signal is E = 80 cos (6.277  108t – 2.092y) az V/m; (b) in an air space within a large
   power transformer where H = 106 cos (377t + 1.2566  10-6z) ay A/m; (c) inside a capacitor where R =
   600 and D = 3  10-6 sin (6  106t – 0.3464x) az C/m2; (d) inside a typical metallic conductor where f = 1
   kHz,  = 5  107 mho/m, R = 1; and the conduction current density is J = 107 sin ( 6283t – 444z) ax
   A/m2. Ans: 0.445 A/m2; 1.257 A/m2; 18 A/m2; 11.13 nA/m2.
3. The unit vector 0.48ax – 0.6ay + 0.64az is directed from region 2 (eR2 = 2.5, mR2 = 2, s2 = 0) toward
   region 1 (eR1 = 4, mR1 = 10, s1 = 0). If H1 = (-100ax – 50ay +200az) sin 400t A/m at point P in region
   1 adjacent to the boundary, find the amplitude at P of: (a) HN1; (b) Ht1; (c) HN2; (d) H2. Ans: 110
   A/m; 201 A/m; 550 A/m; 586 A/m.
4. A coaxial transmission line with



                                                                                                               14
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Description: electromagnet theory basics