Electromagntic field and waves slide 1

1. Course Code: EEE 2215
Course Name: Electromagnetic Fields and Waves
Md Jubayer Faisal
Lecturer, Dept. Of EEE, City University
Department of EEE, City University, Bangladesh

2. 2
Electric field intensity due to a point charge (at the origin)
In order to find the electric field intensity due to q, we draw a
hypothetical spherical surface of a radius R centered at q. Since a
point charge has no preferred directions, its electric field must be
everywhere radial and has the same intensity at all points on the
spherical surface. Applying Eq. (4) to Fig. 3-2(a), we have
Figure 1 Point charge at the origin.
------------(6)

3. 3
Electric field intensity due to a point charge (at the Origin)
Equation (6) tells us that the electric field intensity of a positive
point charge is in the outward radial direction and has a
magnitude proportional to the charge and inversely proportional
to the square of the distance from the charge. This is a very
important basic formula in electrostatics.
Figure 1 Point charge at the origin.
------------(6)

4. 4
Electric field intensity due to a point charge(Not at the Origin)
Figure 2 Point charge not at the origin.
If the charge q is not located at the origin
-------- (7)
where aqp is the unit vector drawn from q to P. Since
-------- (8)

5. 5
EXAMPLE 1 Determine the electric field intensity at P( -0.2, 0, -2.3) due to a point charge
of +5 (nC) at Q(0.2, 0.1, -2.5) in air. All dimensions are in meters.

6. 6
EXAMPLE 1 Determine the electric field intensity at P( -0.2, 0, -2.3) due to a point
charge of +5 (nC) at Q(0.2, 0.1, -2.5) in air. All dimensions are in meters.
Substituting in Eq. (8), we obtain

7. 7
The combination of two equal point charges of opposite sign separated by a small distance (l) is
called an electric dipole or simply dipole and the product (Q.I ) is known as the electric dipole
moment.
Electric Dipole
The potentials due to positive charge ( Q) and negative charge ( - Q) at the point P are given respectively as
𝑉1 =
𝑄
4𝜋𝜀𝑟1
𝑎𝑛𝑑 𝑉2 =
−𝑄
4𝜋𝜀𝑟2
Figure 3 An Electric Dipole

8. 8
Electric Dipole
Hence the total potential P is
V = V1 + V2 =
𝑄
4𝜋𝜀𝑟1
+
−𝑄
4𝜋𝜀𝑟2
V =
𝑄
4𝜋𝜀
(
1
𝑟1
−
1
𝑟2
) -------------(1)
If now the point P is at a very large distance as compared with the separation 1, to that he radial lines
𝑟1 , 𝑟 , 𝑟2 are essentially parallel then,
where r and 𝜃 are as indicated in Fig 3. Hence by putting the value of 𝑟1and 𝑟2 in Equation (1) the
resultant potential at a distance r from the electric dipole is given by

9. 9
Electric Dipole
where r and 𝜃 are as indicated in Fig 3. Hence by putting the value of 𝑟1and 𝑟2 in Equation (1) the resultant
potential at a distance r from the electric dipole is given by
-------------------(2)
• Indicates that potential along perpendicular bisector (i.e 𝜃 = 900) to the dipole axis is 0.
• The Potential increase proportionally with the dipole moment and inversely with the square of the
distance.

10. 10
Electric field due to a continuous distribution of charge
Figure 1 Electric field due to a continuous charge
distribution
Since a differential element of charge behaves like a point
charge, the contribution of the charge 𝜌dv' in a
differential volume element dv' to the electric field
intensity at the field point P is
We have,
or, since 𝑎𝑅 = R/R,
----------------(1)

11. Electric field due to a continuous distribution of charge
If the charge is distributed on a surface with a surface charge density 𝜌𝑠 ( Τ
𝑐
𝑚2 ), then the
integration is to be carried out over the surface (not necessarily flat). Thus,
For a line charge we have,
----------------(2)
----------------(3)

12. Gauss Law and its Application
Statements: Gauss's law states that the total outward flux of the E-field over any closed surface in
free space is equal to the total charge enclosed in the surface divided by 𝜀0.
Applications: Gauss's law is particularly useful in determining the E-field of charge distributions with
some symmetry conditions, such that the normal component of the electric field intensity is constant
over an enclosed surface. In such cases the surface integral on the left side of Eq. 1 would be very
easy to evaluate, and Gauss's law would be a much more efficient way for finding the electric field
intensity.
---------------(1)
Conditions of Gaussian Surface:
(a) The surface is closed.
(b) At each point of the surface D is either normal or tangential
to the surface. (D = Flux density)
(c) D has the same value at all points of the surface where D is
normal Such Gaussian surface

13. Gauss Law and its Application
EXAMPLE 3-5 (DK Cheng) Use Gauss's law to determine the electric field intensity of an
infinitely long, straight, line charge of a uniform density 𝜌l in air.
Note:
• Since the line charge is infinitely long, the resultant E field
must be perpendicular so E along the line cannot exist.
• As the line charges are uniformly distributed so there is no
variation at the 𝜑
• So only exist r at the coordinate system
Solution:
Surface radius = r and surface Length = L
The resulted E field must be radial and perpendicular to the
line charge (E = ො
𝑎𝑟𝐸𝑅).
On this surface, 𝐸𝑟 is constant, and ds = ො
𝑎𝑟𝑟𝑑𝜑 ⅆ𝑧
Hence, Figure 1 Applying Gauss's law to an infinitely long line charge

14. Gauss Law and its Application
There is no contribution from the top or the bottom face of the cylinder because E has no z-component there,
making E • ds = 0. The total charge enclosed in the cylinder is Q = 𝜌𝑙L.
So from equation 1,
2𝜋𝐿𝐸𝑟 =
𝜌𝑙𝐿
𝜖0
where Q = 𝜌𝑙𝐿
Finally,