1. UNIVERSITI MALAYSIA PERLIS
EKT 241/4: ELECTROMAGNETIC THEORY
UNIVERSITI MALAYSIA PERLIS
CHAPTER 3 – ELECTROSTATICS
PREPARED BY: NORDIANA MOHAMAD SAAID

2. Chapter Outline Maxwell’s Equations Charge and Current Distributions
Coulomb’s Law
Gauss’s Law
Electric Scalar Potential
Electrical Properties of Materials
Conductors & Dielectrics
Electric Boundary Conditions
Capacitance
Electrostatic Potential Energy
Image Method

3. Maxwell’s equations Maxwell’s equations: Where;
E = electric field intensity
D = electric flux density
ρv = electric charge density per unit volume
H = magnetic field intensity
B = magnetic flux density

4. Maxwell’s equations Maxwell’s equations: Relationship: D = ε E B = µ H
ε = electrical permittivity of the material
µ = magnetic permeability of the material

5. Maxwell’s equations For static case, ∂/∂t = 0.
Maxwell’s equations is reduced to:
Electrostatics Magnetostatics

6. Charge and current distributions
Charge may be distributed over a volume, a surface or a line.
Electric field due to continuous charge distributions:

7. Charge and current distributions
Volume charge density, ρv is defined as:
Total charge Q
contained in
a volume V is:

8. Charge and current distributions
Surface charge density
Total charge Q on a surface:

9. Charge and current distributions
Line charge density
Total charge Q along a line

10. Example 1
Calculate the total charge Q contained in a cylindrical tube of charge oriented along the z-axis. The line charge density is , where z is the distance in meters from the bottom end of the tube. The tube length is 10 cm.

11. Solution to Example 1
The total charge Q is:

12. Example 2
Find the total charge over the volume with volume charge density:

13. Solution to Example 2
The total charge Q:

14. Current densities Current density, J is defined as: Where:
u = mean velocity of moving charges
For surface S, total current flowing through is

15. Current densities There are 2 types of current: Convection current
generated by actual movement of electrically charged matter; does NOT obey Ohm’s law
Conduction current
atoms of conducting material do NOT move; obeys Ohm’s law

16. Coulomb’s Law Coulomb’s law for a point charge: Where;
R = distance between P and q = unit vector from q to P
ε = electrical permittivity of the medium containing the observation point P

17. Force acting on a charge
In the presence of an electric field E at a given point in space,
F is the force acting on a test charge q’ when that charge is placed at that given point in space
The electric field E is maybe due to a single charge or a distribution of many charges
Units:
F in Newtons (N), q’ in Coulombs (C)

18. For acting on a charge For a material with electrical permittivity, ε:
D = ε E
where:
ε = εR ε0
ε0 = 8.85 × 10−12 ≈ (1/36π) × 10−9 (F/m)
For most material and under most condition, ε is constant, independent of the magnitude and direction of E

19. E-field due to multipoint charges
At point P, the electric field E1 due to q1 alone:
At point P, the electric field E1 due to q2 alone:

20. E-field due to multipoint charges
Total electric field E at point P due to two charges:

21. E-field due to multipoint charges
In general for case of N point of charges,

22. Example 3
Two point charges with and are located in free space at (1, 3,−1) and (−3, 1,−2), respectively in a Cartesian coordinate system.
Find:
(a) the electric field E at (3, 1,−2)
(b) the force on a 8 × 10−5 C charge located at that point. All distances are in meters.

23. Solution to Example 3
The electric field E with ε = ε0 (free space) is given by:
The vectors are:

24. Solution to Example 3
a) Hence,
b) We have

25. E-field due to charge distribution
Total electric field due to 3 types of distribution:

26. E-field of a ring of charge
Electric field due to a ring of charge is:

27. Example 4
Find the electric field at a point P(0, 0, h) in free space at a height h on the z-axis due to a circular disk of charge in the x–y plane with uniform charge density ρs as shown.
Then evaluate E for the infinite-sheet case by letting a→∞.

28. Solution to Example 4 A ring of radius r and width dr has an area
ds = 2πrdr
The charge is:
The field due to the ring is:

29. Solution to Example 4 The total field at P is
For an infinite sheet of charge with a =∞,

30. Gauss’s law
Electric flux density D through an enclosing surface is proportional to enclosed charge Q.
Differential and integral form of Gauss’s law:

31. Example 5
Use Gauss’s law to obtain an expression for E in free space due to an infinitely long line of charge with uniform charge density ρl along the z-axis.

32. Solution to Example 5 Construct a cylindrical Gaussian surface.
The integral is:
Equating both equations, and re-arrange, we get:

33. Solution to Example 5 Then, use , we get:
Note: unit vector is inserted for E due to the fact that E is a vector in direction.

34. Electric scalar potential
Electric potential energy is required to move a unit charge between 2 points
The presence of an electric field between two points give rise to voltage difference

35. Electric Potential as a function of electric field
Integrating along any path between point P1 and P2, we get:
Potential difference between P1 and P2 :

36. Electric Potential as a function of electric field
Kirchhoff’s voltage law states that net voltage drop around a closed loop is zero.
Line integral E around closed contour C is:
The electric potential V at any point is given by:
Integration path between point P1 and P2 is arbitrary

37. Electric potential due to point charges
For a point charge located at the origin of a spherical coordinate systems, the electric field at a distance R:
The electric potential between two end points:

38. Electric potential due to point charges
For charge Q located other than origin, specified by a source position vector R1, then V at the observation vector R becomes:
For N discrete point charges, electric potential is:

39. Electric potential due to continuous distributions
For a continuous charge distribution:

40. Electric field as a function of electric potential
To find E for any charge distribution easily,
where: = gradient of V

41. Poisson’s & Laplace’s equations
Differential form of Gauss’s law:
This may be written as:
Then, using , we get:

42. Poisson’s & Laplace’s equations
Hence:
Poisson’s and Laplace’s equations are used to find V where boundaries are known:
Example: the region between the plates of a capacitor with a specified voltage difference across it.

43. Conductivity
Conductivity – characterizes the ease with which charges can move freely in a material.
Perfect dielectric, σ = 0. Charges do not move inside the material
Perfect conductor, σ = ∞. Charges move freely throughout the material

44. Conductivity
Drift velocity of electrons, in a conducting material is related to externally applied electric field E:
Hole drift velocity, is in the same direction as the applied electric field E:
where: µe = electron mobility (m2/V.s)
µh = hole mobility (m2/V.s)

45. Conductivity Conductivity of a material, σ, is defined as:
where ρve = volume charge density of free electrons
ρvh = volume charge density of free holes
Ne = number of free electrons per unit volume
Nh = number of free holes per unit volume
e = absolute charge = 1.6 × 10−19 (C)

46. Conductivity
Conductivities of different materials:

47. Ohm’s Law Point form of Ohm’s law states that:
Where: J = current density
σ = conductivity
E = electric field intensity
Properties for perfect dielectric and conductor:

48. Example 6
A 2-mm-diameter copper wire with conductivity of 5.8 × 107 S/m and electron mobility of (m2/V·s) is subjected to an electric field of 20 (mV/m).
Find (a) the volume charge density of free electrons, (b) the current density, (c) the current flowing in the wire, (d) the electron drift velocity, and (e) the volume density of free electrons.

49. Solution to Example 6 a) b) c) d) e)

50. Resistance
The resistance R of a conductor of length l and uniform cross section A (linear resistor) as shown in the figure below:

51. Resistance For a resistor of arbitrary shape, the resistance R is:
The conductance G of a linear resistor:

52. Joule’s Law
Joule’s law states that for a volume v, the total dissipated power P is:
For a linear resistor, the dissipated power P:

53. Dielectrics Conductor has free electrons.
Dielectric electrons are strongly bounded to the atom.
In a dielectric, an externally applied electric field, Eext cannot cause mass migration of charges since none are able to move freely.
But, Eext can polarize the atoms or molecules in the material.
The polarization is represented by an electric dipole.

54. Dielectrics
Fig (a) - Eext is absent: the center of the electron cloud is co-located with the center of the nucleus
Fig (b) - Eext is present: the two centers are separated by a distance d
Fig (c) – an electric dipole caused by Eext

55. Electric Boundary Conditions
Electric field maybe continuous in each of two dissimilar media
But, the E-field maybe discontinuous at the boundary between them
Boundary conditions specify how the tangential and normal components of the field in one medium are related to the components in other medium across the boundary
Two dissimilar media could be: two different dielectrics, or a conductor and a dielectric, or two conductors

56. Dielectric- dielectric boundary
Interface between two dielectric media

57. Dielectric- dielectric boundary
Based on the figure in previous slide:
First boundary condition related to the tangential components of the electric field E is:
Second boundary condition related to the normal components of the electric field E is: OR

58. Perfect conductor
When a conducting slab is placed in an external electric field,
Charges that accumulate on the conductor surfaces induces an internal electric field
Hence, total field inside conductor is zero.

59. Dielectric-conductor boundary
Assume medium 1 is a dielectric
Medium 2 is a perfect conductor

60. Dielectric-conductor boundary
Based on the figure in previous slide:
In a perfect conductor,
Hence,
This requires the tangential and normal components of E2 and D2 to be zero.

61. Dielectric-conductor boundary
The fields in the dielectric medium, at the boundary with the conductor is
Since , it follows that
Using the equation, ,
we get:
Hence, boundary condition at conductor surface:
where = normal vector pointing outward

62. Conductor- conductor boundary
Boundary between two conducting media:
Using the 1st and 2nd boundary conditions:
and

63. Conductor- conductor boundary
In conducting media, electric fields give rise to current densities.
From , we have:
and
The normal component of J has be continuous across the boundary between two different media under electrostatic conditions.

64. Conductor- conductor boundary
Hence, upon setting , we found the boundary condition for conductor- conductor boundary:

65. Capacitance
Capacitor – two conducting bodies separated by a dielectric medium

66. Capacitance Capacitance is defined as:
where: V = potential difference (V)
Q = charge (C)
C = capacitance (F)

67. Example 7
Obtain an expression for the capacitance C of a parallel-plate capacitor comprised of two parallel plates each of surface area A and separated by a distance d. The capacitor is filled with a dielectric material with permittivity ε.

68. Solution to Example 7
The charge density on the upper plate is ρs = Q/A. Hence,
magnitude of E at the dielectric-conductor boundary:
The voltage difference is
Hence, the capacitance is:

69. Electrostatic potential energy
Assume a capacitor with plates of good conductors – zero resistance,
Dielectric between two conductors has negligible conductivity, σ ≈ 0 – no current can flow through dielectric
No ohmic losses occur anywhere in capacitor
When a source is connected to a capacitor, energy is stored in capacitor
Charging-up energy is stored in the form of electrostatic potential energy in the dielectric medium

70. Electrostatic potential energy
The capacitance:
Hence, We for a parallel plate capacitor:

71. Image Method
Image theory states that a charge Q above a grounded perfectly conducting plane is equal to Q and its image –Q with ground plane removed.

72. Example 8
Use image theory to determine E at an arbitrary point P (x, y, z) in the region z > 0 due to a charge Q in free space at a distance d above a grounded conducting plane.

73. Solution to Example 8
Charge Q is at (0, 0, d) and its image −Q is at (0,0,−d) in Cartesian coordinates. Using Coulomb’s law, E at point P(x,y,z) due to two point charges: