1. Lecture 3: Static Fields
Instructor: Dr. Gleb V. Tcheslavski
Contact:
Office Hours:
Room 2030
Class web site:

2. 1. Electrostatic Fields 1.1. Coulomb’s Law
Something known from the ancient time (here comes amber): two charged particles exert a force on each other…
Electrostatic (Coulomb’s) force:
(3.2.1)
where Q1 and Q2 are charges,
R –distance between particles,
uR – the unit-vector
In this notation, negative force means attraction, positive – repelling.

3. 1. Electrostatic Fields 1.1. Coulomb’s Law (Example)
Find the magnitude of the Coulomb force that exists between an electron and a proton in a hydrogen atom. Compare the Coulomb force and the gravitational force between the two particles. The two particles are separated approximately by 1 Ångström 1Å m.
(3.3.1)
(3.3.2)
This is why chemical bounds are so strong!

4. 1. Electrostatic Fields 1.2. Electric (electrostatic) Field
Electrostatic field due to the charge Q:
(3.4.1)
An “alternative definition”:
(3.4.2)
What’s wrong with it?
For a system of two charges:

5. 1. Electrostatic Fields 1.3. Superposition
For several charges placed at different locations in space, the total electric field at the particular location would be a superposition (vector summation) of individual electric fields:
(3.5.1)
a vector sum!
(Example): find the EF at P
Q1 = +1C, Q1 = +2C, Q3 = -3C

6. 1. Electrostatic Fields 1.3. Superposition (cont)
Volume charge density:
(3.6.1)
Surface charge density:
(3.6.2)
Linear charge density:
(3.6.3)
(3.6.4)
There is a differential electric field directed radially from each differential charges

7. 1. Electrostatic Fields 1.3. Superposition (Example)
Calculate the electric field from a finite charge uniformly distributed along a finite line.
Linear charge density: (z’)l
We assume a symmetry along z with respect to the observation point. Therefore, it will be a charge element at –z’ for every charge element at +z’. As a result, fields in z direction will cancel each other:
(3.7.1)

8. 1. Electrostatic Fields 1.3. Superposition (Example, cont)
The radial component:
(3.8.1)
Combining (3.4.1) and (3.6.3), we arrive to:
(3.8.2)
which, combined with (3.8.1) and integrated leads to:
(3.8.3)
(3.8.4)

9. 1. Electrostatic Fields 1.3. Superposition (Example 2)
Calculate the electric field from an infinite plane charged with s and consisting of an infinite number of parallel charged lines.
Symmetry leads to cancellation of tangent components.
Utilize (3.8.4) and that The linear charge density:
(3.9.1)
(3.9.2)

10. 1. Electrostatic Fields 1.4. Gauss’s Law
A charge Q is uniformly distributed within a sphere of radius a.
We can assume first that the charge is located at the center. Than, by (3.4.1):
(3.10.1)
By evaluating surface integrals of both sides
(3.10.2)
At the surface of the sphere, the unit-vector associated with the differential surface area ds points in the radial direction. Therefore, and the closed surface integral is

11. 1. Electrostatic Fields 1.4. Gauss’s Law (cont)
Therefore, the integral form:
(3.11.1)
Here Qencl is the charge enclosed within the closed surface.
By using divergence theorem and volume charge density concept:
(3.11.2)
Differential form:
(3.11.3)

12. 1. Electrostatic Fields 1.4. Gauss’s Law (cont 2)
For the charge Q uniformly distributed within the spherical volume
The volume charge density:
(3.12.1)
The total charge enclosed:
(3.12.2)
Gauss Law
a) Outside the sphere: r > a, Qencl = Q
(3.12.3)
(3.12.3)

13. 1. Electrostatic Fields 1.4. Gauss’s Law (cont 3)
a) Inside the sphere: r < a
(3.13.1)
Gauss Law
(3.13.2)
(3.13.3)

14. 1. Electrostatic Fields 1.5. Gaussian Surface
A Gaussian surface is a closed two-dimensional surface through which a flux or electric field is calculated. The surface is used in conjunction with Gauss's law (a consequence of the divergence theorem), allowing to calculate the total enclosed electric charge by exploiting a symmetry while performing a surface integral.
Commonly used are:
Spherical surface for
A point charge;
A uniformly distributed spherical shell of charge;
Other charge distribution with a spherical symmetry
b) Cylindrical surface for
A long, straight wire with a uniformly distributed charge;
Any long, straight cylinder or cylindrical shell with uniform charge distribution.

15. 1. Electrostatic Fields 1.6. Potential Energy and Electric Potential
A charged particle will gain a certain amount of potential energy as the particle is moved against an electric field.
(3.15.1)

16. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont)
Imaginary experiment: compute a total work required to bring three charged particles from - to the shaded region. No electric field exists at - and there are no friction, no gravity, and no other forces. I
There are no forces here, therefore, no work is required! W1 = 0;

17. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 2) xa xb II
We need to overcome the Coulomb’s force, therefore, some work is required.
(3.17.1)
since both charges are positive
V1 is an absolute electric potential caused by the charge Q1
(3.17.2)

18. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 3)
III
We need again to overcome the Coulomb’s force, therefore, some work is required.
(3.18.1)
Totally, for the three particles:
(3.18.2)
Or, for N particles:
(3.18.3)
Here xij is the distance between charges i and j;
(3.18.4)

19. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 4)
Note: the total work in our case is equal to the total electrostatic energy stored in the shaded region.
Note: the total work (and the total energy) do not depend on the order, in which particles are brought.
The electrostatic energy can also be evaluated as
(3.19.1)

20. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 5)
Electric potential difference between points a and b is the work required to move the charge from point a to point b divided by that charge.
We can express the electric potential difference or voltage as:
(3.20.1) or
(3.20.2)

21. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example)
Evaluate the work (charge times potential difference) required to move a charge q from a radius b to a radius a. 1’ 2’ 3’ 4’ 5’ 6’
The electric field is
(3.21.1)
The potential difference between the two spherical surfaces is
(3.21.2)
The potential at r =  is assumed to be 0 and is called a ground potential.
The electric potential defined with respect to the ground potential is called an absolute potential.

22. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example, cont)
Considering the path , we notice that there are only potential differences while going 1  2 and 3  4. Therefore, these are the only paths where some work is required. When moving 2  3, the potential is constant, therefore no work is required.
A surface that has the same potential is called an equipotential surface.
If the separation between two equipotential surfaces and the voltage between them are small:
(3.22.1)
(3.22.2)

23. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 6)
We can modify (3.22.2) as following:
(3.23.1)
Poisson’s equation
Laplace’s eqn. when v = 0
(3.23.2)
An absolute potential caused by a volume distribution that is not at the origin:
(3.23.3)

24. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 7)
The potential energy would be
for R  
(3.24.1)
(3.24.2)
Note that when a charged particle is moved along a closed contour, no work is required
(3.24.3)
Electrostatic field is conservative and irrotational.

25. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example 2)
Find the potential V due to two equal charges that have opposite signs and are in the vacuum. The distance from the point of interest is much greater than the separation. The configuration is known as an electric dipole.

26. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example 2, cont)
Due to superposition:
(3.26.1)
Since r >> d; r, r1, and r2 are almost parallel.
(3.26.2)
(3.26.3)
A vector p = Qd is a dipole moment.

27. 1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example 2, cont)
Electric potential distribution plot of (3.26.3)
plot
polar

28. Self-study 1. Electrostatic Fields 1.7. On numerical integration
When no symmetry can be used to simplify the problem, numerical integrations are quite helpful. Numerical integration = APPROXIMATION.
trapz
quad
Self-study
dblquad
triplequad

29. 1. Electrostatic Fields 1.8. Dielectric materials
A material can be considered as a collection of randomly (in general) oriented small electric dipoles.
If an external electric field is applied, the dipoles may orient themselves.

30. 1. Electrostatic Fields 1.8. Dielectric materials (cont)
We may suggest that an external electric field causes a “thin layer of charge” of the opposite sign at either edge of the material.
This charge is called a polarization charge.
The density of the polarization charge:
(3.30.1)
where P is the polarization field:
(3.30.2)
Here pj = Qdud is the dipole moment of individual dipole, N – number of atoms (dipoles)

31. 1. Electrostatic Fields 1.8. Dielectric materials (cont 2)
Let us add the polarization charge density to the real charge density. The Gauss’s Law will take a form:
(3.31.1)
which leads to
(3.31.2)
where
(3.31.3)
The total flux that passes through the surface
(3.31.4)

32. 1. Electrostatic Fields 1.8. Dielectric materials (cont 3)
Integrating (3.31.2) over a volume, leads to
(3.32.1)
Dielectric materials are susceptible to polarization. Usually, polarization is linearly proportional to the applied (small) electric field. Then
(3.32.2)
where e is the electric susceptibility
(3.32.3)
for linear and isotropic materials
r is the relative dielectric constant
We consider only linear materials here.

33. 1. Electrostatic Fields 1.9. Capacitance A parallel-plate capacitor
(3.33.1)
A parallel-plate capacitor
(3.33.2)
Assume A >> d

34. 1. Electrostatic Fields 1.9. Capacitance (cont) Stored energy:
(3.34.1)
(3.34.2)
(3.34.3)
Stored energy:
(3.34.4)
Assumed uniform field in the capacitor and uniform distribution of charge on plates

35. 1. Electrostatic Fields 1.9. Capacitance (Example)
Calculate the mutual capacitance of a coax cable with dielectric r inside…
From the Gauss’s Law:
(3.35.1)
Potential difference:
(3.35.2)
(3.35.3)
(3.35.4)
HW 2 is ready

36. 2. Magnetostatic Fields 2.1. Electric currents Let’s consider a wire…
The generalized Ohm’s Law specifies a current density:
(3.36.1)
Alternative:
(3.36.2)
where is the electron volume charge density, is an average electron drift velocity, is the mobility of the material.
(3.36.3)

37. 2. Magnetostatic Fields 2.1. Electric currents (cont)
The total current that passes through the wire
(3.37.1)
In our case f the current is distributed uniformly in a cylindrical wire:
(3.37.2)
The power density (density of power dissipating within a conductor):
(3.37.3)
The total power absorbed within the volume:
(3.37.4)

38. 2. Magnetostatic Fields 2.1. Electric currents (Examples)
a) Calculate the current flowing through the wire of radius a; current density:
“skin effect”
(3.38.1)
(3.38.2)
b) Calculate the power dissipated within a resistor with a uniform conductivity . The voltage across the resistor is V, a current passing through is I.
(3.38.3)

39. 2. Magnetostatic Fields 2.2. Fundamentals of Magnetic Fields
Magnetic loops at the surface of the Sun, as seen with the TRACE solar spacecraft. (©TRACE operation team, Lockheed Martin)
?Magnetic monopole?

40. 2. Magnetostatic Fields 2.2. Fundamentals of Magnetic Fields (cont)
Magnetic field lines are continuous, don’t originate nor terminate at a point. There is no “magnetic monopole”…
(3.40.1)
B is a magnetic flux density, [T] = [Wb/m2]
By applying the divergence theorem to (3.40.1):
(3.40.2)
(3.40.3)

41. 2. Magnetostatic Fields 2.2. Fundamentals of Magnetic Fields (cont 2)
As a result, we can split a bar magnet into tiny pieces and all of them will have both north and south poles.

42. 2. Magnetostatic Fields 2.2. Fundamentals of Magnetic Fields (cont 3)
Magnetic field can be created by the electric current:
Ampere’s Law:
(3.42.1)
electric current enclosed within a closed loop
A cylindrical wire caring a current creates a magnetic field.
“Right Hand Rule” (RHR).

43. 2. Magnetostatic Fields 2.2. Fundamentals of Magnetic Fields (cont 4)
By applying Stokes’s theorem to (3.42.1), we arrive to
(3.43.1)
Therefore, the differential
form of Ampere’s Law is:
(3.43.2)
It appears that even very small currents generate magnetic fields…
An axial view of the cortically-generated magnetic field of a human listener, measured using whole-head magnetoencephalography (MEG) – from the journal “Cerebral Cortex”

44. 2. Magnetostatic Fields 2.2. Fundamentals of Magnetic Fields (Example)
A symmetry in the system greatly simplifies evaluation of integrals in (3.42.1).
(3.44.1)
The right-hand side of (3.42.1) for the radius greater than a is just 0I.
(3.44.2)
Assuming a uniformly distribution of current within the wire, the current density inside the wire is
(3.44.3)

45. 2. Magnetostatic Fields
2.2. Fundamentals of Magnetic Fields (Example 2)
The total current enclosed within the circle of radius   a
(3.45.1)
Therefore:
(3.45.2)
We notice that at the edge of the wire, two solutions given by (3.44.1) and (3.45.2) are equal.
Can you further explain the dependence of magnetic flux on the radius?

46. 2. Magnetostatic Fields 2.3. Magnetic Vector Potential
A magnetic vector potential A such that:
(3.46.1)
(3.46.2)
Ampere’s Law:
(3.46.3)
(3.46.4)
In the Cartesian coordinates:
(3.46.5)
(3.46.6)

47. 2. Magnetostatic Fields 2.3. Magnetic Vector Potential (cont)
Magnetic vector potential, magnetic flux, and current element

48. 2. Magnetostatic Fields 2.3. Magnetic Vector Potential (cont 2)
The Magnetic flux density:
(3.48.1)
(3.48.2)
(3.48.3)
If the current is passing through a wire
(3.48.4)
the Biot-Savart Law

49. 2. Magnetostatic Fields 2.3. Magnetic Vector Potential (Example)
Find the magnetic field on the axis perpendicular to the loop of current. Use the Biot-Savart Law.
We identify the terms appearing in (3.48.4):
(3.49.1)
Due to symmetry, the terms with the unit vector u are zero.
(3.49.2)
(3.49.3)

50. 2. Magnetostatic Fields 2.3. Magnetic Vector Potential (Cont 3)
We have learned the following analytical methods to find the magnetic flux density at a point in space from a current element:
Application of Ampere’s Law, which requires considerable symmetry.
Determination of the vector magnetic potential and the calculation of a magnetic flux density. No symmetry is required.
Application of the Biot-Savart law. No symmetry is required.

51. 2. Magnetostatic Fields 2.4. Magnetic forces
If a charged particle is moving with a constant velocity v in a region that ONLY contains a magnetic field with the density B, the force that acts upon the particle is
(3.51.1)
Direction of the force - RHR!
F+ stands for a positively charged particle;
F- represents a negatively charged one.

52. 2. Magnetostatic Fields 2.4. Magnetic forces (cont)
When a charged particle is going through an area with both: uniform electric field and uniform magnetic field, the force exerted on it would be the Lorentz Force:
(3.52.1)
Recall that the work done by a charged particle moving in a field is
(3.52.2)
A differential charge dQ = vdv moving at a constant velocity creates a current. If this current flows in a closed loop:
(3.52.3)
The total magnetic force:
(3.52.4)

53. 2. Magnetostatic Fields 2.4. Magnetic forces (cont 2)
Example: a charged particle entered a constant magnetic field will move along a circular orbit. Find the radius…
Magnetic force:
(3.53.1)
Centripetal force: , m – particle’s mass
(3.53.2)
(3.53.3)
This radius is called the Larmor radius or gyro radius. This effect is used in mass spectroscopy.

54. 2. Magnetostatic Fields 2.4. Magnetic forces (Example)
Evaluate the force existing between two parallel wires caring currents.
B1 will go up at the location of wire 2.
From (3.52.4) force on the wire 2: a)
(3.54.1)
to the left b)
(3.54.2)
to the right

55. 2. Magnetostatic Fields
2.4. Magnetic forces (Example, cont)… ”Alternative approach”
Let’s re-state the force on wire 1 caused by the magnetic field generated by the current in wire 2 (from )
(3.55.1)
From the Biot-Savart law:
(3.55.2)
Finally:
(3.55.3)
This is what’s called as Ampere’s force.

56. 2. Magnetostatic Fields 2.4. Magnetic forces (Example 2)
Consider a current-caring loop in a constant magnetic field B = B0 uz.
We assume the separation between the In/Out wires to be infinitely small.
Parallel wires carry the same current in the opposite direction. Therefore, the net force will be a vector sum of all forces, which is zero!
However, there will be a torque on the loop that will make it to rotate (say, about x for simplicity).

57. 2. Magnetostatic Fields 2.4. Magnetic forces (Example 2, cont)
The torque on the loop is given by
(3.57.1)
Assumptions?
(3.57.2)
(3.57.3)
Finally:
(3.57.4)
magnetic moment

58. 2. Magnetostatic Fields 2.5. Magnetic materials
Two sources of magnetism inside an atom:
1) an electron rotating around a nucleus;
2) an electron spinning about its own axis.
Types of material:
Diamagnetic: 1) and 2) cancel each other almost completely, magnetic susceptibility m 
Paramagnetic: 1) and 2) do not cancel each other completely, magnetic susceptibility m  10-5.
Ferromagnetic: domain structure, very high m (hundreds and higher)
magnetic dipoles oriented randomly
magnetic dipoles in each domain are oriented

59. 2. Magnetostatic Fields 2.5. Magnetic materials: Ferromagnetics
External magnetic field may change dipole orientation “permanently” - HDD.
Total magnetization (magnetic dipole moment per unit volume):
(3.59.1)
There is a current created inside domains (magnetization current):
(3.59.2)
(3.59.3)
We may modify the Ampere’s Law by adding the magnetization current:
(3.59.4)

60. 2. Magnetostatic Fields 2.5. Magnetic materials: Ferromagnetics (cont)
We introduce a new quantity, the Magnetic Field Intensity:
(3.60.1)
Therefore, the Ampere’s circular law is
(3.60.2)
(3.60.3)
Therefore:
(3.60.4)
where r is the relative permeability.

61. 2. Magnetostatic Fields
2.5. Magnetic materials: Ferromagnetics (cont 2)
Example: a magnetic flux density B = 0.05 T appears in a material with r = 50.
Find the magnetic susceptibility and the magnetic field intensity.
saturation
Hysteresis
Magnetic flux density B exhibits nonlinear dependence on the magnetic field intensity H.
Because of hysteresis, magnetic materials “remember” the magnitude and direction of magnetic flux density. They can be used as memory elements.
saturation

62. 2. Magnetostatic Fields 2.6. Magnetic circuits
Just like electrical circuits, we can build magnetic circuits where magnetic flux “flows”.
L – mean length of the iron region
g – length of the gap.
m – the reluctance.
Assumptions:
1) the gap is very small;
2) the cross-sectional area of the gap is identical to the cross-sectional area of the magnetic material.

63. 2. Magnetostatic Fields 2.6. Magnetic circuits (cont)
The Ampere’s circular law leads to
(3.63.1)
where I is the current flowing through the N turns of a wire.
(3.63.2)
Here A is the cross-sectional area of the iron
is the magnetomotive force [A-turns]
(3.63.3)
is the reluctance
(3.63.4)
The Hopkinson’s Law (aka Ohm’s):
(3.63.5)

64. 2. Magnetostatic Fields
2.6. Magnetic circuits (Examples)

65. 2. Magnetostatic Fields
2.7. Inductance (an ability to create magnetic flux)
(3.65.1)
where  is a magnetic flux linkage.
When j = k – self –inductance; otherwise – mutual inductance.
Ex. 1: Let us consider a solenoid of the length d, cross-section area A, and having N turns.
It may also have a core made from a magnetic material.
z is the solenoid’s axis.

66. 2. Magnetostatic Fields 2.7. Inductance (cont)
The magnetic flux density at the center of the solenoid is:
(3.66.1)
The total magnetic flux:
(3.66.2)
The magnetic flux linkage:
(3.66.3)
Therefore, the self-inductance of a solenoid is
(3.66.4)

67. 2. Magnetostatic Fields 2.7. Inductance (cont 2)
Example 2: a self-inductance of a coaxial cable
What is the main difference as compared to a solenoid?
Here, the magnetic flux linkage equals to the total magnetic flux.
mfd:
(3.67.1)
(3.67.2)
Therefore:
(3.67.3)

68. 2. Magnetostatic Fields 2.7. Inductance (cont 3)x
Example 3: a mutual inductance between two circular solenoids, whose individual lengths are d and areas S1 and S2, separated by x; x << d
(3.68.1)
First coil:
(3.68.2)
Assuming that the magnetic flux has the same value in the second solenoid:
(3.68.3)
Therefore, the mutual inductance:
(3.68.4)

69. 2. Magnetostatic Fields 2.7. Inductance (cont 4)
Example 4: consider a transformer with N1 and N2 turns.
correction coefficient
permeance of the space occupied by the flux
Alternative formulas:
(3.69.1)
(3.69.2)
coefficient of coupling

70. 2. Magnetostatic Fields 2.7. Inductance (cont 5)
Solenoids, transformers, etc. can store magnetic energy:
(3.70.1)
for a solenoid:
(3.70.2)
The total magnetic energy stored within a volume
(3.70.3)
Your homework 3 is available through “My Lamar” 