1. ENE/EIE 325 Electromagnetic Fields and Waves
Lecture 4 Electric Flux, Gauss’s Law, and Divergence

2. Electric flux (1)
Electric flux ( ) is the measure of flow of the electric field through a given area.
Electric flux is proportional to the number of electric field lines going through a virtual surface.

3. Electric flux (2)
Consider volume flow rate of water through a rectangle boundary,
the volume flow rate (v=dV/dt) of fluid through the rectangle is when the area of the rectangle is perpendicular to the velocity vector.
What is the volume flow rate when the rectangle is tilted at an angle  ?

4. Electric flux (3) The electric flux through the surface is  = (Vm).
If the area vector makes an angle  with the vector , the flux is zero when  = 90o because the rectangle lies in a plane parallel to the flow and no electric flux flows through the rectangle.

5. Electric flux (4)
Electric flux is proportional to the density of flow.
Electric flux varies by how the boundary faces the direction of flow.
Electric flux is proportional to the area within the boundary.

6. Electric flux (5)
The amount of flux passing through a surface is given by the product of 𝐷 and the amount of surface normal to 𝐷 . Same polarity charges repel one another
Note: = surface vector
Dot product:
for Cartesian coordinates.
Dot product is a projection of A on B multiplies by B

7. Gauss’ Law
The electric flux passing through any closed surface is equal
to the total charge enclosed by that surface

8. History Johann Carl Friedrich Gauss
(30 April 1777 – 23 February 1855) was a German mathematician and physical scientist who contributed significantly to many fields, including number theory, algebra, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and optics.

9. About Gauss’s law
The law was formulated by Gauss in 1835, but was not published until 1867.
It is one of the four Maxwell's equations which form the basis of classical electrodynamics, the other three being
- Gauss's law for magnetism, - Faraday's law of induction, and Ampère's law with Maxwell's correction.

10. Development of Gauss’ Law
We define the differential
surface area (a vector) as
where n is the unit outward
normal vector to the surface,
and where dS is the area of the
differential spot on the surface

11. Flux through an irregular surface
The projection of the area element onto the spherical surface is

12. For the area element of a sphere of radius r and the area element on the larger sphere of radius 2r , the same number of lines of flux pass through each area element.
Gauss’ Law easily shows that the electric field from a uniform shell of charge is the same outside the shell as if all the charge were concentrated at a point charge at the center of the sphere.
Inverse-square law

13. Mathematical Statement of Gauss’ Law
Line charge:
Surface charge:
Volume charge:
in which the charge can exist in the form of point charges:
For a volume charge, we would have:
or a continuous charge distribution:

14. Using Gauss’ Law to Solve for D Evaluated at a Surface
Knowing Q, we need to solve for D, using Gauss’ Law:
The solution is easy if we can choose a surface, S, over which to integrate (Gaussian surface)
that satisfies the following two conditions:
The integral now simplfies:
So that:
where

15. Example: Point Charge Field
Begin with the radial flux density:
and consider a spherical surface of radius a that surrounds the charge, on which:
On the surface, the differential area is:
and this, combined with the outward unit normal vector is:

16. Point Charge Application (continued)
Now, the integrand becomes:
and the integral is set up as: =

17. Another Example: Line Charge Field
Consider a line charge of uniform charge density L on the z axis that extends over the range
 z 
We need to choose an appropriate Gaussian surface, being mindful of these considerations:
We know from symmetry that the field will be radially-directed (normal
to the z axis) in cylindrical coordinates:
and that the field will vary with radius only:
So we choose a cylindrical surface of radius , and of length L.

18. Line Charge Field (continued)
Giving:
So that finally:

19. Coaxial Transmission Line
We have two concentric cylinders, with the z axis down their centers. Surface charge of density S exists
on the outer surface of the inner cylinder.
A -directed field is expected, and this should vary only with  (like a line charge). We therefore choose
a cylindrical Gaussian surface of length L and of radius , where a <  < b.
The left hand side of Gauss’ Law is written:
…and the right hand side becomes:

20. Coaxial Transmission Line (continued)
We may now solve for the flux density:
and the electric field intensity becomes:

21. Coaxial Transmission Line: Exterior Field
If a Gaussian cylindrical surface is drawn outside (b), a
total charge of zero is enclosed, leading to the conclusion that
or:

22. Coaxial Cable or Transmission Line
Coaxial cable, or coax, is a type of cable that has an inner conductor surrounded by a tubular insulating layer, surrounded by a tubular conducting shield.
Coaxial cable was invented by English engineer and mathematician Oliver Heaviside, who patented the design in 1880.

23. Electric Flux Within a Differential Volume Element
Taking the front surface, for example, we have:

24. Electric Flux Within a Differential Volume Element
We now have:
and in a similar manner:
Therefore:
minus sign because Dx0 is inward flux through the back surface.

25. Charge Within a Differential Volume Element
Now, by a similar process, we find that:
and
All results are assembled to yield: v
= Q
(by Gauss’ Law)
where Q is the charge enclosed within volume v

26. Divergence and Maxwell’s First Equation
Mathematically, this is:
Applying our previous result, we have:
div A =
and when the vector field is the electric flux density:
= div D
Maxwell’s first equation

27. Physical example The plunger moves up and down indicating net movement
of molecules out and in,
respectively.
positive indicates a source of flux. (positive charge)
negative indicates a sink of flux. (negative charge)
An integral form of Gauss’s law can also be written as
Divergence theorem

28. Divergence Expressions in the Three Coordinate Systems

29. The Del Operator = = div D
The del operator is a vector differential operator, and is defined as:
Note that:

30. Divergence Theorem
We now have Maxwell’s first equation (or the point form of Gauss’ Law) which states:
and Gauss’s Law in large-scale form reads:
leading to the Divergence Theorem:

31. Statement of the Divergence Theorem

32. Ex C/m2. Given the surface defined by  = 1 m, 0    90 and -1  z  1, calculate the flux through the surface.

33. What about a closed surface that doesn’t include the charge?
The red dotted line represents some fixed closed surface.
In steady flow, it goes in one side, out the other. The net flow across the surface must be zero.
By analogy, if the charge is outside.

34. For which of these closed surfaces (a, b, c, d) the flux of the electric field, produced by the charge +2q, is zero?

35. What about more than one charge enclosed?
Remember the principle of superposition: the electric flux density can always be written as a linear sum of contributions from individual point charges:
and so from
will have a contribution from each ith charge inside the surface.

36. What is the flux of the electric field for each of the closed surfaces a, b, c, and d ?

37. Spherical symmetry (1)
First, a uniform spherical shell, radius r0, of positive charge.
The perfect spherical symmetry means the electric field outside, at a distance r from the center, must point radially outwards. (rotating the sphere doesn’t change anything, but would change a field pointing any other way.) r0 r

38. Spherical symmetry (2)
The blue circle represents a spherical surface of radius r > r0, concentric with the shell of charge.
For this enclosing surface, Gauss’ Law
becomes Coulomb’s law r0 r r r0

39. Field inside a hollow shell of charge
Now let’s take the enclosing surface inside the hollow shell of charge, r < r0.
Gauss’ Law is now
Because there is no charge inside the shell, it’s all on the surface.
The spherical symmetry tells us the field inside the shell is exactly zero – again, not so simple from Coulomb’s Law. r r0

40. Question How will g change (if at all) on going from the Earth’s surface to the bottom of a deep mine? (assume the earth has uniform density.)
A) g will be a bit stronger at the bottom of the mine B) It will be weaker C) It will be the same as at the surface
B) is correct! For uniform density, it will be weaker. The gravitational field strength varies in exactly the same way as the electric field from a solid sphere with charge uniformly distributed throughout the volume.

41. Field inside a solid sphere of charge
Now let’s take the enclosing surface inside the solid sphere of charge, r < r0.
Gauss’ Law is now
From this, since
so the electric field strength increases linearly from zero at the center to the outside value at the surface.
Gauss’ law also works for gravitation and this is the result for a solid sphere of mass.

42. Question
If you could distribute charge perfectly uniformly throughout the volume of a solid spherical conductor, would it stay in place? A) Yes B) No
B) is correct! Because this charge distribution gives rise to a nonzero outward field inside the conductor - the charge would therefore flow radially outwards to the surface.

43. Charges on conductors (1)
Gaussian surface
Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor.

44. Charges on conductors (2)
Under electrostatic conditions the electric field inside a solid conducting sphere is zero.
Outside the sphere the electric field drops off as 1 / r2, as though all the excess charge on the sphere were concentrated at its center.

45. Faraday cage
A Gaussian surface drawn inside the conducting material of which the box is made must have zero electric field on it (field inside a con-ductor is zero).

46. Ex2 A parallel plate capacitor has surface charge +S located underneath a top plate and surface charge -S located on a bottom plate. Use Gauss’s law to find and between plates.

48. In the limit where the plates are infinitely large, the system has planar symmetry and we can calculate the electric flux density everywhere using Gauss’ law as
By choosing a Gaussian “pillbox” with cap area to enclose the charge on the positive plate, the electric flux density in the region between the plate is
So, the corresponding electric field will be

49. Note that a real capacitor is finite in size
Note that a real capacitor is finite in size. So, the electric field lines at the edge of the plates are not straight lines, and the field is not contained entirely between the plates.
This is known as edge effects, and the non-uniform fields near the edge are called fringing fields. In solving the above example, we assume an idealized situation where the field lines between the plates are straight lines.

50. Ex3 Let us select a 50-cm length of coaxial cable having an inner radius of 1 mm and an outer radius of 4 mm. The space between conductors is assumed to be filled with air. The total charge on the inner conductor is 30 nC. We wish to know the charge density on each conductor, and the E and D fields.
Ex3.2 Haytr