2. Michael Faraday - static electric field
- bothering induced electromotive force
- experimented on a pair of concentric metallic spheres

5. The path of electric flux extending from inner sphere to outer sphere
are shown symmetrically distributed and are radial outward from inner
Sphere to outer sphere.
At the surface of inner sphere, ψ coulombs of electric flux is produced
which is distributed throughout the surface area (4πa2) m2. For inner sphere
sphere the radius is a.
The density of electric flux at this surface is ψ/4πa2 c/m2.

6. The elect5ric field due to point charge is

7. The relation between electric flux density and electric field is given by the relatiion
and

16. And
which agrees with the results we already studied.

17. Are there any other surfaces which would have satisfied our two conditions for
Point charge? Such a simple surfaces as a cube ( Cartesian coordinate) or
Cylindrical surface do not meet the requirements.
Considering the uniform line charge distribution ρL lying along z-axis , infinite
in extent. Considering the knowledge of the symmetry of the field , it can be det-
ermined
1 With which coordinate does the field vary.
2 Which components of D are present?
If we cannot show that symmetry exists then we cannot use Gauss’s law to obtain
a solution. For the line charge configuration, only the radial component of D is
Present.

18. A cylindrical surface will be the only surface to
Which the Dρ is everywhere normal and it may
be closed by z = constant surfaces. A closed ci-
rcular cylindrical surface of radius ρ extending
From z = 0 to z = L is shown.
We apply Gauss’s law,

19. Once the appropriate surface has been chosen, the integration usually
amounts only to writing down the area of the surface at which D is
normal.

25. In this example we will not obtain the
value of D as our answer, but instead
receive some extremely valuable
information about the way D varies in
the region of our small surface. This leads
To one of the Maxwell’s four equations.
At a general point p(x,y,z), the value
of D may be expressed in Cartesian
components

26. Integrating the flux density over the closed surface (in this case all six sides)
having side length Δx, Δy, Δz, and applying Gauss’s law will result

27. Adding these two front and back integral, will result

29. Why approximation ?
Because Dx also varies with y and z coordinates and we have taken only
partial derivative with x-component.

33. Considering this equation

38. And the last equation is the obvious result.

39. To achieve the familiarity with the concept of divergence for the
given Gauss’s law

43. Divergence of scalar is a vector .

44. And finally we had a relation,
The divergence theorem is given below
It is stated as.