1. LINE,SURFACE & VOLUME CHARGES
Electric fields due to continuous charge distributions:

2. LINE,SURFACE & VOLUME CHARGES (cont’d)
To determine the charge for each distributions:
Line charge
Surface charge
Volume charge

3. LINE CHARGE Infinite Length of Line Charge:
To derive the electric field intensity at any point in space resulting from an infinite length line of charge placed conveniently along the z-axis

4. LINE CHARGE (Cont’d)
Place an amount of charge in coulombs along the z axis.
The linear charge density is coulombs of charge per meter length,
Choose an arbitrary point P where we want to find the electric field intensity.

5. LINE CHARGE (Cont’d) The electric field intensity is:
But, the field is only vary with the radial distance from the line.
There is no segment of charge dQ anywhere on the z-axis that will give us So,

6. LINE CHARGE (Cont’d)
Consider a dQ segment a distance z above radial axis, which will add the field components for the second charge element dQ.
The components cancel each other (by symmetry) , and the adds, will give:

7. LINE CHARGE (Cont’d) Recall for point charge,
For continuous charge distribution, the summation of vector field for each charges becomes an integral,

8. LINE CHARGE (Cont’d) The differential charge,
The vector from source to test point P,

9. LINE CHARGE (Cont’d) Which has magnitude, and a unit vector,
So, the equation for integral of continuous charge distribution becomes:

10. LINE CHARGE (Cont’d)
Since there is no component,

11. LINE CHARGE (Cont’d)
Hence, the electric field intensity at any point ρ away from an infinite length is:
For any finite length, use the limits on the integral.

12. EXAMPLE 3
Use Coulomb’s Law to find electric field intensity at (0,0,h) for the ring of charge, of charge density, centered at the origin in the x-y plane.

13. SOLUTION TO EXAMPLE 3
By inspection, the ring charges delivers only and contribution to the field.
component will be cancelled by symmetry.

14. SOLUTION TO EXAMPLE 3 (cont’d)
Each term need to be determined:
The differential charge,

15. SOLUTION TO EXAMPLE 3 (cont’d)
The vector from source to test point,
Which has magnitude, and a unit vector,
The integral of continuous charge distribution becomes:

16. SOLUTION TO EXAMPLE 3 (cont’d)
Rearranging,
Easily solved,

17. EXAMPLE 4 SOLUTION TO EXAMPLE 4
An infinite length line of charge exists at x = 2m and z = 4m. Find the electric field intensity at the origin.
SOLUTION TO EXAMPLE 4
Sketch in three dimensions and the cross section:

18. SOLUTION TO EXAMPLE 4 (cont’d)
The vector from line charge to the origin:
Which has magnitude, and a unit vector,
Inserting into the infinite line charge equation:

19. SURFACE CHARGE Infinite Sheet of Surface Charge:
To derive the electric field intensity at point P at a height h above a charge sheet of infinite area (x-y plane).
The charge distribution, is in

20. SURFACE CHARGE (CONT’D)

21. SURFACE CHARGE (CONT’D)
Consider a differential charge,
The vector from surface charge to the origin:
Which has magnitude, and a unit vector,
Where, for continuous charge distribution:

22. SURFACE CHARGE (CONT’D)
The equation becomes:
Since only z components exists,

23. SURFACE CHARGE (CONT’D)

24. SURFACE CHARGE (CONT’D)
A general expression for the field from a sheet charge is:
Where is the unit vector normal from the sheet to the test point.

25. EXAMPLE 5
An infinite extent sheet of charge exists at the plane y = -2m. Find the electric field intensity at point P (0, 2m, 1m).

26. SOLUTION TO EXAMPLE 5 (CONT”D)
Sketch the figure:
The unit vector directed away from the sheet and toward the point P is

27. VOLUME CHARGE
A volume charge is distributed over a volume and is characterized by its volume charge density, in
The total charge in a volume containing a charge distribution, is found by integrating over the volume:

28. EXAMPLE 6
Find the total charge over the volume with volume charge density,

29. SOLUTION TO EXAMPLE 6
The total charge, with volume:
Thus,

30. VOLUME CHARGE (CONT’D)
To find the electric field intensity resulting from a volume charge, we use:
Since the vector R and will vary over the volume, this triple integral can be difficult. It can be much simpler to determine E using Gauss’s Law.

31. ELECTRIC FLUX DENSITY, (D)
Consider an amount of charge +Q is applied to a metallic sphere of radius a.
Enclosed this charged sphere using a pair of connecting hemispheres with bigger radius.

32. ELECTRIC FLUX DENSITY (CONT’D)
The outer shell is grounded. Remove the ground then we could find that –Q of charge has accumulated on the outer sphere, meaning the +Q charge of the inner sphere has induced the –Q charge on the outer sphere.

33. ELECTRIC FLUX DENSITY (CONT’D)
Electric flux, extends from the positive charge and casts about for a negative charge. It begins at the +Q charge and terminates at the –Q charge.
The electric flux density, D in is:
where

34. ELECTRIC FLUX DENSITY (CONT’D)
This is the relation between D and E, where is the material permittivity. The advantage of using electric flux density rather than using electric field intensity is that the number of flux lines emanating from one set of charge and terminating on the other, independent from the media.
We can find the total flux over a surface as:

35. ELECTRIC FLUX DENSITY (CONT’D)
We could also find the electric flux density, D for:
Infinite line of charge:
Where,
Infinite sheet of charge:
Where,
Volume charge distribution:
Where,

36. EXAMPLE 7
Find the amount of electric flux through the surface at z = 0 with
and

37. SOLUTION TO EXAMPLE 7
The differential surface vector is
We could have chosen but the positive differential surface vector is pointing in the same direction as the flux, which give us a positive answer.
Therefore,
…Why?!

38. EXAMPLE 8 SOLUTION TO EXAMPLE 8
Determine D at (4,0,3) if there is a point charge at (4,0,0) and a line charge along the y axis.
SOLUTION TO EXAMPLE 8
How to visualize ?!

39. SOLUTION TO EXAMPLE 8 (CONT’D)
Let total flux,
Where DQ is flux densities due to point charge and DL is flux densities due to line charge.
Thus,
Where,
Which has magnitude, and a unit vector,

40. SOLUTION TO EXAMPLE 8 (CONT’D)
Therefore, total flux:
And
Where,
So,

41. GAUSS’S LAW
If a charge is enclosed, the net flux passing through the enclosing surface must be equal to the charge enclosed, Qenc.
Gauss’s law constitutes one of the fundamental laws of electromagnetism
Gauss’s Law states that:
The total electric flux, through any closed surface is equal to the total charge enclosed by that surface

42. GAUSS’S LAW (Cont’d) (1)- Integral form
How to derive the Gauss’s Law (which is the first of the four Maxwell’s equations to be derived ??
It can be rearranged so that we have relation between the Gauss’s Law and the electric flux.
Thus:
(1)- Integral form

43. GAUSS’S LAW (Cont’d)
Related to Gauss’s Law, where net flux is evaluated exiting a closed surface, is the concept of divergence.
By applying divergence theorem to
the middle term in equations
Comparing the two volume integrals in above equations, results in:
(2)-Differential or point form of Gauss’s law
Which is the first of the four Maxwell’s equations to be derived. Equation (2) states that the divergence of the electric flux density is the same as the volume charge density.

44. GAUSS’S LAW (Cont’d)
Gauss’s Law is an alternative statement of Coulomb’s Law; proper application of the divergence theorem to Coulomb’s law results in Gauss’s Law.
Gauss’s Law is useful in finding the fields for problems that have high degree of symmetry.
Determine variables influence D and what components D present
Select an enclosing surface, called Gaussian Surface, whose differential surface is directed outward from the enclosed volume and is everywhere (either tangent or normal to D)

45. GAUSS’S LAW (Cont’d)
The expression is also called the point form of Gauss’s Law, since it occurs at some particular point in space. For instance,
Plunger stationary – no net movement of molecules
Plunger moves up – net movement where air molecules diverging  air is expanding
Plunger pushes in – net flux is negative and molecules diverging  air is compressing

46. GAUSS’S LAW APPLICATION (Cont’d)
Use Gauss’s Law to determine electric field intensity, (E), for each cases below:
Point Charge
Infinite length of Line Charge
Infinite extent Sheet of Charge

47. POINT CHARGE Point Charge:
It has spherical coordinate symmetry, where the field is everywhere directed radially away from the origin. Thus,

48. POINT CHARGE (Cont’d)
For a gaussian surface, we could find the differential surface vector is:
So,

49. POINT CHARGE (Cont’d)
Since the gaussian surface has a fixed radius, Dr will be constant and can be taken from integration to yield
By using Gauss’s Law, where:
So, which leads to expected result:

50. INFINITE LENGTH LINE CHARGE
Infinite length line of charge:
Find D and then E at any point
A Gaussian surface containing the point P is placed around a section of an infinite length line of charge density L occupying the z-axis.

51. INFINITE LENGTH LINE CHARGE (Cont’d)
An element of charge dQ along the line will give Dρ and Dz. But second element of dQ will result in cancellation of Dz. Thus,
The flux through the closed surface is:

52. INFINITE LENGTH LINE CHARGE (Cont’d)
Where,
Then, we know that Dρ is constant on the side of gaussian surface

53. INFINITE LENGTH LINE CHARGE (Cont’d)
The charge enclosed by the gaussian surface:
We know that,
So,
Thus, as expected:

54. INFINITE EXTENT SHEET OF CHARGE
Determine the field everywhere resulting from an infinite extent sheet of charge ρS placed on the x-y plane at z = 0.
Locate a point at which we want to find the field along the z axis at height h.

55. SHEET OF CHARGE (Cont’d)
Gaussian surface must contain this point and surround some portion of the charged sheet.
A rectangular box is employed as the Gaussian surface surrounding a section of sheet charge with sides 2x, 2y and 2z

56. SHEET OF CHARGE (Cont’d)
Only a DZ component will be present, and the charge enclosed is simply:
No flux through the side of the box, so find the flux through the top and bottom surface

57. SHEET OF CHARGE (Cont’d)
Notice that the answer is independent of the height of the box.
Then we have: or
And electric field intensity, as expected:

58. Example 9 Solution to Example 9 Suppose:
Find the flux through the surface of a cylinder with and by evaluating the left side and the right side of the divergence theorem.
Solution to Example 9
Remember the divergence theorem?
We can first evaluate the left side of the divergence theorem by considering:

59. Solution to Example 9 (cont’d)
A sketch of this cylinder is shown with differential vectors.
The integrals over the top and bottom surfaces are each zero, since:
Thus,

60. Solution to Example 9 (cont’d)
For evaluation of the right side of the divergence theorem, first find the divergence in cylindrical coordinate:
Performing a volume integration on this divergence,
This is the same!