Understanding Coulomb's Law and Electric Field Intensity Theory

1. ELECTROMAGNETICS THEORY (SEE 2523) ELECTROSTATIC FIELD

2. CHAPTER 2 COULOMB’S LAW & ELECTRIC FIELD INTENSITY

3. INTRODUCTION An electrostatic field is produced by a static charge distribution. A magnetostatic field is produced by the moving charges or a constant current flow (DC current) AC current produced the electromagnetic field.

4. 2.1: ELECTRIC CHARGES • Charges are measured in Coulombs (C). • The smallest unit of electric charge is found on the negatively charged electron and positively charged proton. • The electron mass, qe= -1.602x10-19 (C). • So, 1 C would represent about 6x1018 electrons. • In real world, electric charges can be found at a point along a line on a surface in a volume or combination the above distributions

5. 2.1.1 A POINT CHARGE The concept of a point charge is used when the dimensions of an electric charge distribution are very small compared to the distance to neighboring electric charges. Fig. 2.1: A point charge

6. 2.1.2 A LINE CHARGE A line charge denotes as the electric charge distribution along a thin line. • A line charge density : • Total charge along the line : Q l Fig 2.2: A line charge

7. 2.1.3 A SURFACE CHARGE A surface charge is defined as charges distribution on a thin sheet. A surface charge density : Q S Total charge on the surface : Fig. 2.3: A surface charge

8. 2.1.4 A VOLUME CHARGE A volume charge means electric charges in a volume. This volume charge can be viewed as a cloud of charged particles. • A volume charge density : • Total charge in the volume : Q V Fig. 2.4: A volume charge

9. 2.2 COULOMB’S LAW • Coulomb’s law states that the force F between two point charges Q1 and Q2 is : Directly proportional to the magnitude of the both products of the charges and inversely proportional to the square of the distance Rbetween them.  Expressed mathematically, or and

10. Fig. 2.5: Coulomb’s law where is known as the permittivity of free space

11. A unit vector, then the force : • The force, on Q1 due to Q2 :

12. Things need to considered in Coulomb’s law : • >> The force expressed by Coulomb’s law is a mutual force, for each of the two charges experiences a force of the same magnitude, although of opposite direction. • >> Coulomb’s law is linear, if the charge multiply by n, the force is also multiplied by the same factor, n. • >> If there are more than two point charges, the principle of superposition is used. The force on a charge in the presence of several other charges is the sum of the forces on that charges due to each of the other charges acting alone.

13. 2.3 ELECTRIC FIELD INTENSITY DUE TO POINT CHARGE • Consider one charge fixed in position, Q1 and test charge, Qt that exists everywhere around Q1 ,shown in Fig. 2.6. • From the definition of Coulomb’s law, there will exist force on Qtcause by Q1. Fig. 2.6

14. The electric field intensity E is the force per unit charge when placed in the electric field.

15. For n point charges, the electric field intensity at a point in space is equal to the vector sum of the electric field intensities due to each charge acting alone. (Following superposition theorem). Shown in Fig. 2.7. Fig. 2.7

16.  The total field at point P :

17. 2.4 ELECTRIC FIELD INTENSITY DUE TO THE LINE CHARGE • We have : • Thus, the electric field for a line charge :

18. Ex:4. Find the electric field intensity about the finite line charge of uniform distribution along the z axis as shown in Fig. 2.8. Fig. 2.8

19. Solution:

20. Using • Hence : • Hence :

21. = = • If the line is infinite : • We obtain :

22. 2.5 ELECTRIC FIELD INTENSITY DUE TO THE SURFACE CHARGE • We have : • Thus, the electric field for a surface charge

23. Ex:6. Find the electric field intensity along z axis on a disc with a radius a(m) which is located on xy plane with uniform charge distribution as shown in Fig. 2.9. Fig. 2.9

24. Solution: • Using cylindrical coordinate : The component is cancel because the charge distribution is symmetry. Only component exists.

25. For an infinite sheet of charge : In general, it can be summarized to be :

26. The electric field is normal to the sheet and independent of the distance between the sheet and the point. • In the parallel plate capacitor, the electric field existing between the two plates having equal and opposite charge is given by • The electric field is zero for both above and below plates.

27. 2.6 ELECTRIC FIELD INTENSITY DUE TO THE VOLUME CHARGE • We have • Thus, the electric field intensity for a volume charge is

28. Ex:8. Find the electric field intensity for a sphere with a radius a(m) and with uniform charge density, shown in Fig. 2.10. Fig. 2.10

29. Solution: • The electric field at P due to elementary volume charge is • where

30. Due to the symmetry of the charge distribution, the Ex and Ey becomes zero. • Only Ez exists. Thus, • Using Cosine law,

31. It is convenient to evaluate the integral in terms of R and r, so • Differentiating (keeping z and r fixed) then where and

32.  Substituting all the equations into electric field equation yields

33. Because and then electric field at P(0,0,z) is given by • Due to the symmetry of the charge distribution, the electric field at any point can be written as Which is identical to the electric field at the same point due to a point charge Q located at the origin.