Understanding Maxwell's Equations in Electromagnetism

1. Stationary fields We will consider fields constant in time. Then in Maxwell equations: Electrostatic Electric in conducting media with constant current Magnetostatic This fields are described by pairs of vectors:

2. Pairs of vectors characterise the media and are connected with them • D =  E, where  = f1(x,y,z) • J =  E, where  = f2 (x,y,z) • B =  H, where  = f3 (x,y,z) , ,  - media parameters In general case , ,  can be depended on the point coordinates. Assumption: The medium is linear, homogenous and isotropic when, ,  = const. in considered area.

3. Three Maxwell’s postulates concerning divergence of the field: So called „point equations” inform us about a medium in which the field exists: The differential form of Maxwell’s equations

4. The integral form of Maxwell’s equations

5. I Maxwell’s equation - Ampère’s law A magnetic voltage along the closed curve c is equal to the total current – the sum of the conduction current and displacement current passing through the surface which bound is the curve c.

6. II Maxwell’s equation - Faraday’s induction law Emf – induced electromotive force e A voltage along the closed curve c is equal to electromotive force induced by time changes of magnetic flux passing throughthe surface bounded by this curve.

7. Electrostatic field Equations of electrostatic field: Field is irrotational Field is sourced and field source is the charge with volume density ρ where:ε - dielectric permeability ρ - free charge volume density

8. E dS S c dl What does it mean that electrostatic field is irrotational? Let’s use Stokes’s theorem to this term :

9. Integral form – the circulation of vector =0 Def.The line integral of electric field vector is called electric voltage Differential form Conclusion 1: Electric voltage along any closed path equals 0.

10. C A B D Conclusion 2:Electric voltage between any two points is independent on integration way. QED

11. is the vector of polarization l +q -q Electric induction Def. The notion: is called the electric induction, Polarization vector represents the sum of all dipoles moments (of all particles) included in volume tending to 0.

12. ε +q -q +q -q ε ε0 v v Polarization effect Dielectric sample in electric field Ordered dipoles in dielectric electric flexibility

13. +q -q +q -q What for do we introduce the electric induction? In both of this capacitors the induction is the same, but the field intensity is weaker in dielectric then in vacuum.

14. Gauss’s law Gauss’s law in differential form Gauss’s law known from mathematics Gauss’s law for electrostatic field

15. The flux of electric induction through the surface S n D D S n α dS ds c D·ds=D cosα ds Dn

16. E(r) S P q r Gauss’s lawThe flux of electric induction through the closed surface is equal to the charge included in this surface We have free choice of the surface S, so we choose the sphere because the field has a spherical symmetry and the integration over the sphere will be the easiest. Point P is situated on the sphere with the radius r

17. Superposition Principles Having more than one charge (each charge at different location), the total electric field in the space external to the location of these charges would be the vector summation of the electric field coming from each individual charge.

18. Indicates the unit vector assiociated with each individual charge Q to the point where the electric field is to be computed

19. Find the electric field at the origin P(0,0) Example 4.1(superposition) y x Q1=+4C Q2=-2C

20. Example 4.1 Solution y x

21. Example 4.1 Solution y x

22. Find the electric field at the piont P(3,4) Example 4.2 Q3=+3C y x Q1=+1C Q2=+2C

23. Example 4.2 solution Q3=+3C y x Q2=+2C Q1=+1C

24. Superposition differential electric field • To calculate the electric field that is created by any of the distributed charge density distributions, the principle of superposition should be used in the following way: Q1 Q2 Q3

25. To calculate the electric field at the point P, the differential electric fields Ejcaused by the charges in the differential volumes Δvj are added together vectorially. • Assuming that the differential volumes Δvj are very small and the number of these volumes become very large ->> cause summation to become an integration performed over the entire volume Δv where distributed charge density is located:

26. Example 4.3 The electric field from uniformly distributed finite line of charge.

27. 2a • Calculate the electric field from uniformly distributed finite line of charge. • The linear charge density is : • Problem visualization: z’ dz’ p R

28. The radial component of electric field is given in terms of the differential electric field dE: • Magnitude of the differential electric field dEis computed from the charge dq that is contained in the length dz’: • Therefore becomes

29. The total radial component of electric field is given by the summation of all the infinitesimal components dEp • (integration over the length of the charged line) • Using integral tables or performing substitution z’=ptgΘ:

30. Example 4.4 The electric field from infinite charged plane.

31. An infinite charged plane consisting of an infinite number of parallel charged lines

32. Lets consider the infinite plane as parallel array of juxtaposed infinite charged lines . • We can use as the point of embarkation, where the distance

33. The linear charge density of particular line having width dx’ may be expressed as Taking into account the symmetry, the components of electric field that are tangent to the plane will cancel

34. Finding the sum of the electric field that are normal to the plane we find

35. Conclusion • The electric field is independent of the distance tat is above the infinite charged sheet • Remark: alternative integration could be performed with use of differential areas being concentric circular washers.

36. Electrostatic field vectors D andE on the bound of two media We will show which components of the vectors D and E are continuous on the boundary surface between two media with different electric permeability.

37. E1 α1 E2 A B dl1 h α2 D C dl2 a. The tangential component of electric field intensity

38. E1 α1 E2 A B h α2 C D Δl The segment Δl is small enough to replace the integration with the multiplication. Tangential components:

39. The continuity oftangential component of E results in the discontinuity of tangential component of D. Tangential components of the vector induction are discontinuous on the bound of two media with different permeability.

40. n1 D1 dS1 α1 D2 α2 h dS2 n2 We will show that normal components of electric induction (perpendicular to the boundary surface) are continuous Let’s assume that the charge with the surface density σ is situated on the bound .

41. The cylinder height tends to 0, so the integral over the side surface tends to 0. When a charge is situated on the boundary surface, then normal component of induction is discontinuous. When there is no charge on the boundary surface - normal component of induction is continuous.

42. The continuity ofnormal component of D results in the discontinuity of normal component of E Refraction law: hence The refraction law

43. Potential Energy and Electric Potential

44. A charged particle will gain a certain energy as the particle is moved against an electric field. • (work has to be done to overcome the force due to the electric field)

45. If charge is positive  work should be done, energy must be conserved in this process (+ charge will gain energy) • Positive charge’s energy will be increased . • A negatively charged particle will experience a decrease of potential energy (if it followed the same path). • When the force is perpendicular to the the direction of motion  no work

46. How to define the total electrostatic energy stored in a volume ->>gedanken experiment • Assumptions: • All charges initially are in –h • None exists in the laboratory • Each of charge is infinitely far from its neighbor -> no Coulomb force between them • Any electric field originating at x= –h will have decayed to have a value zero in the Lab

47. Calculation of the work required to bring charges from – infinity into defined space

48. POTENTIAL DIFFERENCE BETWEEN TWO CHARGES THAT IS CAUSED BY CHARGE 1 ALREADY PRESENT IN THIS REGION

49. Formula for absolute potential

50. The total work