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CHAPTER 2

ELECTROMAGNETIC FIELDS THEORY. CHAPTER 2. Electrostatic Fields: Electric Field Intensity. In this chapter you will learn. Electrostatic Fields -Charge, charge density -Coulomb's law - Electric field intensity - Electric flux and electric flux density -Gauss's law

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CHAPTER 2

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  1. ELECTROMAGNETIC FIELDS THEORY CHAPTER 2 Electrostatic Fields: Electric Field Intensity

  2. In this chapter you will learn.. Electrostatic Fields -Charge, charge density -Coulomb's law -Electric field intensity - Electric flux and electric flux density -Gauss's law -Divergence and Divergence Theorem -Energy exchange, Potential difference, gradient -Ohm's law -Conductor, resistance, dielectric and capacitance - Uniqueness theorem, solution of Laplace and Poisson equation BEE 3113ELECTROMAGNETIC FIELDS THEORY

  3. Two like charges repel one another, whereas two charges of opposite polarity attract • The force acts along the line joining the charges • Its strength is proportional to the product of the magnitudes of the two charges and inversely propotional to the square distance between them Charles Agustin de Coulomb BEE 3113ELECTROMAGNETIC FIELDS THEORY

  4. F1t F1t Qt Q1 F1t Electric Field Intensity This proof the existence of electricforce field which radiated from Q1 BEE 3113ELECTROMAGNETIC FIELDS THEORY

  5. Electric Field Intensity • Consider a charge fixed in a position, Q1. • Let’s say we have another charge, say Qt which is a test charge. • When Qt is moved slowly around Q1, there exist everywhere a force on this second charge. • Thus proof the existence of electric force field. BEE 3113ELECTROMAGNETIC FIELDS THEORY

  6. Qt1 Qt2 Qt3 Electric Field Intensity Experience E due to Q1 Electric field intensity, E is the vector force per unit charge when placed in the electric field Q1 Experience the most E due to Q1 Experience less E due to Q1 BEE 3113ELECTROMAGNETIC FIELDS THEORY

  7. Electric Field Intensity • According to Coulomb’s law, force on Qt is • We can also write Which is force per unit charge BEE 3113ELECTROMAGNETIC FIELDS THEORY

  8. Electric Field Intensity • If we write E as • Thus we can rewrite the Coulomb’s law as • Which gives the electric field intensity for Qt at r due to a point charge Q located at r BEE 3113ELECTROMAGNETIC FIELDS THEORY

  9. Electric Field Intensity • For N points charges, the electric field intensity for at a point r is obtained by • Thus BEE 3113ELECTROMAGNETIC FIELDS THEORY

  10. Field Lines The behavior of the fields can be visualized using field lines: Field vectors plotted within a regular grid in 2D space surrounding a point charge. BEE 3113ELECTROMAGNETIC FIELDS THEORY

  11. Field Lines Some of these field vectors can easily be joined by field lines that emanate from the positive point charge. The direction of the arrow indicates the direction of electric fields The magnitude is given by density of the lines BEE 3113ELECTROMAGNETIC FIELDS THEORY

  12. Field Lines The field lines terminated at a negative point charge The field lines for a pair of opposite charges BEE 3113ELECTROMAGNETIC FIELDS THEORY

  13. Example 4.1 BEE 3113ELECTROMAGNETIC FIELDS THEORY

  14. Solution BEE 3113ELECTROMAGNETIC FIELDS THEORY

  15. Line charge Imagine charges distributed along a line We can find the charge density over the line as: BEE 3113ELECTROMAGNETIC FIELDS THEORY

  16. Charge density and charge distribution • We can find the total charge by applying this equation: Charge density Length of the line BEE 3113ELECTROMAGNETIC FIELDS THEORY

  17. Surface Charge Similarly, for surface charge, we can imagine charges distributed over a surface We can find the charge density over the surface as: And the total charge is BEE 3113ELECTROMAGNETIC FIELDS THEORY

  18. Volume Charge We can find the volume charge density as: And the total charge is BEE 3113ELECTROMAGNETIC FIELDS THEORY

  19. + ρL + + + + ρS + + + + + + + + ρv + + + + + + + + Continuous Charge Distribution • for line charge distribution: • for surface charge distribution: • for volume charge distribution BEE 3113ELECTROMAGNETIC FIELDS THEORY

  20. + ρL + + + + E due to Continuous Charge Distributions Line charge ρS + + + + + + Surface charge + + ρv + + + + + + Volume charge + + BEE 3113ELECTROMAGNETIC FIELDS THEORY

  21. What About the Field Lines? E field due to line charge E field due to surface charge BEE 3113ELECTROMAGNETIC FIELDS THEORY

  22. E field due to line charge BEE 3113ELECTROMAGNETIC FIELDS THEORY

  23. What about volume charge? BEE 3113ELECTROMAGNETIC FIELDS THEORY

  24. 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 BEE 3113ELECTROMAGNETIC FIELDS THEORY

  25. 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. BEE 3113ELECTROMAGNETIC FIELDS THEORY

  26. 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, BEE 3113ELECTROMAGNETIC FIELDS THEORY

  27. 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: BEE 3113ELECTROMAGNETIC FIELDS THEORY

  28. LINE CHARGE (Cont’d) Recall for point charge, For continuous charge distribution, the summation of vector field for each charges becomes an integral, BEE 3113ELECTROMAGNETIC FIELDS THEORY

  29. LINE CHARGE (Cont’d) The differential charge, The vector from source to test point P, BEE 3113ELECTROMAGNETIC FIELDS THEORY

  30. LINE CHARGE (Cont’d) Which has magnitude, and a unit vector, So, the equation for integral of continuous charge distribution becomes: BEE 3113ELECTROMAGNETIC FIELDS THEORY

  31. LINE CHARGE (Cont’d) Since there is no component, BEE 3113ELECTROMAGNETIC FIELDS THEORY

  32. IMPORTANT!! BEE 3113ELECTROMAGNETIC FIELDS THEORY

  33. IMPORTANT!! BEE 3113ELECTROMAGNETIC FIELDS THEORY

  34. BEE 3113ELECTROMAGNETIC FIELDS THEORY

  35. 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. ρ is the perpendicular distance from the line to the point of interest aρ is a unit vector along the distance directed from the line charge to the field point BEE 3113ELECTROMAGNETIC FIELDS THEORY

  36. BEE 3113ELECTROMAGNETIC FIELDS THEORY

  37. Example of Line Charge • A uniform line charge of 2 µC/m is located on the z axis. Find E in cartesian coordinates at P(1, 2, 3) if the charge extends from • a) −∞ < z < ∞: • b) −4 ≤ z ≤ 4: BEE 3113ELECTROMAGNETIC FIELDS THEORY

  38. Example of Line Charge z dEz 4 dE ρ dEρ P(1, 2, 3) r dl z r’ y x -4 BEE 3113ELECTROMAGNETIC FIELDS THEORY

  39. Solution (a) • With the infinite line, we know that the field will have only a radial component in cylindrical coordinates (or x and y components in cartesian). • The field from an infinite line on the z axis is generally Therefore, at point P: BEE 3113ELECTROMAGNETIC FIELDS THEORY

  40. Solution (b) • Here we use the general equation: • Where the total charge Q is BEE 3113ELECTROMAGNETIC FIELDS THEORY

  41. Solution (b) • Or, we can also write E as: • Where and BEE 3113ELECTROMAGNETIC FIELDS THEORY

  42. Solution (b) • So E would be: • Using integral tables, we obtain: BEE 3113ELECTROMAGNETIC FIELDS THEORY

  43. What if the line charge is not along the z-axis?? BEE 3113ELECTROMAGNETIC FIELDS THEORY

  44. Line Charge z P(x, y, z) ρ R y dl (0, y, 0) (0, y’, 0) x BEE 3113ELECTROMAGNETIC FIELDS THEORY

  45. What if the line is not along any axis?? BEE 3113ELECTROMAGNETIC FIELDS THEORY

  46. Example A uniform line charge of 16 nC/m is located along the line defined by y = −2, z = 5. If ε= ε0. Find E at P(1, 2, 3) BEE 3113ELECTROMAGNETIC FIELDS THEORY

  47. Solution • Draw the line y = −2, z = 5. • Find aρ • Find E BEE 3113ELECTROMAGNETIC FIELDS THEORY

  48. Solution z (1, -2, 5) 5 ρ P(1, 2, 3) -2 y x BEE 3113ELECTROMAGNETIC FIELDS THEORY

  49. Solution • To find aρ , we must draw a “projection” line from point P to our line. Then find out what is the x-axis value at that projection point • Then aρ can be calculated as: BEE 3113ELECTROMAGNETIC FIELDS THEORY

  50. Then finally, calculate E BEE 3113ELECTROMAGNETIC FIELDS THEORY

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