1 / 41

SPECIAL TECHNIQUES

SPECIAL TECHNIQUES. To find the electric field of a stationary charge distribution :. Find the potential of the distribution. To Solve : Poisson’s / Laplace’s Equation. To determine V. Poisson's / Laplace’s equation. +. A set of boundary conditions. Proof ?. Uniqueness Theorem.

Download Presentation

SPECIAL TECHNIQUES

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. SPECIAL TECHNIQUES To find the electric field of a stationary charge distribution : Find the potential of the distribution To Solve : Poisson’s / Laplace’s Equation Dr. Champak B. Das (BITS, Pilani)

  2. To determine V Poisson's / Laplace’s equation + A set of boundary conditions Proof ? Uniqueness Theorem Dr. Champak B. Das (BITS, Pilani)

  3. First Uniqueness Theorem : The solution to Laplace’s equation in some volume is uniquely determined if the potential is specified on the boundary surface. Corollary The potential in a volume is uniquely determined if (a) the charge density throughout the region, and (b) the value of the potential on all boundaries, are specified. Dr. Champak B. Das (BITS, Pilani)

  4. Second Uniqueness Theorem : In a volume surrounded by conductors and containing a specified charge density, the electric field is uniquely determined if the total charge on each conductor is given. (The entire region can be unbound/bounded by another conductor). Dr. Champak B. Das (BITS, Pilani)

  5. SPECIAL TECHNIQUES • The Method of Images • Multipole expansion Dr. Champak B. Das (BITS, Pilani)

  6. METHOD OF IMAGES P ( x, y, z) z q d y Grounded conducting plane x To find out the potential in the region above the plane Dr. Champak B. Das (BITS, Pilani)

  7. Solution of Poisson’s equation: (in the region z > 0 ), WITH • A point charge q at (0,0,d) • Boundary conditions: • V = 0 when z = 0 • V  0 when x2+y2+z2 >> d2 Dr. Champak B. Das (BITS, Pilani)

  8.  one function that meets the requirement Guaranteed by : Corollary of First Uniqueness Theorem The potential in a volume is uniquely determined if (a) the charge density throughout the region, and (b) the value of the potential on all boundaries, are specified. Dr. Champak B. Das (BITS, Pilani)

  9. Z P ( x, y, z) +q d Y d -q X A new problem: Dr. Champak B. Das (BITS, Pilani)

  10. Z = 0 x2+y2+z2 >> d2 Final answer (By virtue of Uniqueness Theorem) Dr. Champak B. Das (BITS, Pilani)

  11. Induced Surface Charge Total induced charge : Dr. Champak B. Das (BITS, Pilani)

  12. Force Force of attraction on q towards the plane Force of attraction on +q towards -q Dr. Champak B. Das (BITS, Pilani)

  13. ENERGY Two point charges and no conductor : Single point charge and conducting plane : Dr. Champak B. Das (BITS, Pilani)

  14. R a q V=0 Another example : A point charge and a grounded conducting sphere : Dr. Champak B. Das (BITS, Pilani)

  15. Image charge : Location of image charge : (to the right of the centre of the sphere) Dr. Champak B. Das (BITS, Pilani)

  16. rs r rs´ b q q' a Two point charges q and q and no conductor Dr. Champak B. Das (BITS, Pilani)

  17. rs r rs´ Prob. 3.7(a): θ z q b q' a  V=0 r = R Dr. Champak B. Das (BITS, Pilani)

  18. Prob. 3.7(b) : Induced surface charge on the sphere : Total Induced surface charge : Dr. Champak B. Das (BITS, Pilani)

  19. Prob. 3.7(c) : Force on q : Energy of the configuration : Dr. Champak B. Das (BITS, Pilani)

  20. rs d' r θ' r' MULTIPOLE EXPANSION To characterize the potential of an arbitrary charge distribution, localized in a rather small region of space Dr. Champak B. Das (BITS, Pilani)

  21. Law of cosines  Dr. Champak B. Das (BITS, Pilani)

  22. Legendre polynomials More on this next sem. in Maths - III Dr. Champak B. Das (BITS, Pilani)

  23. Systematic expansion for the potential of an arbitrary localized charge distribution, in powers of 1/r Multipole expansion of V in powers of 1/r Dr. Champak B. Das (BITS, Pilani)

  24. Monopole term Dipole term Quadrupole term Dr. Champak B. Das (BITS, Pilani)

  25. The Monopole Term: …… is the most dominant term for r >> Potential of any distribution  Vmon , (if looked from very far point) For a point charge at origin, V = Vmon, everywhere Dr. Champak B. Das (BITS, Pilani)

  26. The Dipole Term: …… is the most dominant term if total charge is zero Dr. Champak B. Das (BITS, Pilani)

  27. dipole moment of the distribution Dr. Champak B. Das (BITS, Pilani)

  28. z -q d r'_ +q r'+ y x For a collection of point charges, For aphysicaldipole: Dr. Champak B. Das (BITS, Pilani)

  29. P rs+ +q r  d rs- -q Potential of a physical dipole Dr. Champak B. Das (BITS, Pilani)

  30. Potential due to a point charge ~ 1/r Potential due to a dipole ~ 1/r2 Dr. Champak B. Das (BITS, Pilani)

  31. physical dipole Dr. Champak B. Das (BITS, Pilani)

  32. Potential for a pure dipole (d  0) Physical dipole Pure dipole for d  0, q   , with p=qd kept fixed Dr. Champak B. Das (BITS, Pilani)

  33. z r rs y d O q x Role played by ORIGIN of coordinate system in multipole expansion A point charge away from origin :  Posses a non zero dipole contribution Dr. Champak B. Das (BITS, Pilani)

  34. Dipole moment changes when origin is shifted : d' y r' a x Dr. Champak B. Das (BITS, Pilani)

  35. If Q = 0, then If net charge of the configuration is zero, then the dipole moment is independent of the choice of origin. Dr. Champak B. Das (BITS, Pilani)

  36. 3q a a a -2q -2q a q Prob. 3.27: In the charge configuration shown, find a simple approximate formula for potential, valid at points far from the origin. Express your answer in spherical coordinates. Answer: Dr. Champak B. Das (BITS, Pilani)

  37. Field due of a dipole Potential at a point due to a pure dipole: z r  p y  x Dr. Champak B. Das (BITS, Pilani)

  38. Field due of a dipole (contd.) Recall: Dr. Champak B. Das (BITS, Pilani)

  39. z r  y x Prob 3.33: Electric field in a coordinate free-form : p Dr. Champak B. Das (BITS, Pilani)

  40. z y Field lines of a pure dipole Dr. Champak B. Das (BITS, Pilani)

  41. Field lines of a physical dipole Dr. Champak B. Das (BITS, Pilani)

More Related