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Happy New Year -2006

Happy New Year -2006. Champak Baran Das Physics Group (3242-S) cbdas@bits-pilani.ac.in Chamber Consultation: Friday 5.00 to 6.00 PM. PHYSICS-II (PHY C132). Text Book: PHYSICS, VOL 2: by Halliday, Resnick & Krane (5 th Edition). Reference Books : Introduction to Electrodynamics:

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Happy New Year -2006

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  1. Happy New Year -2006 Champak Baran Das Physics Group (3242-S) cbdas@bits-pilani.ac.in Chamber Consultation: Friday 5.00 to 6.00 PM

  2. PHYSICS-II (PHY C132) Text Book: PHYSICS, VOL 2: by Halliday, Resnick & Krane (5th Edition) Reference Books: Introduction to Electrodynamics: by David J. Griffiths (3rd Ed.) Concepts of Modern Physics: by A. Beiser (6th Ed.)

  3. Electromagnetism Electromagnetism deals with electromagnetic force and field Optics Electricity Magnetism

  4. Electric Field • An electric field is said to exist in the region of space around a charged object. • When another charge object enters this electric field, an electric force acts on it.

  5.  F E = lim q0 q0 0 The test charge qo experiences an electric field E directed as shown. The electric field E at a point in space is defined as the electric force F acting on a unit positive test charge qoplaced at that point :

  6. Test charge should be small not to disturb the charge distribution of the source (a) For small enough qo, the distribution is undisturbed. (b) For a larger qo' , the distribution gets disturbed.

  7. r + q0 q1 Electric force and field • The Coulomb force is F= kq1q0/r2 (where, k = 1/40) The electric field at r = Force per unit charge , => E = F/q0 = kq1/r2

  8. q1   E E q1 Negative source charge Positive source charge E= kq1/r2

  9. Negative source charge Electric Field Lines

  10. Electric Field Lines: a graphic concept as an aid to visualize the behavior of electric field. • Begin on + charges and end on - charges. • Number of lines entering or leaving a charge is proportional to the charge

  11. Electric Field Lines: (contd.) • Density of lines indicates the strength of E at that point • The tangent to the line passing through any point in space gives the direction of E at that point • Two field lines can never cross.

  12. Like charges (++) Opposite charges (+ -) Electric Field Lines .

  13. -Q +Q d Electric Dipole An electric charge dipole consists of a pair of equal and opposite point charges separated by a small distance, d.

  14. +Q -Q Dipole Moment Dipole momentp is a measure of the strength of the dipole and indicates its direction pisin the direction from the negative point charge to the positive point charge. d

  15. To find the electric field E at point P, Electric Field of a dipole At P, the fields E1 and E2 due to the two charges, are equal in magnitude. The total field is E = E1 + E2, E1 = E2 = kq/r2 = kq /(y2 +a2) The y components cancel, and x components add up => E || x-axis |E| = 2E1 cos . cos  = a/r = a/(y2 +a2)1/2 E = k 2aq /(y2 +a2)3/2

  16. Electric Field of a dipole (cont’d) E = k 2aq /(y2 +a2)3/2 If y >> a, then E ~ k p/y3 E due to a dipole ~ 1/ r3 E due to a point charge ~ 1/ r2

  17. y q -q x 2a Electric Field of a dipole (cont’d) To find the electric field at a distant point along the x-axis. The E field at any point x : When x >>> a, then x2 a2 ~ x2  E ~ 4kqa/x3

  18. Ex 26.11:Field due to Electric Quadrupole

  19. Pr 26.4: Field due to Electric Quadrupole To find out E at P:

  20. A Dipole in Electric field The net force on the dipole is always zero. But there is a finite torque acting on it This torque tends to rotate it, so that p lines up with E.

  21. x Dipole in a Uniform Electric Field Torque about the com=t =F x sin q + F(d-x)sin q =Fdsin q =qEdsin q =pEsin q =p x E t = p x E

  22. Work done by external field E to rotate the dipole through an angle 0 to :

  23. Change in potential energy of the system: Choosing reference angle 0= 90° and U(0) = 0.

  24. Ex 26.36: • Dipole: q = 1.48 nC; d = 6.23 µm • E (ext.) = 1100 N/C To find: (a) dipole moment p (b)difference in potential energy corresponding to dipole moment parallel and antiparallel to E. Ans. (a) p = 9.22 ×10-15 Cm (b) U = 2.03×10-11J

  25. Ex 26.37: Dipole: q = 2e; d = 0.78 nm E (ext.) = 3.4 ×106N/C. To find: torque  (a) p  E (b) p  E (c) p is opposite to E

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