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Happy New Year -2006 Champak Baran Das Physics Group (3242-S) Chamber Consultation: Friday 5.00 to 6.00 PM

# Happy New Year -2006 Champak Baran Das Physics Group (3242-S) [email protected] Chamber Consultation: Friday 5.00 to 6.00 PM

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Happy New Year -2006 Champak Baran Das Physics Group (3242-S) [email protected] Chamber Consultation: Friday 5.00 to 6.00 PM Slide 2 PHYSICS-II (PHY C132) Text Book: PHYSICS, VOL 2: by Halliday, Resnick & Krane (5 th Edition) Reference Books: Introduction to Electrodynamics: by David J. Griffiths (3 rd Ed.) Concepts of Modern Physics: by A. Beiser (6 th Ed.) Slide 3 Electromagnetism Electromagnetism deals with electromagnetic force and field ElectricityMagnetism Optics Slide 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. Slide 5 The test charge q o experiences an electric field E directed as shown. E = lim q0q0 q 0 0 F The electric field E at a point in space is defined as the electric force F acting on a unit positive test charge q o placed at that point : Slide 6 Test charge should be small not to disturb the charge distribution of the source (a) For small enough q o, the distribution is undisturbed. (b) For a larger q o ', the distribution gets disturbed. Slide 7 Electric force and field The electric field at r = Force per unit charge, => E = F/q 0 = kq 1 /r 2 + q 0 q1q1 r The Coulomb force is F= kq 1 q 0 /r 2 (where, k = 1/4 0 ) Slide 8 Negative source charge E= kq 1 /r 2 Positive source charge q1q1 E q1q1 E Slide 9 Negative source charge Electric Field Lines Slide 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 Slide 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. Slide 12 Electric Field Lines. Like charges (++)Opposite charges (+ -) Slide 13 Electric Dipole An consists of a pair of equal and opposite point charges separated by a small distance, d. An electric charge dipole consists of a pair of equal and opposite point charges separated by a small distance, d. d +Q -Q Slide 14 Dipole Moment Dipole moment p is a measure of the strength of the dipole and indicates its direction +Q -Q p is in the direction from the negative point charge to the positive point charge. d Slide 15 Electric Field of a dipole To find the electric field E at point P, At P, the fields E 1 and E 2 due to the two charges, are equal in magnitude. The total field is E = E 1 + E 2, E = k 2aq /(y 2 +a 2 ) 3/2 E 1 = E 2 = kq/r 2 = kq /(y 2 +a 2 ) The y components cancel, and x components add up => E || x-axis |E| = 2E 1 cos. cos = a/r = a/(y 2 +a 2 ) 1/2 Slide 16 Electric Field of a dipole (contd) If y >> a, then E ~ k p/y 3 E due to a dipole ~ 1/ r 3 E due to a point charge ~ 1/ r 2 E = k 2aq /(y 2 +a 2 ) 3/2 Slide 17 Electric Field of a dipole (contd) 2a q -q x y To find the electric field at a distant point along the x-axis. The E field at any point x : When x >>> a, then x 2 a 2 ~ x 2 E ~ 4kqa/x 3 Slide 18 Ex 26.11 : Field due to Electric Quadrupole Slide 19 Pr 26.4: Field due to Electric Quadrupole To find out E at P: Slide 20 A Dipole in Electric field The net force on the dipole is always zero. This torque tends to rotate it, so that p lines up with E. But there is a finite torque acting on it Slide 21 p x E Dipole in a Uniform Electric Field Torque about the com F x sin F(d-x)sin Fdsin qEdsin pEsin p x E x Slide 22 Work done by external field E to rotate the dipole through an angle 0 to : Slide 23 Change in potential energy of the system: Choosing reference angle 0 = 90 and U( 0 ) = 0. Slide 24 Ex 26.36: Dipole: q = 1.48 nC; d = 6.23 m 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 = 2.0310 -11 J Slide 25 Ex 26.37: Dipole: q = 2e; d = 0.78 nm E (ext.) = 3.4 10 6 N/C. To find: torque (a) p E (b) p E (c) p is opposite to E ##### Abstract arXiv:2004.10178v1 [cs.LG] 21 Apr 2020Email addresses: [email protected] (Pushpendu Ghosh), [email protected] (Ariel Neufeld), [email protected]
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