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

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Presentation on theme: "Happy New Year -2006 Champak Baran Das Physics Group (3242-S) Chamber Consultation: Friday 5.00 to 6.00 PM."— Presentation transcript:

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 (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.)

3 Electromagnetism Electromagnetism deals with electromagnetic force and field ElectricityMagnetism Optics

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 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 :

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.

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 )

8 Negative source charge E= kq 1 /r 2 Positive source charge q1q1 E q1q1 E

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

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

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

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

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

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

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. This torque tends to rotate it, so that p lines up with E. But there is a finite torque acting on it

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

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 -11 J

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


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