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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 DasPhysics Group (3242-S) [email protected]

Chamber Consultation: Friday 5.00 to 6.00 PM

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.)

Electromagnetism

Electromagnetism deals with electromagnetic force and

field

Electricity Magnetism Optics

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.

The test charge qo experiences an electric field E

directed as shown.

E = lim q0q0 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 qo placed at that point :

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.

Electric force and field

The electric field at r = Force per unit charge ,

=> E = F/q0 = kq1/r2

+ q0q1

r

The Coulomb force is F= kq1q0/r2

(where, k = 1/40)

Negative source charge

E= kq1/r2

Positive source charge

q1

E

q1

E

Negative source charge

Electric Field Lines

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

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.

Electric Field Lines

.

Like charges (++) Opposite charges (+ -)

Electric DipoleElectric DipoleAn An electric charge dipole consists consists of a pair of equal and opposite point of a pair of equal and opposite point charges separated by a small charges separated by a small distance, distance, d.d.

d

+Q -Q

Dipole Moment Dipole moment p is a measure of the strength of the dipole and indicates its direction

+Q

-Q

d p is in the direction from the negative point charge to the positive point charge.

d

dQp

Electric Field of a dipoleTo find the electric field E at point P,

At P, the fields E1 and E2 due to the two charges, are equal in magnitude.

The total field is E = E1 + E2,

E = k 2aq /(y2 +a2)3/2

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

Electric Field of a dipole (cont’d)

If y >> a, then E ~ k p/y3

E due to a dipole ~ 1/ r3

E due to a point charge ~ 1/ r2

E = k 2aq /(y2 +a2)3/2

Electric Field of a dipole (cont’d)

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 x2 a2 ~ x2

E ~ 4kqa/x3

22222 )(

)2(2

)()( ax

xaqk

ax

kq

ax

kqE

Ex 26.11: Field due to Electric Quadrupole

4

0

2

2

23

x

qaE

Pr 26.4: Field due to Electric Quadrupole

To find out E at P:

4

0

2

4

23

z

qdE

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 finitetorque acting on it

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

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

0

.

dW

0

00

coscos

sin

pE

dpEd

Change in potential energy of the system:

)cos(cos 0 pEWU

Choosing reference angle 0 = 90°

and U(0 ) = 0.

EppEU

cos

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

Ex 26.37:

Dipole: q = 2e; d = 0.78 nm

E (ext.) = 3.4 ×106 N/C.

To find: torque

(a) p E

(b) p E

(c) p is opposite to E