PH 1020

PHYSICS - II

Venkatachalam Subramanian

Department of Physics

CONTACT DETAILS

• Class timings at CRC 101

– MONDAY 2:00 – 2:50 PM

– TUESDAY 2:55 – 3:45 PM

– WEDNESDAY 3:50 – 4:40 PM

• My office: HSB 222

– Phone no. 4883

– Email: manianvs@iitm.ac.in

• Proposed examination pattern:

– One Mid-Sem for 40 marks

– End Semester for 60 marks

COURSE CONTENT

ELECTROSTATICS PART - I

• Force and electric field due to continuous charge distribution

• Field lines – Flux – Gauss law – Use of solid angle

• Electric potential – due to continuous charge distribution – equipotential line/ surface

• Poisson’s equation and its solution• Poisson’s equation and its solution

• Workdone in assembling a charge distribution

• Conductors and Capacitors

• Boundary conditions for conductors

• Electric dipole – field lines – Force and torque on a dipole due to external static electric field – Interaction energy between two dipoles.

• Multipole expansion

ELECTROSTATICS

PART - II

Electrostatic fields in matter

• Polarization – restricting to linear, isotropic and homogeneous medium – surface and volume bound charge density

• Electric displacement vector – Dielectric • Electric displacement vector – Dielectric permittivity and susceptibility - Gauss law in materials

• Workdone in assembling free charges in the vicinity of dielectric material

• Complete boundary conditions

MAGNETOSTATICS

PART - I

• Force law – line current, surface current and volume current densities

• Biot-Savart Law – Properties of B

• Magnetic flux – Div B – Magnetic Vector Potential A – Information about scalar Potential A – Information about scalar potential

• Curl B – Ampere’s law

• Properties of A – Multipole expansion –magnetic dipole moment – Force and torque on a magnetic dipole due to external static magnetic field.

MAGNETOSTATICS

PART II

Magnetostatic field in matter

• Magnetization – Bound volume and surface current densitiescurrent densities

• Auxillary field H – Magnetic susceptibility and permeability – general information about para, dia and ferro magnetism

• Boundary conditions

• Force on charged particle under electric and magnetic fields

ELECTROMAGNETICS

• Electromotive force – Faraday’s law of

electromagnetic induction

• Self and mutual inductance

• Displacement current density

• Maxwell’s equation for electromagnetic field

• Derivation wave equation – Plane wave

solution – Poynting vector S.

• Reflection and transmission of electromagnetic

radiation at boundary

SUGGESTED BOOKS

• Introduction to Electrodynamics by Griffiths

• Electromagnetic fields by Wangsness

• Electricity and Magnetism by Mahajan and

RangwalaRangwala

• Elements of electromagnetics by Sadiku

Fields and wave electromagnetics by Cheng

• Electromagnetics by Kraus

CALENDAR

Jan Feb Mar April

3 1 1 6

4 2 2 11T13

5 7T5 7T9 12

10T1 8 8 18T14

11 9 9 19

12 14T6 14T10 20

17T2 15 1517T2 15 15

18 21T7 16

19 22 21T11

24T3 23 22

25 28T8 23

31T4 28T12

29

30

WHY THIS COURSE?

• In our daily life, electromagnetics play a very important role.

• Few examples are:

– It is active in your body

• All your nerve cells (neurons) transmit the signals by passing small amplitude of current.

• Your medicines gets absorbed only due to charges getting separated.

• Your brain gets affected by Earth’s magnetic field lines (Have • Your brain gets affected by Earth’s magnetic field lines (Have you ever experienced of disturbed sleep when you try to sleep along north-south direction with head in the north side?)

– Can you imagine your life without “TV, PC, Cell phone etc.” (Even though they are hazardous to your health)?

– What about fan, electric motors, electric generators?

• WITHOUT TIME VARYING ELECTRIC AND MAGNETIC FIELDS, YOU CAN NOT EXIST!

WHAT DO WE LEARN?

• You learn about the nature/ behaviour of electric charges.

– So, you know how to use them for your purpose or avoiding accidents.

• You learn about the dielectric materials – their behaviour under external

electric field

– So, you know how to make devices – smart devices that changes shape,

gives out electric signals, memorizes your data, photographs you and your

loved ones etc.

• You learn about generating magnetic field not from the natural materials but using piece of wire and moving charges.using piece of wire and moving charges.

– So, you know how a giant magnet picks up metal wastes – magnetic levitation of trains etc.

• You learn about magnetic materials

– So, you know the basics of materials that makes your warship and planes

not detected by enemy RADARs.

• You learn about the moving charges that produce both electric and magnetic

fields

– So, you know how you survive!

For more info: Hyperphysics.phy-astr.gsu.edu

COULOMB’S LAW• Postulated by French physicist Charles Augustin

de Coulomb

• Here, F12 refers to the force on charge 1 due to charge 2.

( )2 11 221 3

0 2 14

r rq qF

r rπε

−=

−

� ��

� �

charge 2.

• The above equation is valid only for stationary charges else relativistic and magnetic effects must be considered.

• Gravitational force always attract but weak in nature – for like charges, electostatic force repels.

• This force is conservative.

– So, you know that curl F is zero.

• Follows super-position principle

– This helps us to obtain the force for continuous

charge distribution.

Fundamental Forces:

Importance of Electromagnetic forces

Force Range Relative strength

Strong

Electromagnetic

Weak

Gravitational

10−15 m

∞

10−17 m

∞

1

10−2

10−5

10−39

Strong and Weak forces are realized in very low dimensions whereas gravitational forces are realized for very large masses

Most of the natural phenomena we see around us, phenomena

involving atoms and molecules and their aggregations in various

phases of matter, all chemical and biological processes, and so on,

are ultimately traceable to electromagnetic interactions in one form

or another.

CHARGE

• In general, any material formation requires

binding of two or more units of basic building

block, the atoms.

• These atoms consist of equal amount of

positive and negative charges positive and negative charges

• These charges are quantized. The value of the

charges are integral multiples of the magnitude

of an electronic charge 1.6 x 10-19 C.

• But “quarks” carry fractional charges like

1/3 or 2/3 of 1.6 x 10-19 C.

• A test charge means a point positive charge of

1 C.

• Any charge will spread its ‘effect’ till infinity.

This is called “FIELD” of the charge – The

electric field.

• If any other charge comes within its field

range, it gets attracted or repelled as per range, it gets attracted or repelled as per

Coulomb’s law.

• By explanation, the field of a charge at a point

P is equal to the force experienced by a test

charge kept at P. That is,F qE=� �

• By virtue of the equation, one immediately

finds that

– Electric field is conservative in nature.

• Curl E is a null vector.

– Electric field obeys super position

principle.

• One can use this for evaluating the field due to

system of charges or continuous charge

distribution.

2

0

ˆ

4

req

Erπε

=�

For system of charges,

Shall we work out some examples to get

familiarized with force and electric field?

1 2

1 1 2 2

3 3 3

0 0 01 2

...

...4 4 4

n

n n

n

E E E E

q rq r q rE

r r rπε πε πε

= + + +

= + + +

� � � �

�� ��

� � �

familiarized with force and electric field?

Electric field due to an isolated point charge.

– The field lines or lines of force comes out radially

for positive charge.

– The magnitude of electric field forms equi-field

surfaces as concentric spherical surfaces.

Divergence of the Electric Field & Dirac Delta Function

Divergence of the function ( )2ˆ rer

Consider the Vector function 2

1r̂e

rΨ =�

v�

is directed radially outward everywhere around the charge

and shows a positive divergence. But if we take divergence for and shows a positive divergence. But if we take divergence for

this vector function we get zero

( )2

2 2 2

1 1 11 0r

r r r r r

∂ ∂ ∇⋅Ψ= = =

∂ ∂

�

From Gauss Divergence theorem we know that

( )V s

d daτ∇ ⋅ Ψ = Ψ ⋅∫∫∫ ∫∫��� ����

But

2

2

2

0 0

1ˆ ˆ( sin )

sin 4

r r

s

da e R d d eR

d d

π π

θ θ φ

θ θ φ π

Ψ ⋅ = ⋅

= =

∫∫ ∫∫

∫ ∫

����

L.H.S ≠≠≠≠ R.H.S. Something wrong with the Gauss-Divergence

theorem ? No! The manner in which the integration is done is

wrong. wrong.

The function v blows up at the point r=0. While the surface integral

on the right hand side is independent of r, the volume integral on the

right hand side includes the point r=0. So the value of 4ππππ must be

arising due to the contribution from the point r=0.

This can not be explained by usual function. Hence we make use of

the Dirac Delta function.

One Dimensional Dirac Delta Function

It is an infinitely high and infinitesimally narrow spike

with area unity.

∫∞

∞−=

=∞

≠=

1)(

0

0,0)(

dxxand

xif

xifx

δ

δ

Technically δδδδ(x) is not a function at all, since its value is Technically δδδδ(x) is not a function at all, since its value is

not finite at x=0. It is a generalized function or distribution.

δδδδ(x)

Area=1

x

Properties of Dirac Delta Fuction

1. If f (x) is some ordinary function (continuous), then the

product f (x) δδδδ(x) is zero everywhere except at x=0.

)()0()()( xfxxf δδ =⇒

Since the product f (x) δδδδ (x) is always zero except at x=0,

we can replace f (x) by its value at origin.we can replace f (x) by its value at origin.

∫ ∫∞

∞−

∞

∞−== )0()()0()()( fdxxfdxxxf δδ

Under an integral the delta function picks out the value of

the function at x=0. The limits of the integral need not be -

∞∞∞∞ and ∞∞∞∞ but can extend from -εεεε and εεεε provided that this

domain extends across the delta function.

2. If the spike is not located at the origin but at x=a

∫∞

∞−=−

=∞

≠=−

1)(

,0)(

dxaxwith

axif

axifax

δ

δ

−=− )()()()(, axafaxxfthen δδ

∫∞

∞−=−

−=−

)()()(

)()()()(,

afdxaxxfand

axafaxxfthen

δ

δδ

3. Although δδδδ itself is not a legitimate function, integrals

over δδδδ are acceptable. It is always intended for use under

an integral sign.

Two expressions involving delta functions D1(x) and D2(x)

are considered equal if

dxxDxfdxxDxf )()()()( 21 ∫∫∞

∞−

∞

∞−=

for all ordinary functions f(x).

4. For any non zero constant k,

)()(

)(1

)(

xx

xk

kx

δδ

δδ

=−⇒

=

)()( xx δδ =−⇒

Proof: For any arbitrary function f (x), consider the integral

∫∞

∞−dxkxxf )()( δ Let y ≡≡≡≡ kx, then x = y/k and dx = 1/k dy

If k is positive, the integration limits are -∞∞∞∞ to +∞∞∞∞ . But if k

is negative then x = ∞∞∞∞ implies y = - ∞∞∞∞ and vice versa, so the

order of limit is reversed.

)0(1

)0(1

)()/()()( fk

fkk

dyykyfdxkxxf =±=±=⇒ ∫ ∫

∞

∞−

∞

∞−δδ

The –ve sign applies when k is negative.

dxxk

xfdxkxxf∫ ∫∞

∞−

∞

∞−

=⇒ )(

1)()()( δδ

)(1

)( xk

kx δδ =∴

The Three-Dimensional delta function

Generalize the delta function to three dimensions

( ) ( ) ( ) ( )zyxr δδδδ =�3

r�

is the position vector

Three dimensional delta function is zero everywhere except at (0,0,0), where

it blows up

Its volume integral is 1

∞ ∞ ∞�( ) ∫ ∫ ∫∫

∞

∞−

∞

∞−

∞

∞−

== 1)()()(3dxdydzzyxdr

allspace

δδδτδ�

generalizing

∫ =−allspace

afdarrf )()()( 3 ����τδ

Delta function picks out the value of the function f

Divergence of is zero everywhere except at the origin, and yet its

integral over any volume containing the origin is a constant

2

ˆ

r

e r

( )π4

)(4ˆ 3

2r

r

er ��

πδ=

⋅∇

)(4ˆ 3

2R

eR��

πδ=

⋅∇or

generallyrrR ′−=���

)(42

RR

πδ=

⋅∇generallyrrR ′−=��

Note that the differentiation is with respect to r, which r/ is held constant

DIVERGENCE OF E

/

2

/

0

ˆ)(

4

1)( dae

R

rrE R

allspace

∫=

�� σ

πε

Noting that r-dependence is contained in

( ) τρπε

′′

⋅∇=⋅∇ ∫ dr

R

eE R

2

ˆ

4

1 ���

rrR ′−=���

( ) τρπε

′′

⋅∇=⋅∇ ∫ drR

E2

04

( )RR

eR��

3

24

ˆπδ=

⋅∇

Thus

( ) ( ) τρπδπε

′′′−=⋅∇ ∫ drrrE�����

3

0

44

1)(

1

0

r�

ρε

=

Gauss’s law

How do we use delta function to denote one, two or

three dimensional charge distributions?

For charge distribution on the z = 0 plane, σσσσ(x,y)δδδδ(z)

For charge distribution on the surface of a sphere of

radius a is: σσσσ(θθθθ,ϕϕϕϕ)δδδδ(r-a).

EXAMPLES FOR DIRAC DELTA

How do we use delta function to denote charge distributions?

For charge distribution on the z = 0 plane: σ(x,y)δ(z)

For charge distribution on the surface of a sphere of radius a is:

σ(θ,ϕ)δ(r-a).

For a charge distribution on a circle of radius R: λ(ϕ)δ(z)δ(ρ-R)

For a charge distribution on a circle of radius R: λ(ϕ)δ(θ-π/2)δ(r-R)

WORK DONE IN ASSEMBLING CHARGE DISTRIBUTION

Now you know that to bring a charge from infinity to

any point requires some work to be done.

If such charge is a test charge, the work done will be

nothing but the potential due to the existing charge nothing but the potential due to the existing charge

distribution. Else, we term this as the work involved

in bringing the charge q from infinity to point P.

Now take the case of ‘first charge’ q1 being kept at

point P1. There is no existing field and therefore no

work is involved to bring this charge.

W qV=

Thus after bringing the first charge q1, there is going

to be electric field and therefore to bring another

charge Q to the point P2, work has to be involved.

The work in this case is

1 1 11 0W q V= =

W q V q V= =where V2=V12 denotes the potential at the point P2

due to q1. Similarly, to bring the third charge q3 to P3,

where V3 denotes the potential at the point P3. V13

and V23 are due to q1 and q2 respectively.

2 2 2 2 12W q V q V= =

( )3 3 3 3 13 23W q V q V V= = +

Now the total workdone is W = W1+W2+W3.

In the general notation for assembling n charges,

( )3 13 23 2 12W q V V q V= + +

1n i

W q V−

=∑ ∑and

1 1

n ij

i j

W q V= =

=∑ ∑

0

1

4

j

ij

ij

qV

rπε=

Therefore to bring one charge after another, one can

follow this procedure. The work involved is none

other than the energy stored in this charge

distribution or ‘Electrostatic energy’.

It may be now noted that the work to bring charge q2

at a point P2 due to the presence of charge q1 at P1

is same as that of bringing charge q at P due to the is same as that of bringing charge q1 at P1 due to the

presence of charge q2 at P2.

Therefore for a given distribution of discrete charges,

take a charge, evaluate the potential at the position

of the charge due to all other charges and obtain the

workdone to bring this charge. Repeat this

procedure for all the charges.

In this procedure, we eventually would end up

evaluating twice the value of workdone. This is

because,

Therefore the general formation would now be,

i ji j ijqV q V=

1 1n n n n q∑ ∑ ∑ ∑

1 1 1 1 0

1 1

2 2 4

n n n nj

n ij i

i j i j ijj i j i

qW q V q

rπε= = = =≠ ≠

= =∑ ∑ ∑ ∑

To assemble a continuous charge distribution, the

work done will be

Take the case of volume charge distribution: Using

differential form of Gauss law,

1 1 1

2 2 2c c

v s

W d V daV dlVρ τ σ λ= = =∫ ∫ ∫

differential form of Gauss law,

( )0

1 1

2 2c

v v

d V E d Vρ τ ε τ= ∇∫ ∫�i

Using the vector identity:

( ) ( )EV E V E V∇ = ∇ + ∇� � �i i i

2E−

( ) ( ) 20 0 0EVd EV d E dε ε ε

τ τ τ∇ = ∇ +∫ ∫ ∫� �i i

Using divergence theorem:

( ) ( ) 20 0 0

2 2 2v v v

EVd EV d E dε ε ε

τ τ τ∇ = ∇ +∫ ∫ ∫� �i i

( ) 20 0

2 2s v

W EV ds E dε ε

τ= +∫ ∫� �i�

Therefore one has to obtain the E(r) and V(r) of the

charge distribution and evaluate the integrals at the

surface and in the volume.

If the surface we choose is at infinity, then the

contribution from electric field and potential will be

zero and hence one has to evaluate only the

volume integral. volume integral.

In this case, one has to evaluate E not only in the

volume but also outside.

20

2allspace

W E dε

τ= ∫

You must note that the electrostatic energy does not

follow superposition principle.

Depending on the situation, one can use any one of

the four equations to evaluate the stored energy.

This energy keeps the total charge distribution

intact.intact.

The electrostatic energy density, energy stored per

unit voume is defined as,

20

2dW E

ε=

METHODS TO REMEMBER

1

1 1

−

= =

=∑ ∑n i

i ij

i j

W q V0

1

4

j

ij

ij

qV

rπε=

1 1 0

1

2 4= =

= ∑ ∑n n

j

i

i j ij

qW q

rπε

20

2allspace

W E dε

τ= ∫

( ) 20 0

2 2= • +∫ ∫

� �

�s v

W EV ds E dε ε

τ

1 1 02 4= =≠

i j ijj i

rπε

Let us evaluate the energy stored in an uniformly charged spherical shell of radius a.

For a spherical shell, the electric field inside it will be zero and outside will vary as 1/r2.

22 220 0 sin

q rW E d dr d d

π πε ετ θ θ ϕ

π ε

∞

= =∫ ∫ ∫ ∫

EXAMPLE – 1

2 2 4

0 0 0

sin2 2 16

allspace r a

W E d dr d dr θ ϕ

τ θ θ ϕπ ε

= = =

= =∫ ∫ ∫ ∫2 2

2 2

0 0

1 4 1 1

2 16 8r a

q qW dr

r a

π

ε π πε

∞

=

= =∫

Let us use ( ) 20 0

2 2= • +∫ ∫

� �

�s v

W EV ds E dε ε

τ

E inside is zero. Hence the volume integral is zero.

E at the surface and V at the surface: sur 2

0

qE

4 a

qV

4 a

=πε

=πε

�

Therefore,0

V4 a

=πε

22

20

3

0 0

14

2 4 8

= =

q qW a

a a

επ

πε πε

Therefore,

EXAMPLE - 2 r

1) Adding little by little:2 2 25

2

0 0 0

r 4 aW Vd 4 r dr

3 15

3q

20 a

ρ πρ= ρ τ = ρ π = ⇒

ε ε πε∫ ∫2) Using surface and volume:

2

2 20 0q q rW 4 a 4 r dr

ε ε ρ= π + π

πε πε ε∫2

0 0 0v

W 4 a 4 r dr2 4 a 4 a 2 3

= π + π πε πε ε

∫2 2 5 2 2 2

00 0 0 0

q 4 a q q 1W

8 a 90 8 a 8 a 5

3q

20 a

πρ = + ⇒ + ⇒

πε ε πε πε πε

In case, this sphere is covered by a spherical shell of radius b (b > a), what would be difference in the energy stored?

CONDUCTORS

-Zero band gap

-Free carriers, electrons, are responsible for

conduction

-These carriers are so mobile that as soon as you

give potential difference, the electrons flow towards give potential difference, the electrons flow towards

the higher potential region so fast to effectively nullify

the difference.

-Therefore a perfect conductor must not have any

potential difference on its surface or volume.

-Conductors form equipotential surfaces or volume.

Within the conductor material, you can not have

electric field.

Why? As soon as there is an electric field, there is a

potential difference and hence electrons flow

instantaneously to nullify this difference. That

means ELECTRIC FIELD WITHIN THE VOLUME IS ZERO.ZERO.

What about the surface?

By the same argument, conductor surfaces are

equipotential and therefore, ELECTRIC FIELD TANGENTIAL TO THE SURFACE IS ZERO. BUT NORMAL COMPONENT OF ELECTRIC FIELD EXISTS.

So, it is very clear that only normal component of the

electric field exists.

If a point positive charge is kept near a conductor, the

field lines touch the conductor surface ‘normally’. The

surface also gets negatively charged and the distribution

of this charge is so arranged that the tangential

component of the electric field is zero. This is calledcomponent of the electric field is zero. This is called

“ELECTROSTATIC EQUILIBIRIUM”

Due to this electrostatic equilibirium, any induced charge

on a conductor spreads only on the surface.

WHAT ABOUT THE SURFACE CHARGE DENSITY?

σσσσ depends on the curvature of the surface.

The charge distribution is such that the net force on every

charge would be zero. In a flat surface, equal distance between

the charges would solve this problem.

Assume a shape as given here. For the

charges A and B, the replusive force is

almost parallel to the surface. But for

charges C and D, the replusive force

will have the parallel component to the

surface. The main effort is to keep the

sum of the parallel component to be

zero. Therefore, this would be possible

if C and D are kept nearer to each other

unlike that of A and B.

Therefore, as the curvature increases, charge density

increases. That is the reason, lightning arrestors have

sharper tips.

If a conducting spherical shell has a point charge at the

center, the inner surface of the shell gets –ve charges and

q+

++

++

+

+

+outer surface gets +ve charges for the

total value of +q.

If the shell is charged to +Q and point

charge q is kept at the center, then the

outer surface of the shell gets a total

-

--

--

-

-

-- ++

+ outer surface of the shell gets a total

charge of q+Q.

If the conducting shell is charged to Q without any charge

in the center, then the electric field inside the shell at any

point will be zero.

DRAW THE ELECTRIC FIELD AND POTENTIAL

FOR THE GIVEN CONFIGURATIONS

Uncharged sphere Charged to +Q Charged to +QUncharged sphere Charged to +Q Charged to +Q

Charged to +Q

+q +q

(a) (b) (c)

(d)(e)

2

04

q Q

bπε

+

2

04

q

aπε

2

04

Q

bπε

|E| variation with distance (r-2)

2

04

q

aπε

2

04

q

bπε

2

04

Q

bπε

04

q Q

bπε

+

04

Q

bπε

V variation with distance (r-1)

04

q

bπε04

Q

bπε

If an infinite conducting sheet has a surface charge

density σ0, then the electric field coming out = ?

PTo evaluate the electric field,

draw the Gaussian surface

as a pill box of height 2z

(extending from –z to+z) and

area A.area A.

( )0 0

ˆ. 2 ;2

z

s

zAE ds E A E e

z

σ σ

ε ε= = =∫

� ��

�

The electric field coming out of a charged sphere:

( )2

2 0 0

0 0

4ˆ4 ;

r

RE R E e

σ π σπ

ε ε= =

�

If two such conducting plates one with charge +q and

another with –q are kept parallel to each other but

separated by a distance d, then this forms a device

called CAPACITOR.

You already know that the field within this parallel

plate capacitor is constant and equal to σ0/ε0 and the

direction being from +ve to –ve plate.direction being from +ve to –ve plate.

For an isolated but charged conducting sphere, can

you calculate the capacitance?

C = Q/V and V is the potential difference = Q/(4πε0R).

Therefore, C = 4πε0R.

LECTURE - 7

Discussion about dipoles

DIPOLE

A dipole is a combination of equal amount of

positive and negative charges separated by a

distance d.

A dipole has a dipole moment defined as p=qd,

where d points from +ve to –ve charge.

A point dipole has this separation d � zero.

Therefore, for this only p is specified.

Let us evaluate E due to this.

+q

-q

d

Let P is the point at which we evaluate E due to this

dipole of moment p = qr and with its geometric

center at the origin.

+q

-q

r

P

O

r1

r2

R

Let OP = R; R >> r

r2 = R-r/2; r1 = R+r/2

2 1r rq qE

= −

� �� -q

2 1

3 3

0 02 1

3 3

0

4 4

2 2

4

2 2

r rq qE

r r

r rR R

qE

r rR R

πε πε

πε

= −

− + = −

− +

�

� �

� �� �

�

� �� �

333 2 22

2 22 2 2 4 2

r r r r rR R R R R

− = − − = + −

� � � �� � � �

i i

As R >> r, we can neglect the term r2. Therefore,

( )33

32 222 2 32

22 1

4 2

r r R rR R R R r R

R

+ − = − = −

�� �� � i�i i( ) 2

2 14 2

R R R R r RR

+ − = − = −

i i

3

23 3

2 2

31 1

2

R r R rR R

R R

−

− − − = +

� �� �i i

After neglecting

higher order

terms.

As the above term lies in the denominator:

Similarly,3

32

3 3

2 2

31 1

2 2

r R r R rR R R

R R

−−

− − + = + = −

� �� � �� i i

Now,

( )

3 2 2

0

3 31 1

4 2 2 2 2

q r R r r R rE R R

R R Rπε

= − + − + −

� �� � � �� � �i i

� � �( )( )3 2 3

0 0

3ˆ ˆ3

4 4

R R rq qE r R R r r

R R Rπε πε

= − = −

� � �i� � � �

i

p qr=� �

AS ,( )

3

0

ˆ ˆ3

4

p R R pE

Rπε

− =

� �i�

This form of the electric field does not depend on

any coordinate system – Coordinate independent

form.

You now know that:

E(r) for a point charge varies as (1/r2)

E(r) for a point dipole varies as (1/r3)E(r) for a point dipole varies as (1/r3)

Special cases:

1. If P lies on the line joining +q and –q (end-on),

[ ]3 3

0 0

ˆ ˆ3 2

4 4

p p R pRE

R Rπε πε

−= =�

ˆ

ˆ

p pR

p R p

=

=

�

�i

2. If the point P lies on the broad side,ie.

perpendicular to the line joining +q and –q.P

3

04

pE

Rπε

−=

��

ˆ 0p R =�i

Electric field

Now about the potential due to the dipole:

On the broad side (on the x or y-axis when dipole is

oriented along êz, V = 0 since both charges are

equidistant from the point. Therefore, x-y plane is the

equipotential surface with V=0.

For the end-on position, the potential can be

calculated ascalculated as

For any other position,

3 2

0 04 4

R Rpdz p

V E dzz Rπε πε

∞ ∞

= − = =∫ ∫� �i

2

0

ˆ

4

p RV

Rπε=

�i

How about the equipotential contours?

0 2 2

0 0

ˆ cos

4 4

p R pV

R R

θ

πε πε= =

�i

2

cosC

R

θ=and

When we keep this dipole in an external electric

field, what will happen?

+q

-q

OEext

F+

Let us take the case

where Eext is uniform.

Due to this external

field, there is going to

be a torque developed

with respect to the

origin.-qF-

origin.

2 2

r rF F r qE p Eτ + −= × − × = × = ×

� �� � � �� � �

The effect of torque is to rotate the dipole such that paligns along the electric field direction. If the initial

direction of dipole is –E, then ττττ = ? And what would

be the final alignment of the dipole?

If the electric field is not uniform?

+q

Oθ

y

∆y

( ), ( , )x x x

F q E x x y y E x y= + ∆ + ∆ −

( , ) ( , )x xx x x

E EF q E x y x y E x y

x y

∂ ∂= + ∆ + ∆ − ∂ ∂

Using Taylor series expansion,

x xE E

F q x y ∂ ∂

= ∆ + ∆ -q

∆x x

x xx

E EF q x y

x y

∂ ∂= ∆ + ∆

∂ ∂

( )x xx x y x

E EF p p p E

x y

∂ ∂= + = ∇

∂ ∂

��i

( )F p E= ∇� � ��

iIn general,

Now let us use the vector identity

( ) ( ) ( ) ( ) ( )p E p E E p p E E p∇ = ∇ + ∇ + ×∇ × + ×∇ ×� � � � �� � � � �i i i

Curl E = 0; Div E = 0 for charge free region

Curl p and Div p = 0 since p does not depend on

position. Therefore,

( ) ( )= ∇ = ∇� � �� �

( ) ( )F p E p E= ∇ = ∇� � �� �

i i

The force is negative gradient of potential energy.

( )F U p E= −∇ = ∇� ��

i

( )U p E= −��iTherefore,

Therefore when non-uniform electric field is applied

to the dipole, the dipole reacts such a way that the

potential energy is minimum.

Hence, the dipole aligns along the electric field and

moves towards the higher electric field region to

attain minimum energy configuration for stability.

If the source of the external electric field is another

dipole, then the interaction energy between the two

dipoles is given by, ( )1 1

2 1 2 3

0

ˆ ˆ3

4

p R R pU p E p

Rπε

− = − = −

� �i�� �

i i

denotes the unit vector from p1 to p2.R̂