ph 1020 physics - iimanianvs/ph102-1.pdf · • displacement current density • maxwell’s...
TRANSCRIPT
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: [email protected]
• 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