Summary of Lectures Sept last week

Inductance

From the Faraday law, we find that a changing current in one circuit induces emf in a nearby circuit. This is because, it produces a changing B-field and hence also produce an E-field. This induced E-field creates emf in the second circuit and drives current in it. The emf is obtained by the usual Flux rule, 
ε = -dφ/dt. The flux itself is proportional to the current in the first circuit and the proportionality constant is called Mutual Inductance between the two circuits.

The Demo of FM radio shown in class was to show how changing current in one coil creates current in the second coil and the magnitude of this induced current (and hence the flux) depends on the geometrical placing of the two coils.

The mutual inductance between the two coils is 
M= ∫∫ (μ₀/4 π) dl₁.dl₂/|(r1-r2)|, though this is seldom used in calculations.

A changing current in a circuit also produces an emf in this circuit itself for the same reason of creation of ionduced E-field. . The flux linked to the circuit is proportional to the current itself and the proportionality constant is called self inductance L. The back emf is ε = -L di/dt.

When a current passes in a circuit, there is an E-field in the wire. The current density J is proportional the the NET E-field existing in the wire at that point, . This net E-field may have contribution from any accumulated charge on the surface of the wire or induced E-field or any other source. J = σ E where sigma is the electrical conductivity.

Ampere-Maxwell Law 

Just as changing magnetic field is accompanied by E-field, changing E-field is accompanied by B-field. Any that creates time varying E-field will also produce B- field. Thus curl of E is related to ∂ B/ ∂ t for the E field corresponding to B-field. Curl E is also related to J. Doing dimensional analysis and using charge current continuity equation, one gets

∇ X  B = μ ₀ J +ε ₀ μ ₀ ∂ E / ∂ t.

This is known as Ampere-Maxwell law and it was Maxwell who worked out this extra term (from a different analysis)

Poynting Vector 

The magnetic field energy is B²/2 μ ₀ per unit volume. Electrical energy is ε₀E²/2 per unit volume. The Poynting vector S is defined as S =(E X B)/μ₀.

The direction of S gives the direction in which EM energy travels and magnitude of S gives energy flowing per unit time per unit area (perpendicular to the direction of flow)

For any volume, ∫ S.da =-d/dt(U_em)-∫ (E.J)

∫ (E.J) gives the loss of EM energy due to work done by the fields on the charges as they move to form current. This appears as the thermal energy if the current is in a conducting wire. You can write 
∫ S.da + ∫ E.J = -d/dt (U_em)

LHS first term gives EM energy going out of the volume through the surface, LHS second term gives energy of EM fields spent on charges (which appears as thermal energy but we are accounting only EM energy and not the total energy). The it gives loss of EM energy in the volume per unit time which is same as the RHS.

Induced E-fields

Consider a bar magnet moving with velocity v in x-direction as seen from S. Does it produce electric field?

Let us look at the situation from S'. The magnet is at rest and we know, it produces only a magnetic field. You also know the magnetic field directions at different points due to this magnet. You can now do a field transformation from S' to S and get the fields in S. Indeed there is an electric field.

Any source that produces time varying B-field also produces E-field. Time varying B-field is always accompanied by E-field. Such an electric field is called Induced electric field.


If a conducting loop, coil or circuit is placed in an induced E-field, a current can be driven by the induced E-field. The emf is given by the Flux Rule.

The equations governing Induced E-field are:

∇ X E = -(∂ B /∂ t), ∇. E = 0.

These equations have the same mathematical structure as

∇ X B = μ₀ J and ∇. B = 0.

Hence induced E can be obtained from ∂ B/∂ t in the same way as B can be obtained from J .

Lecture-2 Sept 17, 2010 Motional emf

Today we continued with E-B fields of a point charge q moving with a uniform velocity v. We found that (1) The E-field at a point at an instant is radial from the instantaneous position of the charge. So if you wish to get the field at P at time t, and the charge is at A at this time t, the field at t is along AP (q positive). (2) The magnitude of the E-field is largest in the direction perpendicular to the direction of motion and smallest in the direction of motion.


Then we started Motional emf. Questions: What is emf?, What is driving force and what is maintaining force? What charge distribution creates electric field in a wire connected to a battery, resistances etc? The most juicy part was the calculation of magnetic field and electric field inside a conducting rod moving with a uniform velocity v in a uniform magnetic field. The rod, velocity and the B-field were taken mutually perpendicular.
Why should there be an E-field inside the rod, where I had only applied B-field? This is because the magnetic force pushes free electrons on one side and there is a charge distribution coming up on the surface of the moving conductor.


This charge distribution also moves with the rod and hence produces a B-field other than the applied B-field. Thus the B-field inside the rod (and in the vicinity of the rod) will the different from the applied field B. The task is to get these fields. And to do this we go to the rod frame which is S'. Clearly distinguish applied fields in S, Applied Fields in S', Inside (rod) Fields in S and Inside fields in S'. At this moment we know Applied fields in S, and wish to get inside Fields in S. Using Direct transformation equations, get the Applied Fields in S'. Though the Applied E-field is zero in S, it is not zero in S'. It is in y-direction in the scheme taken in class. The Applied B field has somewhat larger magnitude in S' though the direction is the same.


Now that we know the applied fields in S', we turn our attention to fields inside the rod in S'. As the rod is at rest and there is an E-field in y-direction, charges will redistribute on the surface of the rod and finally the E-field inside the rod will be zero. Remember we are talking from S' frame. The redistributed charges are at rest and so produce no magnetic field. The magnetic field inside the rod is THEREFORE equal to the magnetic field Applied (all talks in S'). Hence we obtain E and B fields inside the rod in S'.


Since we know E and B fields inside the rod in S', we do Back transformation to get E and B fields inside the rod in S. And that completes the task.


What do we find for the fields? Now I am talking from S. The job of S' is done. All our experiments are in S. We took help of S' only to get the fields inside the rod in S. As the surface charge distribution on the rod was not known, we could not have directly calculated fields inside the rod in S.
1) The B-field inside the rod is larger than the applied B-field by the factor 1/(1-v²/c²).
2) There is an E-field inside the rod. You can check, E = - v x B as expected in steady state.


Hope it had been a challenging session. I was benefited by the 30-m questions from various students after the class. Thank you.

Summary of Lecture‐1 of Part II

Summary of Lecture‐1 of 2nd part  
--------------------------------------------
1. E, B are frame dependent. Examples: (a) Charge at rest in S, B=0 in S and not zero in S’ 
(b) E= E ₀ in y‐direction, B= B₀  in z
direction in S. Charge sent at velocity E/B in x‐direction in S, Keeps moving 
with this velocity along x‐direction. In S’, it is at rest showing E=0. 


2. Equations for E,B transformation stated without proof. 


3. Example: Line charge λ at rest in S. E Field in S written using Coulomb’s law, B=0. Fields in S’ 
obtained from transformation equations. Comparing with Coulomb’s, Biot‐Savart law shows that  linear charge density is  
λ/√(1‐v²/c²).


Idea of length contraction introduced to get increased λ. 
To get such a contraction, x’ =x‐vt modified by the factor 
1/√(1‐v²/c²). y’=y, z’ = z stated. 


4.Example: A point charge moving in S with velocity v. In S’ the charge is static and fields written  using Coulomb’s law, B’=0. Using inverse transformation equations, E‐field obtained in S.  There are questions about length contractions, and modification in coordinate transformation equations. These will be dealt in somewhat more detail, hopefully, in PHY102 under special theory of  relativity. But once you take the E-B transformation equations granted, they follow naturally. 
   

Lecture Schedule for PHY 103 N Part 2. ( Prof. H.C. Verma)

Lecture Plan of PHY103 2010-11 ( Part II)


Lect No    Date           Topic
---------------------------------------------------------------------------------

1    15 Sept 10    Relativistic Transformation of E-B fields      

2    17 Sept 10    Motional  emf, Flux rule, Lenz law      

3    20 Sept 10    Flux rule in Generalized  case, Faraday's law,  
                            Calculation of induced Electric field.

4    22 Sept 10    Inductance, L-R circuit      

5    24 Sept 10    Magnetic  energy density, Poynting theorem      

6    27 Sept 10    Displacement current, B-field from time varying  E-field      

7    29 Sept 10    Maxwell's Equation in material medium, Boundary conditions      

8    1 Oct 10 (GS)    EM waves in free space, Integrating Optics with Electromagnetism      

9    4 Oct 10            Interference of EM Waves, Light, spatial and temporal coherence      

10    6 Oct 10            Fraunhofer Diffraction of Light at a single slit, N-slits      

11    18 Oct 10            EM waves in material medium, conducting and dielectric medium      

12    20 Oct 10            Reflection and refraction of EM waves      

13    25 Oct 10            Low intensity YDSE, Photoelectric effect, Compton Scattering      

14    27 Oct 10            Davisson Germer Expt,  YDSE with electrons, Heisenberg uncertain

15    29 Oct 10           Wave function,  Probability density, Relation with position and
                                momentum distributions, Delta function wave fn and plane waves      

16    1 Nov 10           Operators for observables,  Schrodinger Equation, Definite Energy states      
 
17    3 Nov 10           Deep Square well ,  Finite Square well, hetero junctions      

18    8 Nov 10           Barrier penetration, Nuclear Fusion and Coulomb Barrier      

19   10 Nov 10           Hydrogen atom wave functions      

20    12 Nov 10          Revision and Discussion      
           
Books:
1.    Electrodynamics by D J Griffiths
2.    Optics by P K Srivastava
3.    Quantum Physics by H C Verma

Magnetostatics in Material Media II

Linear Magnetic Materials
----------------------------------
Just like that for Linear Dielectrics we also have Linear Magnetic Materials for which the Magnetization M is proportional to the applied field. The applied field is taken to be the H field because that is the field which is measured in the Lab. This is unlike the case for electrostatics where the E-field is the applied field . So we have M=χ_mH where χ_m is called the Magnetic susceptibility. Since B=μ₀( H + M )=μH where μ is the permeability of the material we have μ=μ₀(1 + χ_m). As in Dielectrics it is also possible to define a relative permeability μ_r=μ/μ₀=(1 + χ_m).


Since in the linear magnetic materials both the H field and the Magnetization M are proportional to the B- field, it may appear that in this case as ∇.B=0 this implies ∇.H=0. This is wrong, for the same reason as in Dielectrics where Curl E=0 does not imply Curl D=0. This is because at the boundary the proportionality changes and a careful examination of the corresponding integral forms of the law show that both M and H have non zero divergences except in the case when the entire space is filed with a single magnetic material in which case there are no boundaries. For linear materials since the bound current J_b=∇ X M and M =χ_m H 
we have J_b=χ_m J_free. 


The most important and widely used magnetic materials are however Ferromagnetic materials. These are non linear magnetic materials containing permanent magnetic dipoles associated with unpaired electrons in odd electron atoms just as in Paramagnetism. However the difference in this case is that there is a very strong
interaction between neighbouring dipoles due to quantum mechanical reasons. For this reason neighbouring dipoles tend to point the same way even in the absence of an external magnetic field. In small regions the dipole orientation is almost 100 % due to this reason. These are called Ferromagnetic domains and the entire region may be described by a single total dipole moment vector called the magnetic domain vector .


There are a large number of such domains in a Ferromagnetic material with randomly oriented domain vectors subject to random thermal vibrations. When a a Ferromagnetic material is subjected to an external magnetic field the domain vectors tend to align
together and this causes domains to merge and grow and for string fields the entire material may be described by a single domain and results in a very strong Magnetization. When the external field is switched off some of the domain vectors stay aligned and gives rise to permanent magnetization. This is called Hysteresis .


Normally random thermal motion determined by the Temperature of the material compete with domain alignments. However at a certain critical temperature called the Curie Temperature the alignment of domain vectors are favored over random thermal motions. For IRON this is T= 770 deg Centigrade above which it is Paramagnetic wit no domain formations and below the Curie Temperature it is Ferromagnetic. The transition between the Paramagnetic phase and the Ferromagnetic phase is thermodynamically alike to a liquid-solid phase transition in materials, like water-ice transition. The properties of Magnetic materials are decided by quantum mechanics and is a subject of frontline research in Condensed mater physics.


This ends our discussion on the Electric and Magnetic effects of static charges and steady currents. The second part of the course will deal with dynamic situations where charge densities and currents are functions of time and we will see
that this would unify the Electric and Magnetic aspects into a single framework of Electrodynamics involving time dependent electric and magnetic fields and described by a set of 4 equations
called the Maxwell's equation. It was Maxwell that unified the 
apparently different phenomena into a single framework. You will also see how Electrodynamics is intimately connected to
the Special Theory of Relativity.


Hysteresis link: 
http://www.tpub.com/content/chemical-biological/TM-1-1500-335-23/css/TM-1-1500-335-23_208.htm




http://en.wikipedia.org/wiki/Ferromagnetism 

Week 7 Lex 3: Magnetostatics in a Material Media I

Having completed Magnetostatics in free space ( vaccuum) we now turn to study the Magnetostatics in Magnetic Materials. This is the magnetostatic analog of Electrostatics in Dielectric Material.


Recall that the source of Magnetism are currents. In magnetic materials the magnetism is due to atomic currents. These currents essentially arise due to two causes. (i) The orbital motion of the electron around the nucleus (ii) The intrinsic ( quantum mechanical) spin of the electron. These small atomic current loops are equivalent to tiny atomic dipoles from a macroscopic ( large scale ) point of view. Normally their effects cancel out due to (i) random orientation (ii) thermal motion of the atoms in a material. However in an applied external magnetic field B these magnetic dipoles tend to either align parallel or anti-parallel to the applied field giving rise to weak magnetic effects called (i) paramagnetism (ii) diamagnetism respectively. For permanent magnetic materials there is a FROZEN-IN or PERMANENT magnetization. Apart from this we also have Ferromagnetic Materials about which we will discuss later.




The torque on a magnetic dipole in an uniform field B is given as N=m X B whereas in a non uniform field the dipole experiences a force ∇ ( m.B) where the dipole moment m=I A where A is the vector area of the current loop. ( For flat current loops this is the usual area vector). This torque tends to orient the dipoles along the applied field direction resulting in a Magnetization. Due to the Pauli exclusion principle paired electrons in atoms with up and down spins cancel each others torques. So Paramagnetism is most readily observed in atoms with unpaired electrons. As paired electrons do not contribute to the magnetic effects Paramegnatism is a weak magnetic phenomena. Diamagnetism on the other hand arises due to the orbital motion of the electrons. The electron due to the magnetic forces speeds up/slows down in an external magnetic fields B. This causes a change in the dipole moment associated with the orbital motion anti-parallel to the field. ( Check Griffiths). The weak diamagnetic effects arise from this incremental dipole moment.


Hence at a macroscopic ( large scale) level a magnetic material has magnetic polariaztion and can be described by a magnetic dipole moment density which is called Magnetization M which is the magnetic dipole moment/unit volume. Just as in dielectrics it is now possible to describe the macrsocopic field due to a magnetized material by a distribution of surface and volume bound current densities. However the equations for the magntetic field being different the bound currents are now given as a surface bound current density K_b=M X η and a volume bound current desnity J_b=∇ X M ( Check Grifiths for the proof which is just like that in Dielectrics but now involving the Stokes Theorem.)


The bound current densities arise just as in dielectrics from now the cancellation between adjacent atomic current loops leving only the boundary surface contributions for uniform Magnetization M when every loop carries the same current. For non-uniform M the currents are different for different loops and only a partial cancellation occurs giving rise to also a volume bound current density inside the material apart from a surface bound current density.


The differential form of Ampere's Law can now be written as 
∇ X B= μ₀ J_tot where J_tot= J_free + J_b is the total current density consisting of the free current density if any in the material and the bound current density due to the Magnetization M. ( We are neglecting the surface bound current density for the same reason as in Dielcetrics. For real magnetic materials the Magnetization goes to ZERO rapidly within a small surface thickness and surface bound currents do not develop in real life.) But for our problems we will consider such IDEALIZED surface bound currents. ( DIPA/IPSA)


It is now easy to define the magnetic analog of the electric displacement D as the H field which satisfies ∇ X H= J_free. So the source for the H field is the free current density only and  
H= (B/μ₀ - M). We can easily define the corresponding integral form of the Amperes law for H as ∫_C H.dl=I^enc_free. When symmetry allows us we can calculate H from knowing the free currents ( see Eg 6.2 Griffths ).


Similar warning as in Dielcetrics for the displacement vector D is also in force with H. H is not quite like B as ∇.B=0 but ∇.H=-∇.M. Only if M is constant or uniform is B like H. In particular do not assume that H is ZERO because there are no free currents. The H field satisfies similar boundary conditions with both normal and tangential components discontinuous at a surface current density. This is obvious from a look at the equations for H namely ∇.H=-∇.M and  ∇ X H=-K_free for a surface current density.

Week 7 Lex 2: Magnetostatics -II ( Magnetic Vector Potential)

Just as in Electrostatics where the formulation of a scalar potential V ( scalar field)  made the problem of determination of the field simpler because E=-∇ V in Magnetostatics also a Potential formulation is possible. The electrostatic potential formulation for V was a consequence of the fact that ∇ X E =0, and this is ensured as Curl ( Grad V)=0 identically for any scalar field.  For magnetostatics ∇. B=0 implies that the magnetic field B=∇ X A where A is a vector field and called the Magnetic Vector Potential since Div( Curl A)=0 identically for any vector field ( proved in the vector calculus part of the course).


So ∇ X B=∇ X ( ∇ X A)=∇(∇. A) -∇² A=μ₀ J. Notice however that there is an inherent non-uniqueness of the Vector Potential A as A + ∇χ where χ is a Scalar field gives the same magnetic field B=∇ X A + ∇ X (∇ χ) and the second term is identically ZERO because Curl ( Grad χ)=0 for any scalar field χ. This transformation of the vector potential by the gradient of scalar field is called a Gauge Transformation. ( gauge transformations
are at the heart of modern fundamental physics and electromagnetism is an example of the simplest type of gauge theories which describe the 3 forces weak, strong and the electromagnetic


So there are an infinite number of vector potential A differing from each other by a gauge transformation, all of which gives the same magnetic field upon taking the Curl of A due to the reason mentioned above. We can use this non-uniqueness of A to fix the divergence of the vector potential as ∇.A=0 9 once this is fixed the vector potential is unique). This is called a Gauge condition or a Gauge choice. There can be many such gauge choices but we will only consider the condition Div A=0. For this gauge choice then the first term of the RHS of ∇ X B vanishes and we have the equation ∇²A=-μ₀ J.


Notice that this equation ( IN CARTESIAN CO-ORDINATES ONLY ) are actually 3 Poissons equations for the three cartesian components of the vector potential A current density J. These equations are exactly like the Poisson equation in electrostatics for the scalar potential V except that the source term on the RHS is a cuurent density J instead of a charge density ρ. So it is obvious that these equations will have the same solution as that for the electrostatic scalar potential provided we identify ( replace) ρ/ε₀=μ₀J_x,y,z whichever component is appropriate for the RHS. 

Note that this holds only for a Caretesian coordinate system because in a curvilnear co-ordinate system it is easy to see that the original equation do not break up into 3 Poissons equations. So even if you use curvilinear co-ordinates for solving teh integrals for the vector potential A, the current density J must first be expressed in Cartesian coordinates ( Check the DIPA problems and Grifiths examples.)


So if we know the solution to a corresponding electrostatic problem we can simply read off the solution of the magnetostatic problem with the replacement as mentioned above. The trick is this is to identify the correct electrostatic problem. Check out exaples in Griffiths and the DIPA problems.


Magnetostatic Boundary Conditions : The magnetic field B follows certain boundary conditions just like electrostatic fields. These follow just as in the electrostatic case from the integral form of the equations ∇.B=0 and the Amperes Law ∇ X B=mu₀ J . Just as in electrostatics the integral form of these equations evaluated over a infinitesimal Gaussian pillbox and an Amperian loop straddling the boundary shows that the B field has a discontinuity at a surface current just as the electric field has a discontinuity at a surface charge. However in the case of B the tangential component is discontinuous which is perpendicular to the direction of the surface current. The normal component of B is continuous. As the vector potential is chosen to have ∇.A=0 and the magntic flux through an Amperian loop of vanishing width is zero, the vector potential A is continuous across the boundary for both its normal and tangential components. However the normal derivative of the vector potential inherits the discontinuity of B ( note that B comes from a derivative of A which is Curl A).

Magnetic Multipole Expansion: As the Vector Potential A is given by an integral of the current density ( line , surface or volume) similar to the scalar potential V ( which is given by an intergral of the charge density) it is obvious that the Vector potential A will also have a multipole expansion just like the scalar potential by expanding the denominator in a power series. As in the electrostatic case the multipole expansion of the Vector potential is also a series in powers of 1/r over simple current configurations. However as there is no free magnetic poles ( no magnetic monopoles in clasical electromagnetism) the magnetic multipole expansion starts from the pure magnetic dipole term. From the expansion it is possible to determine the vector potential due to a pure magnetic dipole in terms of the magnetic dipole moment m of a pure dipole where m=I a and a is the area of the corresponding current loop. The pure dipole is when the area of the loop goes to zero and the current to infinity keeping m=I a fixed. It is now possible to find the magnetic vector potential due to a small loop at the origin and has a similar form to that of the electric field of a pure electric dipole. Check Griffiths for the field line picture of a pure and a physical magnetic dipole.

Week 7 Lex 1: Magnetostatics in free space-I

Having studied the force on a static charge due to a static chrage distribution
elsewhere (source) which was the fundamental problem of Electrostatics we now turn to studying the forces due to charges moving steadily on each other. Steadily moving charges constitute steady currents and Magnetostatics studies the forces affecting such currents. Two such current carrying conductors exert a force/unit length on each other whose direction is dependent on the direction of the steady current I. These forces are not electrostatic forces as current carrying conductors are overall neutral.


Just as in electrostatics, the fundamental problem of magnetostatics is to find the force on a test current placed at the field point due to an arbitrary current distribution elsewhere ( source currents). The starting point of magnetostatics is the Lorentz force law which is a fundamental law the force on a charge q moving with a velocity v in a magnetic field B is given as F=q v X B . For an electric field also present the force is as we know to be F=qE and this must be added to the previous expression. Moreover as this force is always perpendicular to the velocity v, magnetic forces do not do any work. It is now simple to integrate this force over the length of the conductor to find out the total force on a conductor carrying a steady current I to be F=I∫ dl X B


This is the case for what we call as a line current. Of course just like in electrostatics we can move surface charges in steady motion constitute a steady surface current density K which is the current flowing/ unit length perpendicular to the direction of flow. The magnetic force in this case is given as F=∫ ( K X B).da where the integral is now taken over a surface S. We can similarly define a volume current density J which is the current/unit area oriented perpendicular to the direction of flow. The magnetic force is now given as F=∫ ( J X B).dτ where the integral now must be taken over a volume τ.


As ∫ J.da= over a CLOSED surface S is the total current through the surface this can be related to the total change of charge density with time inside the volume enclosed by the surface S because total charge is always conserved. Using the divergence theorem we arrive at the continuity equations which states that ∇ .J is equal to the -ve of the rate of change of charge density ρ with time t. For steady currents, the charge density is constant in time as steady flow of charges should not have charge piling up anywhere. So for steady currents Div J=0


The analog of the Coulombs law for Magnetostatics is the Biot Savarts Law for magnetic fields due to steady currents. Please note that the fundamental law of Magnetism is the Lorenz force law and Biot Savart Law is a consequence of the Lorentz Force Law. Just as in coulombs law the magnetic field B due to a steady current is given by the integral of the cross product of the current vector
I=I dl'cap with the unit separation vector from the source point to the field point divided by the square of the separation vector.
The generalization to surface currents K and volume currents J is as usual.
Please check Eg 5.1 to 5.6 in Griffiths


Its easy to see that the magnetic field of an infinite staright line current circulates around the conductor according to the right hand rule in an azimuthal ( plus/minus  φ cap ) direction. The Curl of B can easily be calculated by seeing that the circulation of B over a loop C is ∫ B.dl=μ₀ I_enc where I is the current ENCLOSED by the loop C. This called Amperes law and is the analog of the Gauss Law for Electrostatics. Now using Stokes theorem its easy to see that 
∇ X B =μ₀ J where J is the volume current density. For surface currents J is to be replaced by K.


The divergence of B is ∇. B=0 . This is unlike electrostatics where Div E is equal to the charge density/epsilon zero. There are no free magnetic charges and the magnetic field lines are always CLOSED. This can be easily proved for a volume current density J using similar techniques to those in electrostatics. ( Refer to Griffiths for the derivation where like before they show that a surface contribution from the boundary is ZERO.)


Please read Section 5.34 Griffiths for a comparison of Electrostatics and Magnetostatics. Also look through the Eg. 5.7-5.10 Griffiths.

Week 6 Lex 1: Dielcetrics II

In this lecture we continue with our discussions about Electrostatics in a Dielectric medium. Having obtained the field at a point outside a dielectric material due to the surface and volume charges developing in the dielectric material due to Polarisation,
we now turn to study the internal microscopic field in the dielectric. For the field outside we used the potential for a pure dipole, however polarisation which is due to charge displacements involve physical dipoles. Outside the material since the field point is far away this does not matter as the dipole potential dominates ( atom is neutral so it has zero net charge and hence no monopole moment. The discreteness of the molecular dipoles is also negligible far away and a continuous Polarisation vector field is justified. This is however not the case inside.


The microscopic field inside the dielectric material is very complex and impossible to calculate. But we can focus on macroscopic or average behavior which is the one observable in experiments. The macroscopic or the large scale field is obtained as an average of the microscopic field over a region is large enough to neglect discreteness of molecular dipoles but small enough so that we do not lose all the field variations. This is roughly of the order of 1000 molecular lengths.


For the macroscopic field inside the dielectric, we consider a spherical surface
of radius R around that point. The field has two contributions  
E=E_out + E_in
(i) E_out from all the charges (molecular dipoles) outside S the average field is the field they would produce at the center. This is given by just the dipole potential for the dipole moment density outside integrated over the region OUTSIDE.
(ii) E_in from all the charges ( molecular dipoles) inside S the average field is the field due to a dipole of dipole moment p which is the total dipole moment within the sphere assuming the sphere S to be small enough so that the polarisation vector P is roughly constant.


Now the term left out of the integral in (i) is due to the field at the center of a uniformly polarised sphere. This is calculated in eg 4.2 in Griffiths and it is exactly the same as that of the average field inside S. So this term puts back in the region that was left out of the integral for V due to outside charges.

So the final relation is a potential V due to all dipoles inside the full volume of the dielectric material and hence the integral for V any now be written as over the full volume of the dielectric material. This analysis justifies the use of the V due to dipoles for the macroscopic field both inside and outside the dielectric material.
This is not matter how complicated the internal ( microscopic) field is, for macroscopic fields we can replace it with a smooth distribution of pure dipoles. The argument is independent of the spherical shape S used over here and holds for any general shape.


Gauss Law in Dielectrics: We now turn to the form of the Gauss law in a Dielectrics. As Gauss Law refers to the total enclosed charge only, the differential form for Gauss Law will now be
∇.E=ρ_free + ρ_bound
where ρ is the charge density due to the free charges and the bound charges respectively. Neglecting the surface bound charges, the volume bound chrage density
ρ_b=-∇.P where P is the Polarisation vector. Rearranging terms we get
∇.[ε₀ E +P]=ρ_free
or ∇.D=ρ_free where D=ε₀ E +P is called the Electric Displacement Vector .


The Displacement vector only refers to the free field charge density. We can now write down the integral form of the Gauss Law in Dielectrics by integrating both over some volume τ bounded by a closed surface S and then using the Divergence Theorem to convert the LHS to a surface integral over the closed surface S. So the closed surface integral
∫_S D. da =Q_free^enc
where Q is the total free charge enclosed within the closed surface S. The surface bound charges are ignored in the edge thickness approximation where the polarisation falls rapidly to zero at the edge of the dielectric and hence no surface charge density.


Although the displacement vector D sees only the free charges it is important to notice that it cannot be used as a substitute for the electric field E.
They are very different from each other,
(i) No Coulombs Law for D unlike that of E
(ii) Curl D is not zero unlike Curl E=0. ∇ x D=∇ ; X P.
So no potential formulation for D as its not a conservative field in general. The integral form is ∫ _C D.dl=∫ _C P.dl
by using Stokes theorem on the differential form.


The electrostatic boundary conditions on the Displacement Vector D can now be easily derived exactly in the same fashion as derived for the Electric Field E by using the corresponding integral forms ∫D.da=Q_free^enc and ∫ _C D.dl=∫ _C P.dl. This is worked out in Griffiths and is left as a reading exercise for the students. Please check Ex 4.8 in Griffiths


A certain class of Dielectric materials offers a simplification of these relations.
In these materials P, D are both proportional to the total Electric field E ( applied field + Polarisation field). So for these materials
D =ε E an P =ε₀χ_e E. Its easy to see that
ε=ε₀(1 + χ_e )
The ratio ε/ε₀=(1 + χ_e )=ε_r is called Relative Permittivity or Dielectric Constant.


However please note that even for Linear Dielectrics the Curl D is NOT ZERO unless the entire space is filled with a single homogeneous dielectric. The reason for the decrease of the Electric Field of a point charge in a dielectric medium as compared to a point charge in free space is now clear. The polarisation charges actually screen the point charge so its actual effective magnitude within the dielectric decreases and hence the corresponding electric field is reduced.

Week 5 Lex 3: Electrostatics in Material Media. ( Dielectrics)

Having covered a number of issues for Electrostatics in free space we now move to the study of Electrostatics in a material media like dielectrics/insulators. The basic difference with conductors in this case is that in a dielectric charges have restricted mobility unlike free charges in a conductor. So the charges cannot move freely but they can be displaced a little from their equilibrium position
which makes the atom neutral.


When a dielectric material is placed in an external electric field a separation of the +ve and -ve charges take place so that there is a field inside the dielectric material which opposes the external field such that the net field inside is reduced. Due to the charge separation surface charges appear on the dielectric.


At the microscopic level the centers of the +ve and -ve atomic chrage distributions are displaced by the external applied field ( like the problem of two spherical charge distributions +ρ and -ρ displaced by a small distance d done in a DIPA). The atom is now said to be polarized and develops a small dipole moment p which is proportional to the applied field E such that p=α E and α is called the atomic polarisability. For a molecule the situation is more complex but once more a dipole moment develops but the polarisability α is now direction dependent and is a rank 2 tensor.


Apart from this certain molecules due to their asymmetric structures have a charge distributions with a FROZEN-IN PERMANENT dipole moment p. These are called POLAR molecules. In an uniform external field E these permanent dipoles feel a torque N=p X E which tends to orient the dipole moment vector in the direction of the external field ( for a non uniform field the dipole experiences a force F=(p.∇) E. Note that (p.∇)is a scalar differential operator which operates on each component of the vector field E. This is the operator version of scalar multiplication of a vector.)


So the microscopic picture that emerges for a dielectric from the above consideration is that a dielectric material either develops induced dipoles or has permanent dipoles ( for polar molecules)
both of which tend to orient along the direction of the applied external field. Some of this orientation will even stay after the field is removed especially for Polar molecules. So in a dielectric material
we have a large number of dipoles oriented in the same direction. This results in a dipole moment per unit volume which is called the Polarization vector P .


Now it is possible to calculate the field due to such a Polarised material by finding the Potential due to a small volume dτ' which has a dipole moment Pdτ' and integrating over the entire volume. Manipulation of this volume integral using product rules and the Divergence theorem show that the potential at some point due to a Polarized material is due to two contributions.

(i) A surface BOUND chrage density σ_b=P. n cap
(ii) A volume BOUND charge density ρ_b=-∇. P


This mathematical result is confirmed by a physical analysis of the microscopic dipoles canceling each other in the volume resulting in a surface bound charge density and a volume bound charge density when the cancellations are not complete in the volume due to unequal dipoles ( non constant dipole moment /unit volume which is non uniform Polarisation). The expression match exactly confirming the mathematical analysis using Vector Calculus.

Week 5 Lex 2: The Method of Images.

The method of images for electrostatic problems is a way to obtain the potential V for a problem by identifying it with the known ( simpler to calculate) potential V of a different electrostatic problem but satisfying the same boundary condition as the original V in the region of interest. Uniqueness theorem then ensures that the two potentials are identical in the region of interest.


Classic Image Problem I:


A point Charge q placed at a distnace d in front of an infinite conducting plane which is grounded, to find the potential at a point above the plane.


The trick to solve this problem is to identify that one of the equipotential at the mid point between two equal and opposite charge +q and -q is a plane and its at V=0. This is identical to the case of a point charge in front of a conducting plane at ZERO potential because the potential satisfies the same boundary conditions V=0 at z=0 taking the z-axis to be perpendicular to the plane.


Thus the image problem is that of two charges +q and -q situated at equal distance from the origin in the +ve and -ve direction of the z-axis. Now the potential due to this arrangement is very simple to calculate by superposition of the potential due to each of the two charges the image charge and the point charge.


Note that the solution is only valid above the conducting plane. Below the plane the situation is completely different but that is excluded from our region of interest. Uniqueness theorem ensures that in the region of interest this is the only UNIQUE solution that satisfies the given boundary conditions. Hence the solution for the potential


Note also that the image charge cannot be placed in the region of interest because that would completely change the original problem we are seeking a solution to. Its important to understand that the method of images is a method to find an alternate simple problem whose potential is identical with that of the original problem due to both satisfying the same boundary conditions in the region of interest.


The solution can be easily extended to that of the isolated  ( ungrounded ) conducting plane.


Classic Image Problem II:


A point charge q is placed outside at the distance a from the center of a conducting sphere of radius Rwhich is grounded, to find the potential everywhere outside. The trick for this is to identify that one of the equipotentials for a system of two unequal charges q and q' is a sphere. So the only thing now is to set the boundary conditions.
Since the original sphere is grounded ( at ZERO potential) in the new problem we will have to adjust q' so that the spherical surface is at V=0. Now its esay to calculate V ( IPSA Problem) for the two unequal charges and see that the image charge must be set at x=R²/a from the center and its magnitude should be q'=-R/a x q to make the potential V(R)=0.


The problem with the grounded sphere may easily be extended to cover an isolated ( ungrounded) sphere. The sphere would be an equipotential now at V=V₀


It is now possible to use these two image problems and superposition of potentials to solve more complex problems which I have mentioned in the lecture and some of which will be discussed in the future tutorials.

Week 5 Lex 1: Uniqueness Theorem and Electrostatic Boundary Conditions

The contour map of the solution to Laplace's equation inside the disk. (Courtesy http://oak.ucc.nau.edu/jws8/classes/461.2009.7/animations.html and Google).

Having seen the multipole expansion as a method to obtain the approximate potential

far away from a localized charge distribution we now look for other methods to obtian the potential exactly. The most direct method is to solve the Laplace or Poisson equation ∇² V=0 or ∇² V= -ρ/ε₀ as the case may be to obtain the poetntial V. However its not easy to solve these equations as they are partial differential equations ( PDE) and also they have infinite number of solutions. The solution specific to a problem is obtianed by using boundary conditions ( values of potential at boundaries, which are determined from the physics of the problem. The Uniqueness theoerm pertains to such boundary conditions and the uniqueness of a solution which satisfies a specific set of boundary conditions.

Solutions to the Laplace equations are called harmonic functions and they have certain special properties.

(i) The value of the potential V(r) at a point r is given by the average of the poetntial over a sphere S of radius R with the pint r as its center.

(ii) Follows from (i) There cannot be any local maxima or minima of V, all extremum values of V occur at the boundaries. ( This is obvious because if V had a maxima

its value would be higher than on a sphere of radius R and hence cannot be the average V over that sphere. Average value can neither be larger nor smaller than the

largest and smallest values in the sample over which the average is being taken. It must lie somewhere in between.)

Proof: Take a point charge q outside a spherical surface of radius R. Its easy to show that the average potential over the surface S is equal to the potential produced

at the center. By superposition this can be extended for any arbitrary discrete or continuous charge distributions outside S, For all such configurations the average potential over the sphere is equal to the net potential produced at the center of S

First Uniqueness Theorem: The solution to the Laplace equation in a volume τ is uniquely determined if the potential V is specified at all boundary surfaces

S.

Proof: Assume two solutions V₁ and V₂ which staisfy the Laplace equation in τ and which have the same value V_S at S. Its easy to see that V₃=V₂ -V₁ also satisfies the Laplace equation and its value is ZERO on S. Since for Laplace equation all extrema must occur at the boundary so the value ZERO is both maxima and minima for V₃.

Which implies that V₃=0 everywhere. So V₂-V₁=0 so

V₂=V₁. So V is unique. this can be easily extended to regions with a non zero charge distribution ρ also by using the Poisson equation.

Corollary: The potential in a volume τ is UNIQUELY determined if the charge density is specified throughout and the potential V is specified on all boundaries

S.

The second Uniqueness Theorem is for conductors and is stated as ; In a  volume τ surrounded by conductors and having a specified chrage density ρ the electric field E is UNIQUELY determined if the total charge Q on each conductor is given. ( Proof: Read from Griffiths).

Electrostatic boundary conditions: The normal component of the electric field is discontinuous across any boundary with a charge distribution σ by an amount σ / ε₀.

The tangential component of the electric field is always continuous across any boundary.

The potential V is always continuous across any boundary but its derivatives are not. In particular the normal derivative ∂ V/∂ n cap =∇ V. $n$ cap is discontinuous by -(σ/ε₀)

A Guide to Preparations.

Dear Students,

I am getting a lot of queries about Mid Sem I. So here is a guide to
preparations. This is only a guide.

1. Dont panic or be stressed. It is never helpful. If you need help
please call the exam helplines ( this will be sent by DOAA) or talk to
your student counsellor.

2. Study the class notes with Griffiths text and clarify all the concepts.
Work out fully all the DIPA and IPSA Problems. Mid Sems will be of the
level of DIPA problems and some short questions both problems and
concepts. There will be no long theoretical questions like derivations.

3. You can try other problems of same level in Griffiths but very
complicated and hard problems are not necessary to be done. Griffiths has
many problems and you should refer to only what I have taught. Everything
in Griffiths is not in the syllabus.

4. I have already announced in class that long expressions for operators
in curvilinear coordinates will be supplied. You dont have to memorize them.
You should know how to use and manipulate them as in the DIPA and IPSA
problems. ( The general derivation is in Appendix A of Griffiths, but that
is only for your interest. Its not included in the syllabus.)

5. Complicated problems with delta function and its properties will not be
asked. Only the problems of the same level as in IPSA and DIPA.

6. It will be useful ( time saving) to remember some standard
expressions/formula for potentials and fields of standard configurations.
Anything that has been derived/worked out in the class or DIPA or IPSA can
be used directly.

6. IMPORTANT: Please take some time to read and understand the question.
Many times people misread the question and then their answers are
completely incorrect. You will have to answer only what has been asked.


I hope this will help.

Best of luck

GS

DIPA SET 4 PDF FILE IN YOUR E-MAIL BOXES.

Please note that DIPA Set 4 PDF file is in your e-mail.

Week 4 Lex 3: Multipole Expansion II

The multipole expansion arises from the idea that the potential due to a point charge q at a distance r goes as 1/r. If we calculate the approximate potential at a large distance from a physical dipole which is a pair of equal and opposite charges +q and -q at a distance d  where the vector d is from the -ve to the +ve charge is given by a binomial expansion to first order ( higher orders neglected) as 
V=(1/4πε₀)[ q d cos θ/r²] and goes as 1/r² to a first order.


This tells us that it may be possible to obtain the approximate potential at a large distance from an arbitrary charge configuration in terms of a series of contributions in powers of (1/r). This is somewhat like expressing a complex waveform in terms of some simple waves of different frequencies ( fundamentals and harmonics), called a Fourier expansion.. What allows this is the Superposition Principle.


When we expand the denominator of the the expression for the potential at a faraway point r due to an arbitrary volume charge distribution ρ(r') given as
V=(1/4πε₀)∫_τ ρ(r')dτ'/r
( note that r is the Griffiths script/curly r. )
in a binomial expansion, we obtain a series of contributions in powers of (1/r) which is the distance between the source point r' and the field point r. The coefficient of each power of (1/r) is an integral that provides the respective moments like the monopole moment ( total charge Q=∫ ρ(r')dτ') then a dipole moment p= ∫_τ r'ρ(r')dτ' and
the contribution to the potential is given as V_dip=(1/4πε₀)p. r cap /r² please note that r cap is a unit vector along r. Note that θ or θ' is simply the angle between the position vectors r and r'. It is not the polar angle unless z axis is chosen along the r' or r direction. Note that this is the dipole moment of a PURE DIPOLE


*For a point charge at origin the monopole term is the only non zero term and the potential due to a point charge is EXACT all higher multipoles vanishing. ( Check it out)


*The dipole moment expression works for a line charge λ, surface charge σ distribution replacing the volume distribution ρ and integrating over
a surface or a line.


For a discrete point charge distribution of q_i at r'_i the pure dipole moment is p=∑_i=1ⁿ q_i r'_i. A physical dipole moment is a special case of this for just two charges +q and -q and is given as p=qd where the vector d is the vector from - to the + charge. This is consistent as it gives the correct potential where
for a physical dipole. But note that this potential is approximate because for a physical dipole of two unlike but equal charges the monopole term is zero as total charge Q=0 but higher multipole terms are not zero. They are smaller than the dipole contribution and hence neglected in the approximation that the field point r is far away from the source.


It should be obvious now that for a single point charge away from the origin ( that is r' not equal to 0) will have all possible moments starting from the monopole moment ( total charge Q=q in this case) and all possible such terms in the expression for the potential V at a field point r.


* The physical dipole reduces to a pure dipole in the limit q
goes to infinity and d goes to 0 with qd=p held fixed. So a physical dipole are finitely separated charges and a pure dipole is more like a point. The pure dipole is the second moment in the multipole expansion. For large distances the field
of a pure dipole and a physical dipole are identical but are very different close to the source. ( See the two figures given in Griffiths).


The multipole expansion is dependent obviously on the coordinate system. The example of a point charge q away from the origin is an example of this where all higher multipoles also contribute although the dominant term is the monopole moment.


For a shift in origin the monopole moment Q does not change. The (pure) dipole moment changes due to the shift except when the total charge Q=0 ( algebraic sum). In that case ( for total charge Q=0) the (pure) dipole moment is unchanged. See the example for a shift of origin by a vector a  in Griffiths at the end of the section which was discussed in class.

For quick reference

http://cr4.globalspec.com/blogentry/2842/Multipole-Expansion

Week 4 Lex 2: Conductors and Multipole Expansion.

Conductors: In this lecture we first examine the issue of charged and uncharged conductors in an electric field. A perfect ideal conductor is characterized by an unlimited supply of free electrons. When a conductor is placed in an external electric field E ₀ the electrons which are very loosely bound to the atom flow in the direction opposite to that of the external electric field. This causes separation of the positive and negative charges within the conductor and this sets up an induced Electric field E_ind inside the conductor in opposition to that of the external field. The charges flow till the induced field completely cancels the external applied field inside the conductor. Hence the field inside a conductor is ZERO and all charges appear on the surface of the conductor. ( inside the conductor the charge density is ZERO. This is because if any chrage density builds up inside the conductor, the resultant electric field with drive the charges till the density will is zero.)


Since the field is zero and the field is derivative of the potential E=-∇V so the conductor must be an equipotential surface with a constant potential that extends into the conductor. So on the surface of the conductor and inside the potential is constant. This also shows that the electric field has a discontinuity ( increases to a finite value from zero discontinuously) just away from the surface of the conductor. This always happens whenever the electric field encounters a surface charge density.


Now since dV=∇V.dl=0 on the surface as V is constant on the surface, this shows that the electric field just away from the surface of a conductor is normal to the surface ( -∇V is perpendicular to dl on the surface to make dV=0 on the surface which is an
equipotential).


So whenever a charge q is brought near a conductor the electric field draws the electrons to the side closer to the charge leaving the positive charges piled up on the side which is further away. Hence an induced charge distribution occurs on the surface of the conductor due to the presence of the charge q. As the conductor was originally neutral and since the field inside the conductor is zero, application of Gauss theorem shows that the induced surface charge is always equal to -q.


The fact that a conducting surface is an equipotential is used in a Capacitor which is an arrangement of two or more conductors at different potentials with charges +Q and -Q. The charge Q=CV where C is the capacitance of the arrangement. Note that the
capacitance is a geometrical quantity and depends on the shape size and arrangement of the conductors. The Capacitor when charged by a battery has electrons removed from the +ve plate to the -ve plate and this process continues till the electric field due to the plate charges exactly balances the electric field due to the battery. The work done to move the electrons is stored in the capacitor electrostatic (potential) energy. This energy will appear when the conductors are shorted through sparking. The energy expression may be calculated and W=1/2CV²


Multipole Expansion: In what we have seen over the last 3 weeks it seems that the fundamental problem of electrostatics which was to find the electric field due to a certain charge
distribution/configuration is far easier to obtain provided we know the potential due to this charge distribution at the field point since E=-∇ V and differentiation is easier than integration. However its not easy to find the potential due to a charge distribution because that also involves solving an integral although the integrand in this case is a scalar. Although possible in simple cases its not easy to do this integral for an arbitrary chrage distribution.


The multipole expansion is a way around this difficulty to obtain at least the approximate potential at a field point far way from the charge distribution. We will see that the approximate potential is a very good estimate and becomes better as more and more terms in the expansion ( series) are considered.


A single point charge is called a monopole and we know the potential due to this. A simple binomial expansion provides us with the approximate potential due to a charge pair called a physical dipole. This provides us with the idea of expanding the potential due an arbitrary localized charge distribution at a field point far away from the charge configuration. A binomial expansion an the approximation that the point at which the potential is required is far away provides us with a systematic expansion of the potential V(r) in terms of a series in powers of (1/r) and specific basic charge configurations like a monopole, dipole, quadrupole, octupole etc. The approximation becomes better and better as more terms in the series are included. Note that each successive term contributes less and less to the potential as powers of (1/r). Schematically the series may be written as
V(r)=1/4πε₀ [ K₀/r + K₁/r ² + K₂/r ³ +...........]
Where the terms denote the monopole, dipole and quadrupole contributions respectively. Note that the dipole term in the monopole expansion is different from the potential due to a physical dipole.
We will have more to say about this in the next lecture.

Week 4 Lex 1: Work and Energy in Electrostatics.

The picture is that of a Plasma Lamp that works using Electrostatic
charges.

After having looked at the Laplace and Poisson equations satisfied by the electrostatic potential V (note that V is scalar field) in this week we start with issues of work and energy in electrostatics.

Since the Curl E of the electric field is zero the field E =-∇ V. This means that the electrotstatic filed is a conservative field and the work done ( which is the line integral of the force F =-QE, -ve as work is done by an external agent against the electric field) to move a test charge Q from point a to point bin the electrostatic field of a fixed charge distribution is path independent and depends only the value of the potential V at the end points W=Q [ V(b)- V(a)]. If the initial point a is taken at infinity where the reference of the potential is taken such that V=0 at infinity and the second point is taken as vector r then the work done to bring in charge Q from infinity to a point vector r, is W=Q V
where V is the potential at the point r.


Using this we can now find out the work done to assemble/create a static discrete charge distribution of N charges by bringing in charges from infinity one by one and then summing over the total work done to bring in each charge one by one from infinity to their respective positions. Care must be taken to avoid double counting by restricting the summation. Note that the work done is stored in the configuration as electrostatic ( potential) energy.


It is now straightforward to generalize this work expression to continuous charge distributions. For a volume distribution ( not necessarily uniform) ρ the work expression reduces to a volume integral over the volume τ over which the charge density ρ is non zero, W= 1/2 ∫ _τ ρ d τ. The volume distribution may be replaced by σ or λ for surface or line charge distributions respectively.


Now since ∇E=ρ/ε₀ the integral may be reduced to W= 1/2 ∫ _τ (∇.E) V d τ. We many now use the result of the product formula ∇ . (V E )=V ∇.E + E.∇V and the divergence theorem to reduce the integral to a volume integral over a volume τ containing the chrage distribution + a surface integral over the closed surface S which is the boundary of the volume τ as W= ε/2 ∫_τ E² d τ + ε/2 ∫_S V E.da. 

Now we can enlarge the original volume τ to include all of space without changing the value of the original volume integral over ρ V because the only non zero contribution will come from where ρ is non zero. So on the RHS also we have to take the two integrals over all space. Note that the volume integral has contributions only from the interior where as the surface integral is over the surface at the boundary of all space which is at infinity. Now the integrand of the surface integral falls off as 1/r so at infinity it is zero. Hence the surface integral is zero but the volume integral which still has contributions from all interior points ( with non zero ρ) is not zero. So finally we have that the energy of a static continuous charge distribution is W=ε/2∫ _{all space} E ²d τ.
NOTE THAT SINCE THE VOLUME INTREGAL GIVES THE TOTAL ENERGY IN THE CONFIGURATION
OVER ALL SPACE THE INTEGRAND ε₀/2 E² IS THE FIELD ENERGY DENSITY.




This energy ( which is the work done to assemble the charge distribution) is contained in the charge distribution. This energy will be obtained if the charge distribution is disassembled. Note that the location of this energy is ambiguous. One may take it as being contained in the charges or in the field. For electrostatics both interpretations are correct. But when non static fields are involved the energy is clearly contained in the electric field E.


Note that the energy contained in a point charge when calculated with this formula gives infinity as the result of the integral as shown in Griffiths. This is our old friend and occurs because Classical Electromagnetism is not valid at short length scales namely r going to zero limit which is the point charge. So here we dont consider the energy contained in a point charge, the point charge is pre fabricated and given to us in electrostatics. Point charges come ready made in Classical Electromagnetism.


Quick reference links

http://farside.ph.utexas.edu/teaching/em/lectures/node56.html 
http://en.wikipedia.org/wiki/Electric_potential_energy

Week 3 Lex 3: Laplace and Poisson Equation.

In this lecture we first saw the application of Gauss Theorem to determine the Electric field E for charge distributions with certain symmetries. Note that to solve the flux integral to determine E one needs to know the direction of E to take the dot product with the area vector element. This is only possible for very simple situations or when the symmetry in the problem allows us to determine the direction of E uniquely.


Having established the divergence ∇.E=ρ/ε₀ we then turned to the Curl of E. For the electric field E due to a point charge q the line integral of E is seen to be path independent from an explicit calculation and hence the circulation over a closed curve C is zero. From Stokes theorem the surface integral of ∇ × E. da over the open surface S whose boundary is the circulation curve C will also be zero as the LHS closed line integral is zero. The surface integral being zero for arbitrary surface S with boundary C implies that Curl E itself is identically zero for an Electrostatic field ∇ x E=0.


The fact that Curl of the electric field vector E is zero implies that the vector field E=-∇ V, where V is scalar field called the electrostatic potential ( scalar potential). Using the boundary theorem for gradients it can be the seen that V=-∫_C E. dl from a reference point O to the field point. The reference point is taken to be at infinity where V is zero. ( Note that this is true only for localized charge distributions restricted to a finite region. For charge
distributions which are infinite the point of infinity is not a correct choice for reference as it would make the integral diverge.)


Using this definition its easy to directly calculate the potential at a field point due to a point charge. It can be seen that the superposition principle also holds for the scalar potential V. This principle can then be used to generalize th result of the point charge to discrete and continuous charge distributions in terms of summations and integrals respectively. This is just like the integral for the electric field except the fact that the integrand is a scalar and not a vector unlike that for the Electric field. This integral obviously easier to solve to find the potential V from which the electric field may be determined by taking the gradient of the scalar potential V.


However even the integral for the potential is not easy to solve for all cases. Using the differential form of Gauss Law ∇. E=ρ / ε₀
and the fact that E=-∇ V provides us with a partial differential equation for V as ∇² V=-ρ/ ε₀. This is the Poisson equation and can be solved for V given the charge distribution ρ. In most cases we are required to find the potential in a region which is outside the charge distrbution and for these regions we have the Laplace equation ∇² V=0. The main problem of electrostatics has been now reduced to finding solutions of these two partial differential equations whichever is applicable. The Laplace equation is one of the most important equations of classical physics and engineering and the same equation is applicable in many different areas of Physics and Engineering.


The links below are for quick reference but again includes a lot more advanced issues than we need in this course.
http://en.wikipedia.org/wiki/Laplace%27s_equation

Week 3 Lex 2 Coulombs Law and application of Gauss Theorem.

In this lecture we started Electrostatics. The basic question of electrostatics is to find the force on a test charge at some (field) due to a source charge or a system of source charges fixed at some (source) point or points.


Superposition principle for the force simplifies ths to finding the force on the test charge due to each of the source charges and doing a vector sum for the total force.


The force between two charges is given by the fundamental law of electrostatics which is the Coulombs Law. We can then define the Electric Field as a vector field E whose magnitude is the force on a unit test charge at that point. Field then maybe expressed as a vector function of the co-ordinates in a region such that any test charge feels a force when placed in that region.


The electric field due to a point charge can easily be calculated from coulombs law. The expression for the field due to a point charge can then be generalized for a system of discrete charges through superposition by summing over the individual electric fields due to each charge and finally to a continuous distribution through an integral over the charge distribution. Note that there can be 3 types of charge distributions; line , surface and volume characterized by the respective densities which are in general functions of the source co-ordinates eg. ρ(r prime) Most often in problems we will deal with constant or uniform charge distributions although in certain cases we may use a charge distribution which is not uniform.


This reduces the determination of the electric field ( force on unit test charge) at some point to solving an integral with a vector integrand. This is not simple to perform in most cases except a few. So we need to seek alternate ways by studying the properties of the electric field.


It is easily seen in the field line picture that the FLUX of the electric field over a closed surface S is a measure of the charge enclosed within this surface. For a single point charge the flux is given by (q/epsilon zero). This result is easily generalized to a discrete system or a continuous distribution of charge by the superposition principle to give the flux of E as
∫ _S E.n da=Q/ε ₀ where Q is the total charge enclosed by S. Using the divergence theorem the surface maybe related to a volume integral over a volume V which is enclosed by the closed surface S of the div E. Comparison of two sides after writing the total charge Q also as a volume integral of a volume charge density ρ over V now provide the relation that'
∇ .E= ρ / ε ₀
These both are the Integral and Differential form of Gauss Law respectively.


Note that Gauss Law is just another expression for Coulombs Law which is the fundamental law. The Gauss law is a consequence of the Coulombs Law.