ElectroMagnetic Theory Concepts & Solved Numericals

- A conductor (such as Copper, Aluminum, gold) contains electrons that are not bounded to the atom and are free to move within the material. - When an isolated conductor is placed in an external electric field E e , the negative charges ( electrons ) starts moving in the opposite direction to that of electric field. - Soon the electrons get accumulated on the surface of one side of the conductor , while the surface on the other side gets depleted of electrons and hence acquire a positive charge (+ve). - These separated negative (-ve) and positive charges (+ve) on the opposite sides of the conductor produces an internal induced electric field E i which opposes the external field E e inside the conductor. - If the conductor is a perfect conductor or a good conductor the internal electric field E i will cancel the external electric field E e . Hence we can therefore say that “A perfect conductor cannot contain an electrostatic field inside it.” - There

CONVECTION CURRENT DENSITY: - Convection current occurs in insulators or dielectrics such as liquid, vacuum and rarified gas. - Convection current results from motion of electrons or ions in an insulating medium. - Since convection current doesn’t involve conductors, hence it does not satisfy ohm’s law. - Consider a filament where there is a flow of charge ρ v at a velocity u = u y a y . - Hence the current is given as: Where u y is the velocity of the moving electron or ion and ρ v is the free volume charge density. - Hence the convection current density in general is given as: J = ρ v u CONDUCTION CURRENT DENSITY: - Conduction current occurs in conductors where there are a large number of free electrons. - Conduction current occurs due to the drift motion of electrons (charge carriers). - Conduction current obeys ohm’s law. - When an external electric field is applied to a metallic conductor, conduction current occurs due to

- Electric field can not only exist in free space and vacuum but also in any material medium. - When an electric field is applied to the material, the material will modify the electric field either by strengthening it or weakening it, depending on what kind of material it is. - Materials are classified into 3 groups based on conductivity / electrical property:           - Conductors (Metals like Copper, Aluminum, etc.) have high conductivity (σ >> 1).           - Insulators / Dielectric (Vacuum, Glass, Rubber, etc.) have low conductivity (σ << 1).           - Semiconductors (Silicon, Germanium, etc.) have intermediate conductivity . - Conductivity (σ) is a measure of the ability of the material to conduct electricity . It is the reciprocal of resistivity (ρ). - Units of conductivity are Siemens/meter and mho. - The basic difference between a conductor and an insulator lies in the amount of free electrons available for conduction of current.

Q.1 Point charges Q 1 = 1 nC, Q 2 = -2 nC and Q 3 = 3 nC and are positioned one at a time and in that order at (0, 0, 0), (1, 0, 0) and (0, 0, -1) respectively. Calculate the energy in the system after each charge is positioned? Ans: Initially the system is assumed to be charge free. The energy required to bring Q 1 into the system is 0 J. After Q 1 : Energy in the system = 0 J The energy required to bring Q 2 into the system is Hence, After Q 2 : Energy in the system = -18 nJ The energy required to bring Q 3 into the system is: W 3 = Q 3 ( V 31 + V 32 ) + Q 2 V 21 Q2. Determine the work necessary to transfer charges Q 1 = 1 mC and Q 2 = -2 mC from infinity to points (-2, 6, 1) and (3, -4, 0) respectively. Ans: No work is done in transferring the first charge Q 1 . However work done to transfer the point charge Q 2 is given as: Q3. A point charge Q is placed at the origin. Calculate the energy stored in region r > a? Ans:

- In case, if we wish to assemble a number of charges in an empty system, work is required to do so. - Also electrostatic energy is said to be stored in such a collection. - Let us build up a system in which we position three point charges Q 1 , Q 2 and Q 3 at position r 1 , r 2 and r 3 respectively in an initially empty system. - Consider a point charge Q 1 transferred from infinity to position r 1 in the system . It takes no work to bring the first charge from infinity since there is no electric field to fight against (as the system is empty i.e. charge free). Hence, W 1 = 0 J - Now bring in another point charge Q 2 from infinity to position r 2 in the system. In this case we have to do work against the electric field generated by the first charge Q 1 . Hence, W 2 = Q 2 V 21 where V 21 is the electrostatic potential at point r 2 due to Q 1 . - Work done W 2 is also given as: - Now bring in another point charge Q 3 from infinity t

Q.1 Determine the electric field due to the following potential: a) V = x 2 + 2y 2 + 4z 2 Ans: = - (2x a x + 4y a y + 8z a z ) = -2x a x - 4y a y - 8z a z V/m b) V = ρ 2 (z + 1) sinφ = - [ 2ρ (z + 1)sinφ a ρ + ρ(z + 1) cosφ a φ + ρ 2 sinφ a z ] = - 2ρ (z + 1)sinφ a ρ - ρ(z + 1) cosφ a φ - ρ 2 sinφ a z Q.2 A point charge of 5 nC is located at the origin. If V = 2v at (0, 6, -8), find a) The potential at A (-3, 2, 6) b) The potential at B (1, 5, 7) c) The potential difference V AB Ans: If V (0, 6, -8) = 2 V In this case x = 0; y = 6; z = -8 Using the scalar relationship between Cartesian and spherical system , we have r = (x 2 + y 2 +z 2 ) 1/2 = (100) 1/2 =10 Hence Electric potential at point r due to a point charge Q located at origin → C = - 2.5 a) Electric potential at point r due to a point charge Q located at a point (-3, 2, 6) is given as: = 3. 929 V b) Electric potential at point r due t