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- ELECTROMAGNETIC THEORY AND TRANSMISSION LINE - ETTL
- Note
- Electronics and Communication Engineering
- B.Tech
- 4 Topics
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Therefore ....................... (1) As shown in the Figure 1 let the position vectors of the point charges Q1and Q2 are given by and . Let represent the force on Q1 due to charge Q2. Fig 1: Coulomb's Law . We define the unit vectors as The charges are separated by a distance of ..................................(2) and . can be defined as Similarly the force on Q1 due to charge Q2 can be calculated and if represents this force then we can write When we have a number of point charges, to determine the force on a particular charge due to all other charges, we apply principle of superposition. If we have N number of charges Q1,Q2,.........QN located respectively at the points represented by the position vectors , ,...... , the force experienced by a charge Q located at is given by,

.................................(3) Electric Field : The electric field intensity or the electric field strength at a point is defined as the force per unit charge. That is or, .......................................(4) The electric field intensity E at a point r (observation point) due a point charge Q located at (source point) is given by: ..........................................(5) For a collection of N point charges Q1 ,Q2 ,.........QN located at field intensity at point , ,...... , the electric is obtained as ........................................(6) The expression (6) can be modified suitably to compute the electric filed due to a continuous distribution of charges. In figure 2 we consider a continuous volume distribution of charge (t) in the region denoted as the source region. For an elementary charge , i.e. considering this charge as point charge, we can write the field expression as: .............(7)

Fig 2: Continuous Volume Distribution of Charge When this expression is integrated over the source region, we get the electric field at the point P due to this distribution of charges. Thus the expression for the electric field at P can be written as: ..........................................(8) Similar technique can be adopted when the charge distribution is in the form of a line charge density or a surface charge density. ........................................(9) ........................................(10) Electric flux density: As stated earlier electric field intensity or simply ‘Electric field' gives the strength of the field at a particular point. The electric field depends on the material media in which the field is being considered. The flux density vector is defined to be independent of the material media (as we'll see that it relates to the charge that is producing it).For a linear isotropic medium under consideration; the flux density vector is defined as: ................................................(11) We define the electric flux as

.....................................(12) Gauss's Law: Gauss's law is one of the fundamental laws of electromagnetism and it states that the total electric flux through a closed surface is equal to the total charge enclosed by the surface. Fig 3: Gauss's Law Let us consider a point charge Q located in an isotropic homogeneous medium of dielectric constant . The flux density at a distance r on a surface enclosing the charge is given by ...............................................(13) If we consider an elementary area ds, the amount of flux passing through the elementary area is given by .....................................(14) But , is the elementary solid angle subtended by the area at the location of Q. Therefore we can write For a closed surface enclosing the charge, we can write which can seen to be same as what we have stated in the definition of Gauss's Law.

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