 This session will see the boundary conditions part 2, myself Piyusha Shedgarh, Electronics and Telecommunication Department, Valtran Institute of Technology, Solapur. These are the learning outcomes for today's session. At the end of this session, students will be able to derive the boundary condition equations for the different fields at two medias, may be the medias are different or the same. They will be able to apply the principles of electrostatics to the solution of problems relating to boundary conditions. These are the contents. So, before going to start today's lecture, you can pause video here for a second and recall that what is electric flux density. So, if you are considering the boundary between any two media, there is you have to find the different field quantities related to the electric field. So, different field quantities are electric field intensity, electric flux density, potential etcetera. So, generally we are finding the equation for the electric flux density as well as the electric field intensity. Therefore, you must know what is the electric flux density. Yes, it is already covered that the electric flux density is the charge per unit area. So, this is given by the equation d bar equal to q upon 4 pi r square. According to the Gauss law, psi is equal to q and therefore, this can be equated to psi is equal to q upon 4 pi r square. So, this is the equation for the electric flux density. So, before going to start that exactly the boundary conditions, some of the definitions are required to define the different parameters. So, first is the permittivity. Permittivity is denoted with the epsilon symbol. So, it is a major of the resistance that is encountered when forming an electric field in a medium. In other words, the permittivity is major of how an electric field affects and it is affected by a dielectric medium. Permittivity is measured in Faraday's per meter and epsilon r is the relative permittivity of the material. So, epsilon is defined by epsilon naught, epsilon r whereas, the epsilon naught is the permittivity for free space and epsilon r is the relative permittivity of the material. Second parameter is dielectric constant. It gives a major of the polarizability of the material relative to free space and is defined as the ratio between the permittivity of the medium to the permittivity of free space. And what is electric flux density is defined as charge per unit area. So, let us define the boundary conditions to solve the problems involving the boundary surface between the different media. We should know the conditions satisfied by this field components at this boundary and therefore, these are known as the boundary conditions. So, when you are considering the boundary what will be the conditions are for different fields? This can be defined with this boundary conditions. Usually the field components specified above in terms of the tangential as well as the normal components of any field vector. When the field exists in a medium consists of the two different media, the conditions of the field must satisfy this boundary so are called as the boundary conditions. For the electrostatic field the following boundary conditions are important. So, you can consider the boundary between the two media that is dielectric, dielectric media. So, here the two medias are same that is media are same that is dielectric and dielectric whereas, the different media that is conductor and dielectric and again the third may be considered for the conductor and the free space. Now, here we are checking the boundary conditions for the dielectric medium. So, for dielectric medium the what you should know that what is the conductor and dielectric medium or any other medium. So, for that define the conductor so you are familiar with the conductor definition in conductor the valence electrons are present valence electrons that is the conduction electrons or the free electrons the electric field intensity within the conductor is 0. So, what is a dielectric medium? In dielectric medium there are no any free charges and therefore, conduction current is not flowing through this dielectric medium. Boundary conditions can be defined into components boundary conditions can be defined in terms of the E that is electric field intensity and in terms of the D that is electric flux density. So, again the electric field intensity is defined in terms of the tangential component and the normal component whereas, for the D also D bar also you can consider the tangential component and the normal components. So, this is the general figure for the boundary between the two mediums. So, here the medium 1 is considered and the medium 2 medium 1 having suppose the mu 1 epsilon 1 and the sigma 1 that is permeability, permittivity and the conductivity for the medium 1 whereas, for the medium 2 mu 2 epsilon 2 and the sigma 2 parameters are considered and this is nothing but the boundary is along the x y plane whereas, this is nothing but the z axis. Now, when any field is passing from this medium 1 towards this medium 2 suppose this vector is b 1 bar which is passing through this medium 1 to medium 2. So, b 1 bar is one of the vector used in the medium 1 whereas, the b 2 bar is one of the vector or any field in medium 2. So, alpha 1 is making an angle with the z axis for in this medium 1 whereas, the alpha 2 is the angle made by this b 2 field with z axis. So, this is considered for the dielectric-dielectric medium. So, when the dielectric medium is used for the medium 1 also and medium 2 is also the dielectric medium. So, medium 1 having the epsilon 1 whereas, the medium 2 having the epsilon 2. So, in that case suppose this is the box is considered which having the length along the z axis as a delta x. This length horizontal length is the delta y whereas, the height for this is considered as the delta z. So, by applying the Gauss law d bar dot e s bar can be written as a charge enclosed by that surface. So, this surface is considered with the different lengths along the x and y and z axis as delta x delta y and delta z. So, in previous slides we have covered the tangential components. So, here in this lecture we will see the normal components, how to calculate the normal component for the given fields for the given boundary conditions. So, normal component for that area of the plane has the surface charge density is given by rho s. Rho s is nothing but the surface charge density. Area of the plane enclosed by this box delta x and delta y so, area for this is given by delta x into delta y. So, according to the Gauss law charge enclosed is nothing but the d bar dot d s bar. Therefore, this is charge enclosed is given by surface charge density multiplied with the area that is rho s into delta x delta y whereas, the delta x and delta y are the differential lengths are considered along with the x and y axis. Now, integration d bar dot d s bar is equal to rho s delta x delta y. Now, this is the box is considered and therefore, for this you can consider the different surfaces for this box. So, it having the total 6 surfaces top surface, bottom surface, left, right, front and back. So, right this all surfaces for this d bar dot d s bar calculation. So, total d bar dot d s bar can be calculated by considering the each surface. So, the height for the box is vanishingly small that is height for the box is considered as a delta z. It is very small therefore, vanishingly very small. So, it is neglected and only the 2 surfaces are remaining. So, this equation becomes total integration d bar dot d s bar is equal to integration d bar dot d s bar for top surface and integration d bar dot d s bar for bottom surface. So, for the top surface the area is related to the length only the delta x and delta y. Therefore, d s bar for top surface is delta x delta y a bar z and for the bottom surface the area is same, but it is in opposite direction of the a bar z and therefore, minus a bar z. So, d s bar is equal to delta x delta y minus a bar z. Now, you can write this d bar dot d s bar for the top surface as d n 1 normal component of d bar is d n 1 for the medium 1, normal component of d bar for the medium 2 is d n 2. So, here the direction of the a bar z in positive direction therefore, d n 1 delta x delta y for the top surface and for the bottom surface d bar dot d s bar is minus d n 2 delta x delta y. So, by putting these two values in the above equation to get the total d bar dot d s bar. So, d bar dot d s bar is equal to d n 1 delta x delta y minus d n 2 delta x delta y. It can be equated to the RHS side that is rho s delta x delta y. If you are comparing these two sides you are getting d n 1 minus d n 2 equal to rho s. So, from this equation you can say that the normal component of d bar is discontinuous across the surface by the amount of surface charge density. If the charge is not present at the boundary that is when rho s is equal to 0 in that case d n 1 is equal to d n 2. So, that means, the normal component of flux density is continuous across the charge free boundary between the two dielectrics. E n 1 epsilon 1 is equal to E n 2 epsilon 2 this can be written by using the relation between d bar and e bar d bar equal to epsilon naught e bar this is the relation. So, you have to find out the equation for the electric field intensity also therefore, put the value for E n 1 in this equation. So, ratio you are getting E n 1 by E n 2 is equal to epsilon naught 2 by epsilon 1. Again you can say that this is again normal component of the field is again discontinuous. So, the tangential component for the dielectric dielectric medium e are E t 1 equal to E t 2 d t 1 by d t 2 is epsilon 1 by epsilon 2 or you can write d t 1 by epsilon 1 is equal to d t 2 by epsilon 2. And the normal components we have calculated in this lecture d n 1 minus d n 2 is equal to rho s when rho s is equal to 0 in that case d n 1 is equal to d n 2 and epsilon 1 E n 1 minus epsilon 2 E n 2 is equal to rho s. So, these are the boundary conditions or the tangential components and the normal components for the dielectric dielectric medium. These are the references for today's session. Thank you.