 everyone. In this session we will see the next topic boundary conditions for the electric field. Myself Piyusha Shedgar, ENTC department, Valchin Institute of Technology, Solapur. These are the learning outcomes. At the end of this session students will be able to derive the boundary condition equations for the different fields at the two media. They will be able to apply the principles of electromagnetics or electrostatics to the solutions of problems relating to the boundary conditions. These are the contents. So, before going to start the boundary conditions you can recall here what is electric flux density. By considering the different boundaries you have to solve these boundary conditions for the different quantities electric field such as the electric flux density, electric field intensity, etc. So, before that you should know what is electric flux density. Yes, the electric flux density is nothing but charge per unit area. Therefore, it is denoted with the letter D, capital D is equal to q upon 4 pi r square is nothing but the electric flux density. Now, before going to start this all boundary conditions when you are considering the two different media before that recall some of the definitions which is required to define these boundary conditions. So, here epsilon is nothing but the permittivity. What is the epsilon? It is the major of a resistance that is encountered when forming an electric field in a medium. In other words the permittivity is a major of how an electric field affects and it is affected by a dielectric medium. It is denoted with the letter epsilon. This permittivity is major in Faraday's per meter whereas, epsilon r is nothing but the relative permittivity of the material. Next is dielectric constant. This dielectric constant gives the 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. What is electric flux density? So, as you have discussed we have discussed in previous slide as it is nothing but the charge per unit area. It has the same unit as that of the dielectric polarization. Now, consider the boundary conditions. So, the problems related to these boundary conditions can be considered with respect to the two different media or the same media. So, we should know the conditions satisfied by these field components at this boundary. So, these are nothing but the boundary conditions. Usually the field components specified above in terms of the tangential and the normal components of the field vectors. When the field exists in a medium consisting of two different media, the conditions of the field must satisfy are called the boundary conditions. So, for the electrostatic field the following boundary conditions are important. So, the two boundary the boundary is the separation between this one of the dielectric medium and another is also the dielectric medium. Other medium is considered as a conductor and the dielectric that is the boundary is between this conductor and dielectric medium and the third may be considered amongst this conductor and the free space. So, let us check the boundary condition stepwise. So, here first define what is the conductor and what is the dielectric. These are the mediums. What is the conductor? In conductor the valence electrons are present. Valence electrons are nothing but these are called as a free electrons and these are used for the conduction of the current and the electric field intensity within this conductor is 0. This is the property of the conductor. What is dielectric? In dielectric materials there are no any free charges are considered and therefore, as there is no any free charges the conduction current is equal to 0, no conduction current is considered. Boundary conditions can be defined for E bar or either in terms of E bar or in terms of the D, where E is nothing but what the electric field intensity and D is the electric flux density. So, boundary conditions can be calculated in terms of the two components, tangential components and normal components. So, E bar can be find out in terms of the tangential component as well as in normal component whereas, D bar is also can be calculated in terms of tangential and the normal components. So, consider these are the field vectors which is crossing this boundary, this is the boundary, x y is the boundary, z is the axis here one of the medium is considered as a one and the another medium is the medium two, medium one having mu 1, epsilon 1 and the sigma 1. So, these are nothing but what the permeability, permittivity and the conductivity is considered for the medium one. Similarly, the medium two which having mu 2, epsilon 2 whereas, the sigma 2 and the boundary is the separation between these two mediums. So, when one of the electric field line is passing from this boundary 1 towards this boundary 2, it may gives the angle alpha 2 in a medium 2 whereas, in medium 1 it makes an angle equal to alpha 1 whereas, B 1 bar is nothing but one of the field, B 1 bar is one of the field for the medium one whereas, B 2 bar is one of the field for the medium two. Let us consider the dielectric-dielectric interface that is medium one and medium two having the dielectric and the dielectric medium. So, medium one which having epsilon 1 whereas, medium two having the epsilon 2. Consider this is the rectangle the electric field is passing from this medium one to medium two, this is nothing but the boundary between these two. So, here we know that e bar dot dl bar is equal to 0, e bar dot dl bar is equal to 0, apply this to this rectangular path, rectangular path is a, b, c and d denoted with this arrow direction. So, it can be written separately for each path that is a, b path, b, c path, c, d path and d a path. So, separate the integration e bar dot dl bar for a, b path plus b, c plus c, d and plus d a path. So, thus you can say that this total path is consider a, b, c, d, a. So, integration of this closed rectangular path a, b, c, d, a e bar dot dl bar is equal to 0. Now, if you are considering this figure this a, d and b, c are height of these rectangle consider is vanishingly very small therefore, these two are cancelled out. So, the side b, c and d a are very vanishingly very small and therefore, these are equated to 0. So, by equating it to 0 only you are getting integration of this closed path a, b, c, d, a e bar dot dl bar is equal to integration a, b e bar dot dl bar plus integration c, d e bar dot dl bar is equal to 0. Now, a, t, 1 this e bar for the a, b path is nothing but the tangential component for the medium 1. So, a, t, 1 dl bar is the length delta x, dl bar is nothing but this is the length consider for the medium 1 as a delta x. It is in x direction therefore, it is delta x. E t 1 delta x minus E t 2 delta x equal to 0. Now, this can be equated to E t 1 delta x equal to E t 2 delta x. This is the boundary condition for the electric field for the tangential component, but you know that d bar is equal to epsilon naught e bar. This is the equation you are getting in terms of e bar. Now, you have to get the equation in terms of d bar by putting the relation between d bar and e bar as d bar equal to epsilon naught e bar. So, here d t 1 epsilon 1 is equal to d t 2 epsilon 2 or it can be written as in terms of the ratio d t 1 by epsilon 1 equal to d t 2 by epsilon 2. So, d t 1 by d t 2 is nothing but epsilon 1 by epsilon 2. Since the in this case you know that epsilon 1 is not equal to the epsilon 2 that is the permittivity is for medium 1 and medium 2 are different. So, the as these two are different, d t 1 is not equal to d 2. So, d t 1 having whatever is the value is not equal to the value for d t 2. Thus from this what you can say that the tangential component of this d bar are not continuous across an interface, across an boundary is considered between these two medium. So, we will see the normal components for the boundary condition in next PPT. These are the references for this session. Thank you.