 Dr. Dupali Shalke working as an associate professor in electronic department at Walshchian Institute of Technology, Sholapo. In this video lecture we are going to discuss with the electric flux density which is denoted by D bar. At the end of this video the students are able to derive electric flux density also determine electric flux density due to point charge, line charge, shield charge, volume charge, distribution. The contents which we will be covering in these videos are Faraday's experiment, electric flux density, how to measure the electric flux density, a relationship between the electric field intensity and electric flux density, electric flux density due to the point charge, line charge, shield charge and volume charge distribution. There will be few references. To define the electric flux density the Faraday's has performed the experiment. In his experiment he has considered the sphere and this sphere we call it as an inner sphere and one more sphere is to be placed across the inner sphere which is an outer sphere. Now these two spheres are being separated by some dielectric material. The inter active space is filled with some dielectric material and he has placed some positive charge at the inner surface and after few time or after some duration of time it is been observed that these charges are creating the electric field intensity across them and due to this electric field intensity there is a charge created at the outer surface of the surface and this the amount of charge created at the outer surface are defined by the electric flux density. So the Faraday's has defined the electric flux density from his experiment. These charges are called as a displaced charges from the inner surface to the outer surface. From this experiment the Faraday's has defined the term that is electric flux density. The electric flux density is defined as a amount of charge displaced per unit surface area. The mathematically the statement is given as an electric flux density is equal to flux by area. If you see that a flux is represented by the symbol psi upon area but for the electric field this psi is nothing but a charge q because we are considering the charges in the electric that's why it will be q upon area. Now if you observe that the as the Faraday's for his experiment he has considered the sphere. Therefore if you consider r is a distance from the inner mode sphere to the outer surface then if and it is represented by capital R then for this sphere the radius is r and for this sphere then the area will be 4 pi r square. The area will be 4 pi r square and the electric flux density as I had already told you it is denoted by d. d is equal to q upon 4 pi r square. From this figure you can observe that the flux what to which are being created they are also in the same direction as that of the electric field intensity. So the direction of the electric field intensity and the direction of the electric flux density is going to be same. Therefore this d this is giving you the magnitude and the direction of this d bar d will be d bar which will become a vector quantity d bar is equal to q upon 4 pi r square into ar bar. As we are considering it as a direction radius r therefore this will be the ar bar. This will give you the unit direction as well as the magnitude of the d bar. Here a bar is nothing but a unit vector direction which is given defined as a vector upon magnitude. Now let us see a relationship between the electric field intensity and electric flux density. The relationship between the electric field intensity and density is given by in the previous slide we have seen that the electric flux density is given by d bar is equal to q upon 4 pi r square ar bar. We will mark it as an equation 1. This is a electric flux density due to the charge point charge. Similarly a electric field intensity we will recall from the previous videos it is given by e bar is equal to q upon 4 pi r square ar bar. Now if you compare these two equations from this equation 2 if you replace this epsilon 0 towards the LHS side. At the RHS side only you remain with the q upon 4 pi r square ar bar. If you compare this equation with the equation 1 this term is nothing but equal to the d the value of d bar. So we can state that epsilon e bar is equal to d bar epsilon e bar is equal to d bar. In general this equation is given by d bar is equal to epsilon e bar which is equal where epsilon is nothing but epsilon 0 into epsilon r. For the free space with these both terms epsilon 0 and epsilon r are equal to 1 but for another dielectric material the values goes on changing. Now let us discuss with the electric field intensity and electric field density, electric flux density due to the different distribution components. Now for the point charge if the there is a point charge distribution then the electric field intensity due to the point charge is given by q upon 4 pi epsilon r square. From this equation suppose we move this epsilon 0 over the LHS side then that value will give the dq bar which is dq bar is nothing but a electric flux density due to the point charge distribution which is given by q upon 4 pi r square ar bar. For the line charge distribution the q is replaced by rho L, rho L is nothing but a line charge distribution. The charges are distributed over the line which is defined as a charge per unit length. The then electric field intensity due to the line charge which is a e bar e L bar is equal to rho L upon 2 pi epsilon r ar bar. Similar to this equation if we replace this epsilon 0 toward the LHS side that will give the dL bar. The dL bar is nothing but a electric flux density due to the line charge distribution which is given by rho L upon 2 pi r ar bar. Next is a sheet charge or which is also called as a surface charge. The surface charge distribution function the electric field intensity is given by es is equal to rho s upon 2 epsilon an bar. Then this epsilon when you consider toward the LHS side then it will give you the ds that is a surface electric flux density due to the surface charge distribution which is given by rho s upon 2 an. This n may be in any direction. If it is a Cartesian coordinate the value of n may be x y or z. Similarly now for the volume charge distribution the electric field intensity over the volume is given by integral over the volume q upon 4 pi r square ar bar. Again here if you replace this epsilon 0 epsilon 0 towards the LHS then that value will give you the electric flux density due to the volume charge distribution which is given by integral over the volume q upon 4 pi r square ar bar. Now these equations are derived by the different derivations which we can see in the videos previous videos and this is how we can by the relation what we are seen in the previous slide you can determine the value of dq, dl and ds and dv. These are the few references which we are considering for the data. Thank you.