 Good morning everyone, today in this session we will see the Gauss law, Myself Piyusha Shedgarh, Valchin Institute of Technology, Solapur. These are the learning outcomes, at the end of this session students will be able to define Gauss law, they will be able to derive the expression for Gauss law and also they will be able to apply Gauss law in different applications, these are the contents. So before going to start today's lecture, you can pause video here and recall that what is electric flux density? To define the Gauss law, this knowledge of electric flux density is required. So pause video here and you can think on what is the electric flux density? Yes, the electric flux density is given by this equation D is equal to psi is upon 4 pi r square, psi can be equated to qa. So D bar is nothing but q upon 4 pi r square. So the Gauss law was formulated by the Karl Friedrich Gauss in 1835, but was not published until 1867. It is one of the four Maxwell's equations which is in the form of the classical electrodynamics. Gauss law which studies the electric charge along with the surface and the topic of the electric flux. So there are the different Maxwell's equations are defined. So one of the important equation is considered amongst this Maxwell's equation as a Gauss law. Gauss law states that the total electric flux passing through any closed surface is equal to the total charge enclosed by that surface. So in short what you can say that the total electric flux is nothing but the total charge enclosed by that surface. Gauss law gives the total electric flux out of any closed surface is equal to the charge enclosed divided by its permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field. So according to the statement of the Gauss law, this is the mathematical form of the Gauss law. Electric flux is nothing but the integration of the differential amount of that electric flux. It can be equated to integration D bar dot ds bar. So this is the relation between D bar and E bar as D bar equal to epsilon naught E bar. So you can put this epsilon naught E bar in this equation as a D bar. So by putting that you are getting psi is equal to epsilon naught integration E bar dot ds bar. So it is nothing but the q enclosed divided by permittivity epsilon naught. Whereas this epsilon naught is nothing but the permittivity of the free space. So here the psi is given as the total flux within that any closed surface is considered. Whereas it is proportional to the total charge enclosed by that surface. So mathematically Gauss law can now be expressed as charge enclosed is equal to integration D bar dot ds bar. So what is the proof for this Gauss law? So we will see the proof how to prove this q enclosed is nothing but the D bar dot ds bar is equal to the electric flux. So for that consider this is the figure which having the sphere which is placed the center of this sphere is placed at the origin and the charge is also placed at this origin which having the radius is equal to r. So for this sphere consider the small amount of the area ds bar differential amount of this area is denoted with this ds bar. Whereas D bar is normal to that area D bar is normal to that area. So first point is consider the q is situated at the center of any imaginary sphere is considered. So the total flux passing through this sphere is given by integration of d psi, integration of d psi according to the above differential area equation it can be represented as a D bar dot ds bar. Since the point charge is at the center the flux density is defined as D is equal to psi and psi can be replaced with the q upon 4 pi r square. So this is nothing but the scalar quantity if you want to represent in vector notation. So D bar can be represented with along with the unit vector. So as you know that this is the unit vector in spherical coordinate system in r bar direction and therefore D bar can be written as q upon 4 pi r square a bar r. The differential area on the spherical surface is given by for r equal to constant surface as r square sin theta d theta d phi a bar r. So whereas psi is equal to integration of d psi it can be represented or it can also be written in D bar dot ds bar form. Hence by putting this D bar and ds bar values here this is nothing but the D bar and this one is the ds bar. So take the dot product between these two whereas the variation of the r is sorry the variation of the theta is 0 to pi whereas the phi variation is 0 to 2 pi. So put the limits of this integration r square r squares are cancelled out only remaining q upon 4 pi sin theta d theta d phi. So all these integration differentiate it with respect to theta and phi. As you know that a bar r dot with a bar r it becomes equal to 1 and therefore psi is equal to if you are solving this equation thus you are getting q upon 4 pi into 2 into 2 pi. So here 2 into 2 pi is 4 pi this 4 pi and this 4 pi is cancelled out and what the remaining psi is equal to q is remaining. So here you can say that the Goff's law is proved the total electric flux crossing through any closed surface is equal to the charge enclosed by that surface. So if the charge is not placed at the origin in general the charge can be represented in terms of the volume charge density. So that means when the volume is considered the charge can be calculated by integration rho v dv. Then the Goff's law can be defined as q equal to as you know that the q in terms of the differential area d bar dot ds bar and in terms of the differential volume it is rho v dv. So you can equate these two equations q equal to integration d bar dot ds bar equal to integration rho v dv. Whereas this d bar dot ds bar is nothing but the total flux crossing the closed surface and rho v dv is nothing but total charge enclosed by closed surface. Goff's law can be used to derive the Coulomb's law and vice versa. In other words what you can say that the Goff's law states that the two net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. If you apply this Goff's theorem to a point charge enclosed by a sphere you will get back the Coulomb's law easily. Therefore it can be says that one of the applications of the Goff's law. Goff's law can be applied to derive the Coulomb's law. It can also be applied to derive the electric field due to line charge. So as you know that the Coulomb's law is nothing but q1 q2 upon 4 pi epsilon naught r square a bar r that is it is directly proportional to the product of the charges and inversely proportional to the square of the distance between any two point charges. So derivation of this Coulomb's law is one of the application of the Goff's law. So Goff's law can be used as you know that the Goff's law is nothing but it is used to define the charge equation. So force is equal to q dash e. So force is given by this equation. So this is force equation that means it is nothing but the Coulomb's law. Whereas the Goff's law also can be used to find out the electric field due to line charge. This electric field due to line charge is related to the line. When the charge is distributed throughout the line, the line charge density is considered and by using that you can find out the electric field. Another application of the Goff's law is electric field due to an infinite plane sheet of the charge. It can be considered with respect to the surface charge density. Whereas electric field due to charged solid sphere is also calculated by using this Goff's law. So these are the different applications of the Goff's law. These are the references for this session. Thank you.