 Dr. Rupali Sherke, working as an associate professor at Walsh and Institute of Technology in the electronic department. In this video, we are going to discuss with the electric field intensity. Learning outcomes. At the end of this video, students are able to define electric field intensity as well as derive the electric field intensity due to the point charge. The contents what we are going to cover is a electric field intensity, magnitude and direction of electric field intensity, electric field intensity due to the point charge, field charge distribution and references. As we are familiar with the electric fields, electric fields are nothing but an imaginary lines which we feel or we experience. The electric field intensity at a point is defined as a force experienced by the point charge placed at that point. This electric field intensity is an vector quantity and it is denoted by E. The formula for electric field is given by force per unit charge that is F by Q. For explaining the electric field intensity, let us consider a charge Q which is a positive charge placed in free space and there is one more charge Q. Now if the polarity of these two charges are opposite, then they are attracted towards each other. So the distance between the, suppose it is a R distance, this R distance goes on changing. If it is both are opposite charge then they are attracted. If they are negative charge, if they are the similar charge then the distance R goes on increasing, they ripple with each other. Now if the, suppose we will assume this as a positive charge capital Q and this is a negative charge. The force between these two charges, the force between these two charges is recalled by the Coulomb's law. What the Coulomb's law gets? The force acting between these two charges is given by K into Q, product of Q into capital Q into small Q divided by R square. That is force is directly proportional to the product of the two charges and inversely proportional to the square of the distance and K is the proportionality constant. Now consider one fixed positive charge Q and we will move this Q charge in the field of the, field of this capital Q. This is a charge Q placed and we are moving in the, the arrows are representing the electric field across this. When we move this another charge Q in this field then it experience a force. This charge experience the force due to this point charge Q. That force is given by the Coulomb's law and the electric field intensity due to this point charge Q is given by force experience by that charge that is force per unit charge. Now when we know that what is the formula for force that is Q upon K into Q upon R square and this Q is the charge which we have then this Q and this Q will get cancelled out and only remain with the E is equal to K upon Q upon R square. Now what is this Q is the positive charge Q what we are consider and R is the distance between this two charges and K is nothing but a proportionality constant. When you substitute the proportionality constant it will be 1 upon 2 pi epsilon Q upon R square. Now let us find the magnitude and direction of this E for that again let us consider a positive charge Q is placed the direction of the electric field for this positive charge is outward. If this Q is a negative then the direction of that field will be towards the charge as it is a positive it is outside for negative it is inside. If you consider one more charge at some distance R from this positive Q then this Q will also this charge will experience a force and the direction of the experience force will be in the same direction as that of the field the whatever the field electric field direction is the same the force the direction will be for this Q also. So the force is given by Q upon 4 pi R square AR bar direction of is a multiplied with the unit vector that is AR AR is nothing but R upon magnitude vector upon magnitude vector is a distance vector of this between the two charges. Now when you substitute the value of the force in this equation then the equation for E bar will be equal to 1 upon 4 pi epsilon Q upon R square AR bar AR bar is nothing but a unit direction in a R direction unit vector in a R direction. Now if you substitute the value for 4 pi epsilon that is 8.85 into 10 raise to minus to where for epsilon 0 then it will be reduced to the E bar is equal to 9 into 10 raise to minus 9 Q upon R square AR bar this is a as the unit for this is a force per unit charge it will be Newton's per coulomb or it will be volume per meter this is a electric field due to the point charge. Now let us see one example suppose the point charge is placed at a location a point 2 comma minus 3 comma minus 3 minus 1 comma minus 3 this is a coordinates in a represented in a Cartesian form and they ask you to find out the electric field intensity due to this point charge 5 nano coulomb at a origin first we will draw the diagram we will consider the three axis we will place that charge in the free space which is a 5 nano coulomb and a is the point where it is being placed now the desired point is a origin which is given by 0 0 0 coordinates and let us first join these two and we will find out the distance R between them R is found out by the head minus tail formula which will give you the R bar is equal to minus 2 X bar A Y bar plus A Z bar then we will find out AR bar and substitute in this equation R we know AR we know then in the above equation we will get the E bar for this at origin that is minus 1.717 A X bar plus 0.8587 A Y bar plus 2.576 A Z bar now let us see the electric field charge distribution over the if these charges are distributed over the different surfaces for this if it is distributed over the line the number of charges distributed over the line then it is given by rho L which is called as a line charge distribution over a line it is distributed that is called line charge distribution and if it is across the surface then it is called as a surface charge distribution and suppose it is over the volume it is a volume charge distribution then the electric field intensity for this different components will change for line charge it will consider instead of q it will be now rho L dL the value will be a rho L what is rho L rho L is nothing but charge per unit length which is given in a coulombs per meter this is for a small length if you calculate the dL it is a rho L into q dq by dL dq will be rho L into dL therefore the q will be integral of the dq charge will be a q charge in the above equation will be rho L dL similarly for the surface charge it is a charge per unit area that is why it will be meter square and this dS this rho S will be dq by dL if you consider here a small surface dS and for from that if you are considering it is a dq charges are flowing then it will be dq will be rho S into dS rho S into for the entire surface q will be integral of dq just substitute over the dq where rho S dL this will be double integration as we are considering the surface for the surface we are considering the two components similarly for the volume volume charge distribution is given by rho V is equal to charge per unit volume in coulombs per meter square what will be here rho V is given by dq by dV rho V will be dq will be rho V into dV then the total charge q will be integral over the integration of the dq just substitute the value of the dq it will rho V dV this will be as the volume x axis y axis and z axis all the three axis we are considering that is why it will be a integral over the three surfaces now this is this are the references which we are saying thank you.