 Good morning everyone. Today in this session we will see the topic electric dipole. These are the learning outcomes for this session. At the end of this session, students will be able to derive the expression for electric potential and electric field due to electric dipole. Also, they will be able to derive the expression for potential energy. These are the contents for this session. Before going to start the actual topic electric dipole, you can pause video here for 2 seconds and recall that what is electric potential? So, the electric potential of any point P with respect to the point B is defined as the work done per unit charge. It is also given by this equation VAB is equal to minus integration from B to A e bar dot dl bar. Whereas e bar is the electric field intensity and dl bar is the differential length. Now, define the electric dipole. The electric dipole is 2 equal and opposite charges are separated by a distance which is small as compared to the distance of point at which we desire the value of electric field and the potential. This electric dipole is also called as doublet. Now, consider this is the figure for the electric dipole. The point of interest P where we want to calculate the value of electric field and the potential. This is the electric dipole that is plus Q and minus Q are separated by the distance small d. It is lies on the z axis. So, from the center of the dipole plus Q is placed at the distance of d by 2 and minus Q is placed from the center of the dipole at distance minus d by 2. And this point P is placed at the distance r1 from plus Q and it is placed at the distance r2 from minus Q. And the point P is placed at the distance r from the center of the dipole. In the view of the symmetry with the dipole around this z axis, it is convenient to use the spherical coordinate system and therefore the point P is defined in spherical coordinate system as r theta phi. Now, you know the equation for the potential at point P due to charge plus Q is given by plus Q upon 4 pi epsilon naught r1. Whereas, the potential at point P with respect to the minus Q charge is given by minus Q upon 4 pi epsilon naught r2. Now, calculate the total potential by adding these two potential. We are getting this equation v equal to plus Q upon 4 pi epsilon naught r1 minus Q upon 4 pi epsilon naught r2. So, this is the equation of the potential. If the point P is in xy plane, then consider the r1 distance is equal to the r2 distance and therefore we are getting the value of the potential is equal to 0. If the point P is far away from the two charges, then r1 and r2 will seem to run parallel as shown in figure 2. Now, if the point P is far away from the dipole, then r1 is equal to r2 consider as equal to the point P at a distance from the center of the dipole that is small r. And the difference between the two distances r2 minus r1 is equal to d cos theta. Now, for that refer this figure. Now, consider that the point of interest P is far away from the dipole. Then in this case the r1 is seen to be parallel with the r2 value and in that case r1 becomes equal to equal to small r. Now, the additional part can be written as r2 minus r1 is equal to d cos theta. Now, by replacing this v is equal to Q upon 4 pi epsilon naught 1 pi r1 minus 1 pi r2. Here you are getting the difference between r2 minus r1 can be replaced with Q d cos theta upon 4 pi epsilon naught r square. Therefore, you are getting the equation of the potential due to dipole. Thus, the above equation which having the four factors involving dipole moment, the angle, the distance and a constant. Whereas, dipole moment is Q d is nothing but the dipole moment cos theta is the angle. 1 upon r square is the distance and 1 upon 4 pi epsilon naught is the constant. Now, if you know the equation for the electric potential then you can easily find out the equation for the electric field. Thus, you can find the equation for the electric field due to dipole at the point of interest P. Now, you know the relation between the electric field and the electric potential. So, again pause the video for a second here and write down the equation for the electric potential and the electric field. Yes, the electric field E is the negative gradient of the potential that is in mathematical form it can be written as E bar is equal to minus gradient of V. Now, here we are using the spherical coordinate system. Therefore, define del operator in spherical coordinate system as gradient of V is equal to dou V by dou r A bar r plus dou V by r dou theta A bar theta plus dou V by r sin theta dou phi A bar phi. Now, here you can take the partial differentiation of the electric potential with respect to r theta and phi and put these values in the equation of E bar. So, by putting the values of partial differentiation of V separately with respect to of 3 axis r theta phi we are getting the equation for electric field due to dipole is given below. So, E bar is equal to minus of minus q d cos theta upon 2 pi epsilon naught r cube A bar r minus q d sin theta upon 4 pi epsilon naught r square into 1 by r A bar theta. And finally, by solving this equation you are getting the equation of E bar as q d upon 4 pi epsilon naught r cube 2 cos theta A bar r plus sin theta A bar theta. So, this is the expression for the electric field due to dipole. Now, what is the dipole moment? The dipole moment is assigned by the symbol P and it is equal to the product of charge and separation between the two charges. As you know that the electric dipole is nothing but the two equal and opposite charges are placed at the distance d is given by the separation by the small d. Then the dipole moment is defined as P is equal to q into d. Whereas, you know that any dot product of the two vectors A bar dot B bar is equal to A B cos theta. According to that equation you can write the dot product for D bar dot A bar r. Here the D bar which having the magnitude is D whereas, A bar r is the unit vector which having the magnitude is 1. Therefore, this equation can be written as D into 1 into cos theta that is D bar dot A bar r is becomes equal to D cos theta. And P bar dot A bar r can be written as as P bar is nothing but q into d. So, it can be written as q and D bar dot A bar r is equal to q d cos theta. Now, replace this q d cos theta with P bar dot A bar r in equation of the potential and the electric field to get the equation for the dipole moment and the electric potential and the electric field. Now, the electric potential due to this dipole moment can be written as P bar dot A bar r is equal to P bar dot A bar r upon 4 pi epsilon naught r square. And electric field equation is P upon 4 pi epsilon naught r cube 2 cos theta A bar r plus sin theta A bar theta. Thus, the dipole moment expression simplifies the potential field equation. Now, what is the potential energy? Now, consider the empty universe. In that empty universe to bring a positive charge from infinity into the field of another positive charge work is to be required. The work is done by the external source that moves the charge into position. If the source released its hold on the charge, the charge would accelerate, turning its potential energy into kinetic energy. The potential energy of a system is found by finding the work done by an external source in positioning the charge. So, in mathematical form, how to calculate the potential energy? It is denoted with W e. Now, positioning the first charge q1 in empty universe, there is no any work is to be required. That is, no field is present there. Now, positioning more charges in the presence of the charge q1, it is required the work is to be done. Now, the total positioning work can be written as the potential energy of the field is equal to q2 times the potential of q1 charge with respect to q2 charge. So, it can be written as q2 v2 1 plus q3 v3 1 plus q3 v3 2 like that it is continued. Manipulate this expression to get the equation for the potential energy as W e equal to 1 by 2 q1 v1 plus q2 times of v2 plus q3 times of v3 and again it is continued. Thus, you can calculate the potential energy by the product of the charge and the potential. These are the references for this session.