 Good evening. Myself Piyusha Shedgar, Assistant Professor, ENTC Department, Valchan Institute of Technology. Today, in this session we will see the topic Electric Potential. 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 they will be able to differentiate between electric field and electric potential. These are the contents. Now before going to start the bit electric potential, you can pause video here for 2 minutes and think on what is the work done. Yes, in previous lecture we have seen that work done is given by the equation force into distance that is the work done is force multiplied by a distance. In other way, you can define the work done as energy expended is nothing but the work done. The work done is a scalar quantity. Although the work done is the multiplication of the two vectors, work done is not a vector quantity, it is the scalar quantity. Now the next bit is the again work done. Work done by an external source. In moving a point charge q from one point to another point in an electric field E bar is given by the equation W equal to minus q integration E bar dot dl bar while the limits are from initial point to the final point. Work done is measured in joules. Now the next is electric scalar potential. It is denoted with the letter V. The electric potential V at a point in an electric field is the work done to bring a unit positive charge from infinity to the point. That is work done per unit charge is nothing but the electric potential. Now what is the unit of the potential? The unit of the potential is hold. It is also known as the potential difference. Thus you can define the electric scalar potential as it is the potential difference between two points in a conductor is one hold if one joule of energy is converted from electrical to other forms when one coulomb of positive charge flows through it. Thus mathematically it can be written as V equal to W upon q or V equal to E upon q whereas W is the work done and E as energy. Now the potential of point A with respect to point B is defined as the work done per unit charge. As VAB is the potential of the point A with respect to point B is given by integration limits are from B to A negative of E bar dot dl bar. This is also a potential difference between points A and B with point B is considered as a reference point. Now you know that the potential difference is also expressed as VAB, VAB can be written as VA minus VB. Thus it gives a difference between the two potentials whereas VA is the potential at point A and VB is the potential at point B and VAB is the potential at point A with respect to point B. Now how to calculate the potential due to point charge? So consider a point charge q is placed at the origin and these are the electric field lines. Then the electric field due to charge q is given by E bar is equal to q upon 4 pi epsilon naught r square A bar r. Now suppose this field is the radial field and it is surrounded the point charge q. Then take the two points R2 and R1 whereas the R2 is considered as the greater value than the R1. We can find the potential difference by considering the VAB formula only the AB point is replaced with 1, 2 thus V1, 2 becomes equal to V1 minus V2 or in forms of W upon q this can be written as minus integration E bar dot dl bar whereas E bar is the electric field intensity and dl bar is the differential length. As you know that the field is the radial field thus we are considering the differential length in spherical coordinate system. So differential length in spherical coordinate system is given by dr A bar r plus r d theta A bar theta plus r sin theta d phi A bar phi. Now V1, 2 can be written as the product of E bar dot dl bar. So E bar is q upon 4 pi epsilon naught r square A bar r and differential length is defined in spherical coordinate system. You know that the unit vector which having the same coordinate that is A bar dot A bar dot is unity and therefore the integration is only with respect to dr. Solve this integration thus you are getting the equation V1, 2 equal to q upon 4 pi epsilon naught in bracket 1 upon R1 minus 1 upon R2. So in previous slide we have seen that R2 is greater than R1. So if you are putting the value of this R2 and R1 in this equation V1, 2 is positive that is V1 is greater than V1, 2. So from this equation you can see that q upon 4 pi epsilon naught R1 is the V1 whereas q upon 4 pi epsilon naught R2 is V2. From this analysis it is clear that as test charge is moved against the electric field there is rise in potential. If in above equation R2 is considered in infinity position in that case V2 becomes equal to 0 and we can rewrite the equation for V1 to equal to V1 is equal to q upon 4 pi epsilon naught R1 holds. So this potential is called as absolute potential. So in this case we are placing the charge q at the origin. In general the potential V at distance R from a point charge q is given by V is equal to q upon 4 pi epsilon naught R and the unit for the potential is the holds. It having only magnitude and no direction therefore it is often called as the scalar potential. Some of the important points related to the electric field general points for either you are considering the positive charge or the negative charge the potential increases if you move in the direction opposite to that of the electric field and the potential decreases if you move in the same direction as the electric field. The electric potential is the potential energy of a unit charge that is associated with a static as well as time invariant field. The electric potential is expressed in holds or joules per coulomb. Electric field is a vector quantity while the electric potential is a scalar quantity thus you can differentiate between electric field and the electric potential. Electric field is a negative gradient of the scalar potential. The potential difference is the work done per unit charge against the electrostatic force to move a charge from position A to position B. Hence in equation negative sign is used. However this equation is valid only for static electrostatic field. Now the concept of the V that is electric potential with respect to electric field. So if you want to move the charge Q from one position to other position in the presence of the electric field intensity so these are the red lines the electric field intensity. So in both of figure if you move in the direction of E the potential V decreases and if you move in the same direction of E the potential V decreases. These are the references thank you.