 Good morning everyone. Today we are going to see the topic electrostatic field part 2. These are the learning outcomes for this session. At the end of this session, students will be able to derive the expression for work done. They will be able to explain the work done for different paths. These are the contents for this session. Before going to start the topic work done, you can pause video for 2 seconds and recall that what is the electric field and electric field lines. Yes, the electric field is an effect in the space due to which a charge experiences a force. And electric field lines are the graphical representation of the electric field. These are imaginary lines which do not have any real existence. We know that electric field is a vector quantity. Now the work done, some of the examples of work done are given. Actually work done is energy expanded is nothing but the work done. So in first figure, atlas holds up the earth. And in second figure, Garkhan picks up the tray. So these are the examples of the work done. So some force is to be applied to do the work. In this figure, it shows the definition of the work. So if you want to move this block from initial position to the final position, this is denoted with the distance dist. If it shows with the direction, that means in which direction this block is moved. So the direction is given by this angle theta. Thus you can say that the work is nothing but the force into displacement. So here the force is also in vector form and the displacement is also in vector form. But note that work is a scalar quantity, not a vector quantity. Although the force and displacement both are the vector forms, but the work is in scalar quantity. Because any dot product of the two vectors is nothing but the scalar quantity. Now the work is defined in two ways, positive work and the negative work. So again note the first figure. In first figure, charge moves uphill. That is work is done against the field. That means the applied work is positive. As the work is positive, potential energy increases. And in second figure, the charge moves downhill. That is the work is in the direction of the field. So the work is negative. As the work is negative, the potential energy decreases. Now this is the work done. So if you want to move the charge Q in an electric field, that is the force is to be applied. So as the force is to be applied is opposite and equal to that of the exerted field. If the work charges move from this initial position to the final position, the distance is given by DL. When a charge Q is moved at a distance DL in an electric field E bar, then force on charge Q due to an electric field is given by Fe bar equal to Q E bar. Whereas Fe bar is the force on Q due to an electric field E bar. The component of this force in the direction DL bar is given by Fe equal to Fe cos theta. As you know the dot product A bar dot B bar is given by AB cos theta. This equation can be written as Fe bar dot AL bar. Now Fe L is equal to Q E bar dot AL bar. When we are moving a charge Q a distance DL, then the force which we apply must be equal and opposite to that force due to electric field. Therefore applied force is equal to the negative force of E L. Therefore the above equation becomes minus Q E bar dot AL bar. Work done is force multiplied by distance. So it is considered as a differential work. Due to differential length is given by DW. Whereas DW is known as the differential work or incremental work is given by force applied into DL. So by putting the above equation in this equation, you are getting minus Q E bar dot AL bar into DL. Any vector which having the magnitude as well as the direction denoted with this DL bar. Therefore DL AL bar can be written as a DL bar. Whereas DL bar is the differential length in vector form. So differential works becomes DW equal to minus Q E bar dot DL bar. This work equation is the differential work equation. So as you want to find out the total work, then integrate this equation with respect to initial and the final position. Thus the final equation of the work becomes W equal to minus Q E bar dot DL bar and the integration is with respect to initial position to final position. The work done is measured in joules. This differential length can be represented in three different coordinate systems as given below. As you know that in Cartesian coordinate system, the differential length is varied with respect to x, y, z coordinates. In cylindrical coordinate system, it varies with r, phi and z. And in spherical coordinate system, it varies with respect to r, theta, phi. Now the next bit is work done along the different path. Suppose on x axis two points are given x1 and x2. And if we want to move the point charge from x2 position to x1 position in electric field E bar. So different paths are possible to travel from x2 to x1 as shown in figure. So different paths are given as path 1, path 2 and path 3. But the work done is to be same for any path because the initial position and the final position for these paths are same. That is work done is independent of the path. Next work done in a loop. Consider now the circular loop as shown in figure. Again on x axis, x3 point is given. So for this figure, x3 point is the initial point. And the same x3 denotes the final position of the charge movement. Therefore the work done is to be zero around any closed path. Because the total work is calculated with respect to initial position and final position. So the equation w becomes equal to zero that is minus q E bar dot dl bar is equal to zero. This can be written as integration E bar dot dl bar equal to zero. A small circle on the integral sign indicates that the closed nature of the path. The above equation true for static fields. Such a field is called as conservative field or irrotational field. This equation is also known as integral form of Maxwell's equation derived from Faraday's law for the static fields. These are the references for this session. Thank you.