 So, let us try to find out, did you all understand the utility of finding the field, once you know the field, the force at a particular location, the magnitude of the force at a particular location will be straight forward, you just have to multiply in mass with that field to get the force, fine let us try to do that, do the further calculations now, suppose you have, okay it will be mathematically a little bit involved, so okay let us do the simpler one first, you have a ring, okay you have a ring, let us first take this type of ring, which is in the plane of your screen, this ring, all right, can you tell me what is the field at the center of this ring, total mass is m and radius is r, what is the field at the center, see here you can draw lines like this, now you can see that the field because of this point mass, which is located over here, will be in which direction, this direction, right, due to this point mass, the field will be in this direction, these two fields will cancel out each other, similarly, anywhere you will see, if you try to find the field, there will be a point, diametrically opposite point will be there for which the field will be equal and opposite, so all these fields, all these fields will cancel out and total field will come out to be 0, getting it, so due to symmetry, clear, so if you place a mass at the center of the ring, it will not experience any gravitational force, because field is 0, so mass into field which is force that is also 0, clear all of you, okay now try to find out the field because of the sector of the circle, okay let us say you have, this is a center, let us say radius is r and this angle is theta, okay mass is m, total mass is m, it is a ring, sector of a circle, sector of a circle, what does it mean? Oh no, sector of a circle means part of disc, so it is an arc, okay, it is a circular arc which makes angle theta at the center, find out, okay, Niranjan has got some answer, others, Niranjan that is dimensionally incorrect, okay let us see how we can do this, first make full use of symmetry, okay, divide it into two equal parts and then take this small mass over here and take another small mass symmetrically located over there, okay, you will see that the field because of these two will be in this direction, okay, so these two point masses will have field along these two point masses direction, okay, let us say this is dm, this is also dm, alright, I am taking this as phi, okay, so this is also phi, I am taking symmetrically located dm, right, let us see whether your answer is correct, these are the two fields, so you can see that horizontal component of these two fields get cancelled out, okay, only vertical component remains, integration is a scalar addition, you cannot have a vector addition when you integrate, okay, so that is why you can integrate only components along vertical and along horizontal direction in this case, okay, so the horizontal component of the field get cancelled out and every point mass you consider over here, you will have a symmetrically located point mass at the other part of the ring, other part of the arc, so your field will get cancelled out horizontally every time, okay, so you will have only net field along y direction, okay, so that will be e times cos of phi, alright, so let us say I am considering small field due to the small mass dm, okay, so if I take two masses at a time, the total field dey will be 2 times de cos phi, so de cos phi because of this and de cos phi because of that, okay, de is what, g times mass is dm divided by radius square that into cos of phi, now the problem is, problem is that now you have two variables, m and phi, so I need to write down mass in terms of phi, okay, so if you zoom it a bit, you will see that this dm will occupy the small length of the arc because mass is distributed on the arc, right, so if you consider a small mass, it will occupy small length also, so let's say this length is dl, okay, so dm will be equal to mass per unit length which is m divided by the arc length is what, r into theta, right, so this into dl is dm, okay, and from the definition of angle, dl should be equal to r into d theta, all of you understanding, right, is there any doubt, so dm is why you are retracting the messages one by one, it's okay to be wrong, so m by theta into d theta, actually it is d phi, d phi, m by theta into d phi, okay, because I am taking phi as a variable, okay, so this will be equal to 2g dm which will be equal to m divided by theta that into r square to cos of phi d phi, okay, so total field will be integral of this, are you able to see the screen, no, you are able to see now, now tell me the limits of integral from where to where should I integrate, what will be the limits of integral, 0 to theta by 2 you have to integrate, okay, because when you say 2 times, every time you are considering one mass here, one mass there, okay, so if you go from here to here, you are automatically considering corresponding masses from here to here, you are taking two masses at a time, okay, so you will integrate from 0 to theta by 2, but if you don't do two times, then you can integrate from minus theta by 2 to plus theta by 2, okay, because when you take 2, multiply with 2, you are saying that I am taking two masses at a time, so the gravitational field will be integral of cos phi will be sin phi, okay, so gravitational field will become equal to 2 times gm divided by r square theta into sin of theta by 2, right, so if I keep a mass m over here, if I keep a mass small m over here, the force experienced by the small m will be m into e, okay, once you know the field, any doubts, anything you want to discuss, quickly tell me this field is along j direction, so it will be j cap, see even if you are understanding 60%, that is good enough right now to start with, so nobody will understand everything right from scratch if you are listening this for the first time, so all you have to do is after the class sit with whatever we have done today and you will be able to understand everything, you did not understand how we took the limit of integration, there go, this dm you have to take here, then here, then here, everywhere you have to take dms, then only you have to cover entire arc right, so that is why you need to go from this point to that point along this line to cover the entire mass, okay, now what you are doing is when you are considering mass over here, you are considering another mass over there as well, so by the time you travel from here to there, you have already covered automatically all these masses, okay, so the limits will be whatever angles this point mass over here makes to whatever angle this point mass makes from the vertical, total angle is theta, this angle is theta by 2, this angle, angle with the vertical because this vertical line divides it into equal parts, so this will be theta by 2, so dm starts from 0, it goes to theta by 2, sorry theta, the angle phi starts from 0 and goes to theta by 2, okay, clear now, you have a ring, okay, you have a ring, this ring half of it is inside your screen and half of it is coming out of your screen, okay, so basically the plane of the ring is perpendicular to your screen, okay, this distance is z, from this point the center of the ring, the distance is z, okay, the radius of the ring is r, mass is m, okay, you need to find field at this point because of the ring, point is along the axis of the ring, the point is along the axis of the ring, okay, the ring is perpendicular to the plane of your screen, okay, all of you try it, should I do it? I will do it now, all right, so let us do this now, so the field because of this let us say dm will be along this line, right, this field is let us say de, okay, this angle is theta, let us say, okay, now can I say that angle theta will be same for whatever mass you take along the ring, along the ring whichever point you go, angle theta will be same, is this correct statement, okay, and is it also correct that the component, let us say de is component along this axis, along the y axis will be de sin theta, this one will be de cos theta, now if you join this one, okay, that once will also have a same magnitude of electric field, sorry the gravitational field, so it will have a vertical component downward, so the only component that remains will be along the axis, is that also correct? The only component de should be equal to de cos theta because sin theta component gets cancelled anyways, when you sum it up, only component which is along the axis will remain, is this statement correct, okay, so de is what? de is g times dm divided by this distance square from here till this point, okay, that is z square plus r square, okay, this into cos of theta, now cos of theta can be written as z divided by r square plus z square raise to power half, till now you are comfortable, till now whatever we have done, you are okay, okay, now you will see that wherever you go, wherever this dm goes along the ring z is constant, r is constant, g is anyway constant, so all this expression comes out of integral, this thing is equal to gz to z square plus r square raise to power 3 by 2, okay, integral of dm and integral of dm is total mass only, okay, so 0 to capital M total mass, fine, so this will be equal to g times m z divided by r square plus z square raise to power 3 by 2, okay, and it will be along the axis towards the center, what happened to cos theta, they go, drop a perpendicular from here, okay, this will become right angle triangle, this one, okay, so cos of theta in this right angle triangle will be what? z divided by habitanus, habitanus is this, which is using Pythagoras theorem, you will get z square plus r square raise to power half, all of you understood how we get the gravitation field because of the ring, now if you place a point mass at that point, let us say m you place over there, the force will be m into field, because field is force per unit mass, okay,