 Hello, I am Varun and right now we are planning to the electromagnetism tutorial after completing the quantum mechanics tutorial in the morning. So the basic idea is that these are the questions from your tutorial and what you are going to do is, I will just highlight a few questions right now, you will get some time, you can try to solve the questions, I will look at the discussion in the chat itself and after say 2 minutes or 3 minutes, I will discuss the solution as well as different approaches to solve the problem ok. So the first problem that we are going to tackle right now is question number 2, calculate the flux of the vector over the surface of a right circular cylinder of radius R bounded by the surface is z equals 0 and z equals h, calculated directly as well as by use of the divergence theorem. This problem is actually quite simple, it should not I mean take you much time, so have a couple of minutes try it out and the basic idea is that when you calculate it directly as well as when you calculate it using the divergence theorem the answers should match since you especially since you do not have any singularities. So just try out the question for a couple of minutes and let me then discuss the solution ok. So let us start solving the problem, I hope you have already finished it. So the answer if you just want to check it directly it is pi R square h square and so what is the basic idea behind solving the problem? You draw the surface of the cylinder, you get an intuition for what the surface normal is and then you can just compute the flux directly. So the value of the vector field is given as x power 4 y i hat plus sorry this is a minus sign 2 x cube y square j hat plus z square k hat. Now when you try to draw the surface say right now the cylinder that I am drawing it is oriented along the z axis and the base is on the x y plane. So let me draw the cylinder, in this case the right handed system does not matter but typically you should try to draw a right handed system which is important. So now what are the three surfaces, there is one base at the bottom, there is another base at the top and there is the curved surface along which you need to find out the integral. So let us look at the two circular surfaces first. In case of the top circular surface f is given as there is some component along i hat plus some other component along j hat which depends on x and y of course and there is some component which is the height of the cylinder given as h which is h square k hat. Why I am not writing the i hat and j hat components because they are not relevant when you are computing the flux, the flux is always computed in the normal direction. So the flux along the surface will just be f dot k hat times the area of the surface because in this case the value of f dot k hat is the same over the entire area otherwise you would need to integrate. So the area of the surface in this case is just pi times r square and f dot k hat is given by h square. So already what we have is starting to look like the answer. So what we need to check is whether the flux computed along the other two surfaces sums up to 0 or not. So that is what we need to check. Now let us check at the bottom surface by the similar argument as earlier the z component of f. So let me just write down here this is for the top surface and for the bottom surface f dot k hat for bottom is 0 because we took the surface to be along z equals 0. So f dot k hat on the bottom is 0. So in that case flux for the bottom part has to be 0. Now the next thing to do is we need to compute it along the circular part. So in this case we need to find out the normal component and then try to find out the flux. So in this case we can try to do the integral and cylindrical coordinates that would be much easier because in case of cylindrical coordinates in case of cylindrical coordinates the normal for the curved surface will be along the row hat direction or you can also call it the s hat direction depending on what coordinate used to denote it. So that is the idea. Now in cylindrical coordinates x can be written as in this case since you just have a normal cylinder. So you can use the radius directly. So x equals r sin theta and y equals r cosine theta and z prime equals z basically the cylindrical and Cartesian z are the same right. So this is the substitution to be made and you also need substitutions for i hat and j hat. So in that case what you can write down is it is fairly simple but say you have your row hat direction and you have a theta angle in this manner. So row hat equals i hat times cosine theta plus j hat times sin theta. So in this case again you can compute the integral it should not I mean so the integral you can check that I mean it does not make sense to work out all the steps here but you can check that the integral becomes this quantity which I write down cos power phi theta sin theta minus 2 cos cube theta sin cube theta d theta d z where theta integral goes from 0 to 2 pi and the z integral goes from 0 to h. So a good way of solving these kind of integrals is to use the substitution sin theta d theta with a minus sign equals the differential of cos theta and you can check that this integral turns out to be 0. So finally what you get as the so this is what you get finally and using the divergence theorem it is much much simpler if you try to solve it. So this problem becomes much simpler using the divergence theorem so the divergence must be a scalar and this is what the divergence turns out to be 2 z because the other terms cancel out and so what you just have here is according to the divergence theorem the volume integral of the divergence over the volume must be equal to the surface integral of the flux. So this is what you have already computed you need to compute this and double check. So this integral is in this case you can choose whatever coordinates you desire but cylindrical coordinates would be much easier so let us do it in cylindrical coordinates. So you have a volume element which is rho d rho d theta times d z and then you have 2 z and now you need to write down the integration limits. So rho goes from 0 to capital R theta goes from 0 to 2 pi and z goes from 0 to h this integral is quite easy to double check so this turns out to be the theta part turns out to give 2 pi times there is an h square term arising because of the z part and there is an r square by 2 coming from the rho part. So hence you again get your pi r square h square as you have seen I mean this part computing the integral using the divergence theorem is much simpler instead of directly computing the flux. So I mean this is just a simple example but you can think up of complicated ones in the case where computing this quantity is much easier since you are just integrating a scalar field instead of finding out a vector field then the dot product and surface might not be very nice to integrate over as well. So that is the first question done. So the next question that we look at is we look at question number 5 so this again is a relatively simple problem in this case you have a line charge of length L with the charge density of lambda. So the question is to find the electric field at a distance d along the perpendicular bisector of the line. So if you notice in the question the word uses perpendicular bisector but by symmetric considerations the perpendicular bisector will not actually be just a segment it will be a plane. So but at all different points in the plane if they are at a distance d then you should be able to find out the electric field in a similar manner since the situation has cylindrical symmetry. So why do not you try this question for a couple of minutes and then we look at solution. So we are looking at the problem of the line charge. So in this case let me not just discuss the answer first let me just draw a picture. So you have a line charge of length L and there is some charge density lambda. So in this case you do not know the sign of lambda but that is not quite relevant. So let me just draw the perpendicular bisector so these two lines are equal and what we need to do is find out the electric field at this point. So by symmetric considerations again you can point out the direction for the electric field. So what should be the correct direction? First let us look at the three components in cylindrical coordinates. So this one component which is the radial direction in cylindrical coordinates. So radial that is possible because the charges could be giving field away or towards but what will happen to the angular components and the z component. In case of charges since the charge density is equal the electric field from a point on the left hand side will point say assume for now that the charges are positive sign then the field will go towards the right and for the charges on the right side the field will go towards the left. So the only net component that will survive is in the radial direction. So say z direction so field in the z direction is not allowed where I took this as the z direction and for the angular part again since the situation is cylindrically symmetric you cannot get any field in the angular direction as well. Again I mean symmetry simplifies the problem quite a bit. So the next thing to do we just use Coulomb's law and there will be some integration that we need to perform because we need to look at each line element and then integrate over the entire line. So the field due to a point charge is k q by r square times r hat. So where k equals 1 by 4 pi epsilon naught. So in this case if you write down the field you only need to consider the radial component and not the angular z component. So let us just write down this expression. So the electric field must be I am not writing any vector sign because you already know what component I am talking about. So this is 4 pi r square say this is along some point x then in that case the distance will be this is the distance d then the distance of this point from here you can obtain it using Pythagoras system. So there is a 4 pi epsilon naught here there is a times q. So q is the elemental charge at the point. So the charge at the point will be given by lambda dx. So lambda is the charge per unit length and dx is the length. Now you divide by r square. So r square must be d square plus x square plus x square and the thing to note is since you have to take a component say you are considering x in this direction. So since there is some angle alpha here. So you need to multiply by a factor of cosine of alpha. So what is cosine of alpha? It is just quite so there is an integral needs to be performed. So x goes from minus l by 2 to plus l by 2. This is an integral that needs to be performed and cosine alpha is easily obtained. So this will be 1 by 4 pi epsilon naught lambda dx and cosine alpha will be x by there will be another term appearing in the denominator because of the hypertenuse. So this is the integral sorry it is a constant d not an x because this is the length along which you are taking the cosine. So this is what needs to be done. Just do not confuse this d and this d this is the differential and x going from minus l by 2 plus l by 2. So in this case again there is some nice substitution which needs to be made in order to solve the integral quickly. So in this case since you have a d square plus x square term appearing in the denominator you should try the substitution d equals x equals d tan theta. So you will have a sec cube theta coming on in the denominator and a sec square theta term coming in the numerator. So this integral will become l by 2. So x x 1 to l by 2. So theta will go to tan inverse l by 2 d and tan inverse minus l by 2 d. So do not forget to correct the limits otherwise you will probably not get some sensible answer and then you have a lambda you have a d. So this substitution will give you there will be a d square term in the numerator you will have a sec square theta because of the differential of tan theta and then you will have a sec cube theta term with a d cube. So you have some cancellation going on there is an integral this will simplify to an integral of just cosine theta and then you can plug these values in. So I mean the integration part is not very difficult the key part to note is that these components have to be taken correctly. So that you get the correct direction for the electric field I mean right now I am not written the sign but it is understood that the vector field E is E times rho hat. So the final expression I mean you can verify it yourself now that the expression turns out to be lambda by 2 pi epsilon naught times d and l by 2 by square root of d square plus l square plus d square plus d square by 4. So one important thing to note here in this expression is that I mean the expression itself does not seem quite remarkable but when you look at I mean this is not a part of the question but I think you should try and examine this as l goes to infinity this expression becomes lambda by 2 pi epsilon naught d. So this is the correct limiting expression that you obtain using Gauss's law but in case of Gauss's law since you need symmetric considerations you also need that the charge be infinite and that is exactly the limit that we have taken. So that is the point I would like to mention. You have to explain how to find out the value of electric field the radial point using Coulomb's law. But one thing I want to mention that we may use Gauss's law also here because you have mentioned last line but l tends to infinity that is for infinitely long charge length lambda it becomes lambda by 2 pi epsilon naught d and you mentioned this is Gauss's law but this is the expression of electric field applying Gauss's law for infinitely long charge line. Correct. But you may use Gauss's law in general to find out electric field also here. So if you think that you can. Should I apply it to find out force, force of interaction in a conventional manner and Gauss's law is employed to find out electric field. Actually you know it is basic level. So what let me just point out a small thing like Coulomb's law as you correctly pointed out Coulomb's law gives the force between two interacting particles. What happens is when you drop off the term where so this is what Coulomb's law is k q 1 q 2 by r square where r is given by r 2 minus r 1. So you have these vector signs and you have a mod r square term and there is an r hat expression going on. So I mean if you want to look. So this is what the expression for Coulomb's law is but in this case what happens is if you just drop off the term for q 2 what you get the remaining part is the electric field. The force on a particle is just q times E right. So the force on particle q 2 is q 2 times the electric field due to q 1 at that point. So I mean what I meant by Coulomb's law in this case is just computing the electric field without the computing the force on the particle. You cannot use Gauss's law in the situation of such a finite charge because what happens is that say you have a finite charge and you try to use Gauss's law. So obviously the answer will not turn out to be correct but the reason for that is the electric field at this point and the electric field at this point they will have a different magnitude. What we have computed here is only at the perpendicular bisector. We have not computed it at a general point which you can try to do and you will check that the answer is different. So in case of Gauss's law what you do actually is you take a shortcut when you are doing this integral you say that you know that the value of E in magnitude is constant and it somehow normal to d s or say it is something else and you take this out of the integral and you just integrate over the surface because you know that the value is constant. But in this case if you try to draw a Gaussian surface it is the value of E is not constant. So you cannot use Gauss's law to solve the situation when the line charge is just finite. I hope that answers your question. If you consider in our various textbooks it is given that in different applications of Gauss's law where linear distribution of charges have a distribution of charges like that. I have seen for linear distribution of charges they use Gauss's law to find out a particular radial point. The next point is an infinitely long charge conductor. So you use this concept here also. Yeah so that is what I said that you need symmetry. If you do not have the symmetry like in the case of an infinite charge you have full symmetry along the z axis that is why Gauss's law is useful. But in the case of a finite charge it is not useful. You do not have the symmetry. No infinity is a special case. Forget it. Just consider a symmetric charge conductor we mentioned earlier. So here we may use Gauss's law. Whatever be the answer but our concept should be concentrated in a Gauss's law because we are calculating electric field not the force. So I mean what you are going to do is you cannot as far as I understand using this integral form of Gauss's law you cannot directly compute E because you do not know the value and it is different at different points. You need to know there is some in homogeneity right. In some situations yes. But finding out electric field. Yeah. Consider Gaussian surface at any point on the surface we will calculate electrical at the point. That is correct that is what I am saying but it is only applicable in some situations because of symmetry. It is not applicable in all situations. It will be applicable. I mean you can apply it. I mean it is not incorrect. The thing is you cannot solve the integral if there is some in homogeneity that is what I am saying. You cannot solve this integral because you do not know the value of the expression itself. If you are trying to find out the magnitude of E then in that case you need to know that E is constant. You take E out of the integral and then you find out E because you know the surface integral. If you do not have such a symmetry you cannot find out E explicitly using Gauss's law. I hope that answers your question. Thank you. Center 1299. Dronacharya College. Equally in this very questions the problem is that we cannot apply the Gauss law directly because the Gaussian surface is not well defined. If we take the infinite line of charge then we can define the Gaussian surface and we can apply the Gauss law. Until we define the Gaussian surface how we can apply the Gauss law because the charge distribution is finite and at the boundaries the nature of the Gaussian surface will change and we have to apply the condition boundary conditions to solve the problem. Right. And these boundary conditions are not well defined. That is why we are applying the Knomb's law to calculate the electric field. That is one way to put it. What I just tried to say is that I mean in principle you could always draw this surface. The problem is not with drawing or defining the surface. The problem is with taking E outside the integral since because of boundary conditions as you pointed out or by symmetry you do not have E to be uniform. So, that is why you cannot simplify the integral. That is what I am trying to say. I understand your point. Do you have any other specific question? Rest is alright. Thank you please. What is the limitation of Coulomb's law? So what is the limitation of Coulomb's law? So in principle it does not have any limitations. So I mean at least for electrostatics and you do not have any moving charges. So Coulomb's law applies in situations where all of your charges are static. That is one restriction. Otherwise Coulomb's law is not strictly correct but you can still apply it in the limit of slowly moving charges. That is one limitation. Apart from that since you have a nasty integral to be performed every time when you are solving it, I mean you could think of that as a limitation but that is not a limitation of the law but a limitation of your calculation prowess. Otherwise there is no other limitation for Coulomb's law. And for Gauss law? For Gauss's law is also correct. Or Gauss law? All the things are correct. The problem with Gauss's law is that it can be used to compute the electric field only in a situation where you have such a high symmetry that when you draw Gaussian surface, you find out that the magnitude of E on the surface is constant and in that case you can find out the value of E. But in case you have inhomogeneities, especially when you have finite charge distributions like the one I have drawn, in that case you cannot actually use Gauss's law. You can check Gauss's law if you find out the electric field to check whether it is satisfied but you cannot compute E using it. So that is the limitation of Gauss's law. Does that answer your question or did you have something else in mind? Thank you. 1192 MGM's college. Sir, my question is you taken the cylindrical coordinates limits from 0 to R and for the theta 0 to 2 pi. Why you taken the 0 to 2 pi? Theta varying from 0 to pi. So what is the reason behind that? So in case of cylindrical coordinates, what you have is say you take a cylinder. So this is of radius R. So when theta in cylindrical coordinates, if you look at it from top, then theta goes from this is theta. So theta goes from 0 to 2 pi. It is just your ordinary circle. In case of spherical coordinates, I think which you are confusing, in case of spherical coordinates what you have is theta, when you measure theta it is measured from the z axis. In this case we are measuring theta from the x axis. This is cylindrical and this is spherical. So in case of spherical you are measuring theta from the z axis. So and you have phi, the angle phi goes from 0 to 2 pi. Sometimes the notation changes, but phi goes from 0 to 2 pi for spherical and theta goes from 0 to pi. You can check that 0 to pi. You can check that as you keep going down, the phi will keep circling around and as theta goes from 0 at the z axis and as theta goes to pi, you will have get the points around the negative z axis. So that is why theta goes from 0 to 2 pi only for spherical, but in this, in the cylindrical case you will have 0 to 2 pi. Sometimes some authors write theta or some authors might write phi in cylindrical, but I hope you get the idea. Basically the angle I am representing. Does that answer your question? Sir, my answer is that sir, theta is from cylindrical and as well as spherical 0 to pi. But in both the cases the meanings are different. In case of what the notation I am using here, the angle theta in case of spherical is angle between from the z axis and downwards. In case of cylindrical, the notation that I am using here involves the angle from the x axis around the circle entirely. So that is why to cover the circle you need theta goes from 0 to 2 pi for the spherical coordinates. So according to your calculation sir, I am not getting the pi r square h cross square r. Sir, what are you getting? Your problem, second number problem is there. For that I am not getting the that answer pi r square h cross square. You mentioned that answer is pi r square h cross square. So I mean we have already checked the calculation. Let me just show you the calculation. So you have already seen the calculation done using one part. So in this case, do you have a problem with this calculation? Yes sir. Okay, so there is some problem with this one. As far as I can see, there does not seem to be any problem. But I mean you could check it with your colleagues and if something still comes up, I mean you can double check it. Otherwise if your colleagues also agree that there is some mistake here, please post the question on Moodle so we can discuss it later. I mean please post your solution. Thank you sir. Yeah, 1178. Sir, my question is what is the physical significance of Coulomb's law? What is the physical significance of Coulomb's law? So the physical significance of Coulomb's law at least the way it was formulated is that say when you are trying to find out say you have a charge q1 and you have a charge q2, the most important thing to find out is what is the force of interaction? What is f12 or alternatively or f21? If one is known, the other is known by Newton's third law. So what Coulomb's law does is it gives you the force. What the electric field, the abstraction introduced of the electric field is that if you have a particle f and if you have say with some charge q2, then the force on it f divided by the charge q2 must be the electric field at that point. So Coulomb's law gives you the force, if you just divide it by the value of the charge at the point and you think that the charge is very small. So it need not be small, but if you divide it by the value of charge then what you get is the electric field. So you can think about it in two ways. In the original form Coulomb's law gives you the force. If you think of it in this way, then you can think of it as something which gives the value of the electric field at the point and the force is just the electric field times the value of the charge that you place at the point. So basically if you have some charges, you need to quantify the interaction between the two. Coulomb's law is one way to quantify that interaction. Does that answer your question? Sir, actually practically where you can use it? Practically when you can use it. So let me try to give a, I mean if you want a very direct example then say if I do not know if you have heard of the Vandy graph generator, the spelling might not be correct, but you can check it up. So in this case you have a charge sphere and you need to be able to understand the electric field distribution around charge sphere to for different so that you can understand how the particles, different charged particles will move. I mean Coulomb's law in its simplest form, I mean if you do not have Coulomb's law then you cannot compute electric fields and if you cannot compute electric fields then lots of applications say when you have electron guns or something else. Basically when you are trying to move charged particles using an electric field, you need Coulomb's law. So any case where you are trying to accelerate or decelerate a charged particle in principle. Does that answer your question? Yes sir. Okay, thank you sir. So let us go ahead, try to solve a few more problems. Let us look at the 8th question right now. So in this case the question is find out, obtain an expression for the potential due to a long charged wire by using the expression for the electric field. Since we already talked about the situation for a long charged wire, I mean this is just sort of one step ahead. So you can find out the electric field using Gauss's law as pointed out and then the step is to find out the potential and there is some interesting thing to be noted when you try to find out the potential. So just try to do this computation, it should not take much time. So have a couple of minutes and try to compute the potential for a long charged wire and find out the electric field using Gauss's law. Not that the wire must be, in this case we are taking the wire to be straight. So let us look at the solution. So in this case you have a long charged wire of charged lambda and as I pointed out you can use a Gaussian surface. So again by symmetric considerations what will happen is that the electric field on these surfaces, on the circular surfaces will go to 0 and the magnitude of the electric field must take constant and the electric field must be pointed out. I would think that this question has already been discussed. So let me not go into more details. But so this is the first expression since the question is asking you to start out with this expression. Let me just, so this is the expression that you start out. So this is the electric field and I mean this is just sort of a plug-in question. I mean there is nothing, so this is the relation that you have E equals minus del V. But I mean this expression is correct but the problem is with this expression is that you need to somehow invert it to find out V using E right. I mean there is no fundamental issue with the expression itself. Problem is that you need to write down V in terms of E to compute V if you know E. So if you invert this expression this is what you get. So this is the value of the potential V and if you try to do this integral say from one, you just try to do this integral in an indefinite manner instead of indefinite integral. So what you typically do in such problems is that you compute the definite integral where you take some point at r equals infinity and V equals 0. That is what you do. You put the potential V equals 0 at some infinite point. But in this case since you have an infinite charge if you try to do this thing then you will end up with some singularity because as you integrate over rho you will get a logarithm term and you will generate an infinity. So if you just do the indefinite integral so if you do not put the limits then this is what you will get. So you should not let me just write rho. So in this case you should not be putting rho equals infinity otherwise the potential will blow up and you will have to put up an infinite constant to get a 0 potential at infinity. So the thing to I mean point to emphasize is that not in all problems can you put V equals infinity at rho equals infinity usually you can do this. But in the case of infinite charges it is not so obvious that such a thing should be allowed and this is just a simple example which demonstrates that if you try to do something like this then you cannot write down the potential correctly. You can choose some other point say at some distance r say you can put V equals 0 at r. But you cannot choose r to be infinite otherwise you cannot write down a proper expression for the potential. It does not actually matter whether the potential is say if there is some potential V and you write it down you write down a new potential V plus V 0 equals V prime where this is some constant then both of these things represent the same physical electric field. So it does not matter but in this case the point which you usually take it for granted that V equals infinity at rho equals infinity is not true in this situation. I mean you cannot do the same thing when you have an infinite charge. So let us look at a few other questions. Let us look at question number 10 a conducting sphere of radius r has a charge q calculate the force that is exerted on the northern hemisphere of the sphere by the southern hemisphere. So let me just draw a diagram of the problem. I mean it is not too complicated but so there is some net charge q and say q is greater than 0 for simplicity it does not matter. So the northern hemisphere will be repelled by the southern hemisphere. So this is the force on the northern due to the southern and it will be in the upward direction. I mean obviously since it is the same charge sign I mean q greater than 0 does not matter even if q is less than 0 they will repel. But I have just taken this for now. So you have to find out this repulsive force. I have already given you a hint by pointing that f should be along the plus z axis but you should try and argue why it should be so and you need to compute in this question you need to compute the magnitude of this force. So this question is slightly long. So again you can take 2 to 3 minutes to try to solve the problem and then we will discuss. So let us now look at the solution. So the first hint I have already given it to you that the force will be along the z axis. We will use cylindrical coordinates for this problem. So we will use cylindrical coordinates but still you can talk about the z axis of course and so the next thing to do is you try to find out the force. So I am sure you already talked about this in class that you can find out the electrostatic pressure. I mean this concept is quite useful right now at least for this problem. So electrostatic pressure outside the surface of the sphere will be sigma square by 2 epsilon naught and once you have this pressure next thing to do is you look at some surface element say at some angle theta and again you have your another angle as say there is another angle which you call as phi. So this some angle phi and so the surface area of this element over here will be r square sin theta d theta d phi. So this is the area and the electrostatic pressure because the electric field is parallel to if you represent this you can write this as some elemental area where you have some normal vector going out. So in this case the normal vector has to be the radial vector. So e has to be parallel to r hat which is parallel to d s. So the force, force is just pressure times area since the force is normal to the area and in this case you will have sigma square by 2 epsilon naught times r square sin theta d theta d theta d phi. Again the key point to note is that since this spherical symmetry what will happen is that say let me just redraw the figure say you fix some phi angle and this is the theta angle. So in case say this point corresponds to phi equals phi 0 then this the opposite point corresponds to phi equals phi 0 plus phi and the force over here this will be the value of the force will be in this direction and in this case the force will be in this direction. So the only value you need to compute is the component of the force along the z direction because the component of the force in the plane will cancel out and this component this is the angle theta. So you just need to multiply by a factor of cosine theta. So the magnitude of the force the net force if you try to compute it. So this will just be say you write this as f. So this will just be f times cos theta. So this is say d f because you have a d f and you have an integral from again theta goes from 0 to pi because in this case we are again using spherical coordinates and phi goes from sorry 0 to pi by 2 because the force is only being computed on the upper northern hemisphere not on the southern hemisphere. So use theta equals 0 to pi by 2 please note this and phi goes from 0 to 2 pi. So the only thing that needs to be done is to compute this integral and sigma you can find out sigma using the charge since q is given. So sigma into 4 power square which is the area of the sphere will give you the charge q. So this integral you can rewrite this in terms of q as q square by 32 pi square epsilon non r square d phi. Since there is no dependence on phi the integral for d phi will just separate out easily and then you have theta term again note that the limits of integration are from 0 to pi by 2 because you need to find the force only on the northern hemisphere that is what ask in the question. You have a sin theta term appearing from here then a cosine theta from here and d theta. So again you can check that this integral again you can simplify it using the trick I showed earlier d of cosine theta is minus sin theta d theta and the expression that you will finally get is q square by 32 pi epsilon not r square intuitively if you would have done this integral from 0 to pi then you would probably end up with some wrong answer which does not make much sense because the only thing you are trying to find out is the net force on everything. What you should correctly do is the integral from 0 to pi by 2 since you want to find out the force on the northern due to the southern hemisphere 1176 university. If you find the electric field you can assume both gas line, column line same condition is it right sir question number 5. Question number 5 by symmetry you only need the radial component yes. Radial component not in the transverse component sir you have to apply both the gas line, column line in their one condition is it right sir. Now what I did was I only applied Coulomb's law to find out the electric field in that situation what I said is that if you take the limit where the line charge becomes infinite in that case the answer matches with that for Gauss law with infinite charge that is what I said I did not use Gauss law. I am just saying that the two agree when you are taking the limit for L tending to infinity you cannot apply Gauss law to solve the problem in case of finite charge this is what I pointed out. You can solve it for the infinite charge you can solve it for finite charge also for Coulomb you can also solve it for infinite charge using Coulomb but what I did was solve using finite charge take the limit and check that it matches with Gauss's law. One more what is the significance of physical significance Coulomb's law? I already I think I answered it maybe you did not get it so the physical significance is that say in any situation if you want to find out the distribution of electric field say I gave one example of the Vandy graph generator or say anywhere where you are trying to accelerate or decelerate charge particles say your cathode ray tubes. So you need to find out the proper distribution for the electric field and in that case you need to resort to Coulomb's law. Thank you. 1, 3, 3, 4. Siligree Institute. First of all would you please explain that electrostatic pressure thing how did you derive that and secondly was it done the question number 10 was it done for a hollow sphere or solid sphere will do the same. Sorry if I did not clarify it so in this case the sphere is hollow sorry for not clearly specifying it the sphere is hollow and for the expression for electrostatic pressure if you have not already discussed this in class. So one way to think about it is that I mean right now I am just giving you a shorthand way of thinking about it is that so when you have the surface of the sphere. So near very close to any point the electric field outside and inside what will they be you can if you go very close to the surface in the infinitesimal limit you can sort of apply symmetric considerations and you can write down the electric field over here a sigma by epsilon naught and you know that the field inside must be 0 correct. So because how did I write this down so for this case the electric field E dot the elemental area ds must be equal to the charge which is sigma ds divided by the electric epsilon naught. So in that case the idea is that very close to the surface this must be sigma by epsilon naught must be the value of the electric field and inside it must be 0. Yeah I was more sort of confused because we did not mention hollow sphere. Okay so I was assuming that this is already done but if I am not discussed it I mean this is sort of a shorthand explanation of it otherwise you in principle you could also solve the problem using Coulomb's law it just becomes very complicated. This P is also related to the energy density so if I mean that is another way to look at it but I am sorry I was not sure if this wasn't done in class so I apologize for that. Anyway thanks. Yeah okay 1 2 1 5. So this is going slightly off topic but the question asked here is say in Coulomb's law the constant that is used is 1 by 4 pi epsilon naught and this is what you write down I mean just for the magnitude of the force so this is the magnitude of the force. So in this case the constant actually depends on your system of units what we use is usually mks units meter kilogram second in this case the constant turns out to be 1 by 4 pi epsilon naught and you know the value of this constant is roughly equal to 9 into 10 power 9 and there are some units attached in case of Gaussian units or what you call as CGS units you can look these up online in that case Coulomb's law the magnitude of force is just written as Q1 Q2 by r square there is no constant beside it so the value of this constant is purely due to the what units you are using in this case the units of Q1 and Q2 are in Coulomb's but in this case the units of Q1 and Q2 are in ESUs the units of r are in centimeter so depending on your system of units the constant which appears beside the Q1 by Q2 by r square term it will change. Okay sir my second question is that say what is the physical significance of F7 okay so as I just showed you in different unit systems the value of this number I mean it does not fix you could very well think that you changed the value of epsilon naught and you made it 1 by 4 pi or something you can think of it as a mathematical manipulation. The point of epsilon naught is that when you consider different epsilon's say you have thought of you have been thought dielectric media so in case of dielectrics you have a different epsilon which is not equal to epsilon naught and if say Q1 and Q2 Q1 and Q2 are interacting in say some medium field with dielectric say you have some dielectric then the force that they experience is affected by a factor which is depends on the dielectric constant kappa. So in this case basically the force changes right by putting in the dielectric so what epsilon naught can be thought of as it is just the property of the vacuum since there is no other material inside alternatively you can think of it as just appearing because of the chosen system of units if you choose a different system of units then there is no epsilon naught coming. So there is two ways to think about it if you will. But the question is that permittivity of the free space what is the mean of permittivity of electric field? So permittivity since it is appearing in the denominator you can think of it that if say if epsilon naught whatever the current value is if epsilon naught were much greater than that then F would be much smaller. So if you just want some physical intuition you can think of epsilon naught or the corresponding epsilon is how much it allows the electric field to spread or how it diminishes the strength of the electric field that is one way to think about it. If that is. Then my last question is divergence of any vector field cell what is the physical mean of divergence between the graphs? In case of the divergence of a vector field say let me just sketch out a vector field I mean for simplicity. So in this case what you have is that there is no notion of any sources or any sinks appearing in the vector field contrast that with a scenario where you have some spread like this. In this case the flow of the vector field shows you that there must be some point or some area inside in this region where you have a flow outside or you can also have a flow inside. So what divergence measures is the increase or decrease in the flow coming from that point. If there is no divergence then you will just have flow lines coming in and out but there is no source or sink for the vector field inside. Okay sir thank you sir. So let us look at question number 21. So part of a long current carrying wire is bent in the form of a semicircle of radius r calculate the magnetic field at the center of the semicircle. Again in this problem I mean as has been a recurring theme in the problems I have been trying to solve is that there is some kind of symmetry. So you have got to try to figure that out and find out in this case it is a question of magnetic field instead of electric field but so instead of Coulomb's you have Biosauer law but the basic idea is the same that you have some configuration in this case of current instead of charge and you got to find out some dynamical quantity. So you can find out the so just ignore this line for a moment. So we need to find out the value of the magnetic field at this point right. So using Biosauer's law you know that when you try to find out B you have an expression like this mu naught by 4 pi there is an integral over the line and there is dl cross r by r cube. So the important thing to notice that since there is a cross product for these lines the contribution at this point will be 0 and for the curved part the value of the magnetic field will be into the plane so that we denote by a cross. So since current is moving in this direction so you use the right hand rule but so you already know the direction of the magnetic field and it is not I mean this problem is not very difficult. What I just like to say that instead of doing this integral this integral actually very simple you can think of one alternative way to solve the problem is that since there is no x or y component there is only a z component which is going inside what can be done is you rotate this system by pi radians so that you get an opposite and what will happen is that the magnetic field of this combined system if you take an infinitive systemally small distance between these two is that the magnetic field will just double because the only the z component is getting added there is no x and y component. So if you find out the magnetic field for this system then the magnetic field for the half system will be exactly half since there are no other components and in this case as you take this distance to be infinitely small this system is actually just equivalent to a current loop where you have a current going around of magnitude i at some distance r in that case you already know what the expression should be I mean it is quite straight forward to compute it using Biosawa's law since in this integral r does not change only the vector sign changes but the magnitude does not change. So in this case for a current loop you have B equals mu naught by mu naught i by there will be an expression of 2 r and there will be a minus z hat sign but so in case of this system since we only have half the system this is just an imaginary part right so which I added and I said that the field would get doubled. So this is for the circle so for r system the magnetic field will be just half of this which is mu naught i by 4 r times z hat. So this sort of trick only worked because there was no field along the x or y direction if there was some field say you would have some other configuration where you would have some field along the x and y direction then in that case what will happen is that those fields will not get added parallely you will need to compute a vector sum and those might cancel out so you will end up with an incorrect answer. In this case the net field was along the z direction by rotating again the new part also had the field along the z direction and the field due to both of them is the same that is why we can use the symmetry of the problem to find out the field at this point. Otherwise also using Biosawa's law is quite trivial in this problem but you could in principle have a slightly complicated configuration here in case but if you still figure out that the x and y components are 0 then you can use this trick to find out the magnetic field at the center 1092 KIT college. My question is about Gauss theorem as it is actually a scalar product of electrical intensity and surface area where surface area is generally considered to be as a scalar quantity but in that law or in that expression we show it as a vector. So how a surface area will it be considered as a vector? It is because of only it is very small we considered or is there anything other? So that is a good point that you have raised. So let me just try to illustrate. So one thing you have asked is that how can surface area be given some a vector because a vector has some direction right that is what the difference between a vector and a scalar is that a vector has some direction associated with it. So one thing as you correctly pointed out is that on some surface patch I mean the idea of the normal as the area direction of the area the normal will differ at different points. So it does not make really much sense to give an area vector to a combined thing you can of course you can integrate over small parts and that is the catch. So when you are writing in Gauss's law you have the expression integral e dot ds where you only take an infinitesimal surface area patch and for that patch you have a proper normal which makes sense. So that normal can be assigned as the direction for the area. Another catch that Professor Gauss mentioned in his lecture is that the surface must be orientable. This is another requirement if you want to use this because in the case of a mobius strip which you can look up in the lectures is that the mobius strip if you travel around it then I am sorry if my drawing is not that good. So if you travel around it then the normal will change direction and you cannot define an unambiguous normal even when you take an infinitesimal surface patch. In case of say a simple object like a sphere you can take an infinitesimal surface patch you can go around do whatever you please but you can still have a consistent direction for the normal but not so for a mobius strip. So those are two things that you need one of course the surface area needs to be infinitesimal so you assign it a direction which is the normal at the point and the other thing you need orientability which is key. So you mean to say that the electrical field vector that should be always normal? No no no no no the normal what I mean is a surface normal the electric field need not be normal what we are finding with the dot product is the normal component of the electric field. What I am talking about is n hat the surface normal vector you can always take a unit vector which is perpendicular over here you can also think of tangent vectors inside the plane right. Say you have T 1 and T 2 which are non-parallel tangent vectors then in that case n hat will be parallel to T 1 cross T 2 because T 1 and T 2 correspond to tangent vectors in the plane and n corresponds to the normal which is perpendicular to both of them. I am not saying E must be one more question I have sir that is about the magnetic field and electrical field whenever we show we always consider it is tangential to the lines of force the direction of those. So is it because only convention or there is something? No that is precisely the way you talk about the field lines the field if you draw the electric field say you have an electric field configuration and then you sketch out the electric field that field lines say this is the field then the field lines themselves are sketched out by joining these points it is not that you do not draw the field lines out of air what you understand intuitively is that the field should be in this direction you draw it at every point and then you draw the field lines. So by construction it is true that the field must be tangent to the field lines 1 1 0 0 m e s p like sir can you sketch the magnetic field for question number 21. So this was the current configuration that we had okay and the current was traversing in this direction. So I mean the general magnetic field it might not be that obvious what direction it would be since it is a three dimensional I cannot possibly draw the three dimensional part but what one can do at least for the z component one can visualize it in a simple manner because you have these currents. So the field at these points it should I am talking only about the z component not talking about the entire field. So at these points the field should be somewhat like this as you would find for an individual thing but this is not the entire field configuration because it will since it is a vector field you have a three dimensional thing. So visualizing it is not easy and for the symmetry we chose the particular point which was the center of the circle which was drawn I have only drawn the z component over here okay thank you.