 Those who are online, please type in your names. Are you guys able to hear me? Okay, only two people online. Who all are online? We'll wait for some time for others to join in. Okay, you can see a question in front of your screen. Okay, start solving. Welcome everyone. Sanjana, you did not tell me your percentiles? I got many of yours. What happened? Which subject was the least percentile, Sanjana? See guys, if your least percentile is chemistry, then it is really sad because chemistry is something which you can improve very quickly and chemistry is something which is very easy. So you can improve chemistry very quickly. So don't lose out on chemistry at least. Maths and physics, I can understand it takes time to improve. But if your chemistry is something which is worst hit, then you need to pay attention there. Last year, by this time, many students have stopped coming in. Just like four or five students used to attend the classes, those who were in the advanced batch. So because of this online medium, you are not forced to travel and save a lot of your travel time because towards the end, it's all about problem solving. So me sitting here and solving problem with you remotely or in front of you, it doesn't matter. Anyone got the answer? Mr. got the answer. When you put the charge, the distance between the masses will increase. Today I have taken between mains and advanced level questions. Last time we took very easy questions. So I am making it a little difficult. A few of you have answered. Let me check the final answer. Meanwhile, all of you please attempt this one. Okay, let's solve this one. So we have two metallic blocks resting on a frictionless horizontal surface. They're connected by a light metallic spring of spring constant 100 and untratchable length of 0.2. Total charge is placed on the system causing the spring to stretch to equilibrium length of this much. The value of charge Q is what? Now, you can see that both masses are like similar. You know, both the pieces, both the metallic pieces, they are similar in shape and size and conductive and everything else we assume to be same. So the whatever charge you put over here Q, okay, will be divided among them equally. So it will be Q by two and Q by two will be gone because of the charge getting spread out equally. They are same same sign charges. Because of that, the repulsion will happen and the spring will be stretched. Okay, so if you just look at one of the masses, you know, we just look at one of the masses, it will experience a Coulomb force. That will be equal to K Q by two into Q by two divided by the distance between the charges, that is 0.3 square. Fine, this is the Coulomb force and in the opposite direction, there will be a spring force on it. That will be equal to K times delta X. Okay, extension in the spring is 0.3 minus 0.2, which is 0.1. Okay, this K and that K is different. This is a spring constant and that is one by four times from naught. So in a hurry at times, we tend to cancel out similar expressions, but they are not same, this K and that K. So spring constant is given at 100. So this is 100 into delta X, that is 0.1. Unstretched length is 0.2 and the stretched length is 0.3. Okay, people are getting different answers. So this force, spring force, should be equal to K Q and Q two by R square. So K is nine into 10 raised to the power nine. Okay, this into Q square divided by four into 0.3 square. Okay, so, okay, so here you'll have two zeros gone. This is 10 raised to the power seven and then you have four into 0.3 into 0.3 should become 0.09 into 0.1. This is equal to nine into 10 raised to the power seven into Q square. Okay, so you can clearly see that a factor of two will come, isn't it? Q square, see this nine and 0.09 will cancel out and the square is four. So when you square root it, it could become two but then it also depends on 10 raised to the power. Okay, so just be careful. It will be 0.01. So this is 10 raised to the power minus three. This is 10 raised to the power seven. So Q square will be equal to four into 10 raised to the power minus three into 10 raised to the power minus seven. So yes, Q is equal to two into 10 raised to the power minus five coulomb or 20 micro coulomb. So that's our option C is correct. Okay, so if you hurry up, you will get it wrong. Understood all of you. Any doubts? I'll call you later. Okay, I'm going to the next question now. This one, screen is blurry. So do one thing. You know how to increase the resolution of your YouTube video? See, you do it like this. Any YouTube video would like that. Suppose you want to watch this one. Okay, so you can click this. Okay, this gear thing you can click and you can increase the resolution. So all of you please increase the resolution up to 720p. Okay, do not watch it on lower resolution. Okay, fine. So I'm getting different, different answers. Should I attempt now or you guys are in between now? What will happen if you bring these two metal close to each other? You're having equal and opposite charge densities. You're bringing them close to each other. What will happen because of that? Any charge distribution will happen. When you bring them close to each other, you will have a situation like this. They'll, how they'll neutralize until as you touch them, they won't be neutralizing. But the positive charge on this plate will attract negative charge on that plate. So positive and negative charge will come equal and opposite like this. So it will become like a capacitor. Okay, now we need to find electric field over here, here and there. We know that electric field outside the capacitor is zero. Okay, so there are no charges on this side and that side. So it is exactly like capacitor over here because of this positive charge density, electric field will be like that. And because of this negative one, it will be like this. Okay, because of the positive one, it will be sigma divided by 2 epsilon naught and because of negative, it will be sigma divided by 2 epsilon naught in opposite direction. So it will get neutralized over here, electric field. Similarly, electric field will get neutralized over here. But over here, because of the positive one, it will feel like this inside. As in between the plates, electric field because of the negative one will be like this and because the positive one, it will be like that. So they will get added up and hence it will be C. Clear? I think this similar question we have done many times. We'll move to the next one now. Feel free to use the books or any reference material you have for recollecting the formulas or any concept which you want to know. Although I am expecting that you don't need that since we are towards the end of the preparation. Anyone? D. Just visualize how the dipoles are placed. You'll be able to do that. So draw x and z plane. One dipole is at the origin. Like this, it is oriented along the z axis. So this is p k. This is 0 comma 0. And the other dipole is 1 comma 2. So 1 comma 2 probably over here. This is p by 2 k. And we need to find the electric field at 1 comma 0 comma 0. So we need to find over here the electric field, this point. Now try this. You'll see that this point 1 comma 0 comma 0. This point is along the equatorial line of this dipole and along the axis of that one. No one? Look at the formulas. Just add up to electric fields. Electric field due to the first dipole and electric field due to the second dipole. Tell me, what is the electric field along the equatorial plane for a dipole? How much it is along the equatorial plane? Do you know this? Yes, minus of k p by r cube. So this is along the equatorial plane. And e2. Along the axis, the electric field is k 2p by r cube. This is axial. So for this dipole, this point is equatorial. And this point for that dipole is axial. So e1 is minus of k into p. p is along k. So it will be like this. Divide by r cube. r is 1. So 1 cube. This is e1. e2 is k into 2 times p by 2 unit vector k divided by what is this distance from here to here? It is 2. So 2 cube. This 2 is gone. So when you add up e1 plus e2, so you will get k into p common 1 by 8 minus 1 k. That is minus 7 kp by 8 k times. So that is why option B is correct. k is 1 by 4 pi epsilon 0. Any doubts? No doubts? Yes. That's how the formula is. When I say electric field is this along the equatorial plane, I mean along all the points. I don't specify it. And when I say electric dipole moment is this, it is along the p vector throughout the axial line. And in case you have any doubts, you can always fall back onto the basics. I don't know why. Don't have that confusion. If there is any confusion, please fall back on basics. Don't try to memorize. See, this is plus and this is minus. This is dipole moment vector. Take a point over here. Electric field due to minus will be like this. Electric field due to plus will be like that. But due to minus, it will be stronger. So net net electric field will be like this. In the direction of dipole moment, over here, electric field is due to this, due to positive and in that direction due to negative. But this point is closer to the positive charge. So net net electric field direction is along the direction of dipole. This one. Whenever an object is moving in a circle, the first equation you write is net force in the direction of the center should be equal to mv square by r. Okay. Okay. Good that many of you are saying the same answer. Electric field due to the infinite wire is what lambda divided by 2 pi epsilon not r. Right? This is electric field. So because of this electric field, the force, the negative charge will feel towards the center is q into e. Okay? So q lambda divided by 2 pi epsilon not r should be equal to mv square by r. Okay? So r goes off. v is equal to q lambda divided by 2 pi m epsilon not to the power half. Okay? So time period is what? 2 pi r by v. All right? So when you do all this, you will 2 pi r divided by v. So, okay. So k, k is 1 by 4 pi epsilon not r. So you'll see that option a is correct. Okay? Next one now. Solve this. Should I give you a hint? Or you guys are in between? See. Nobody else? You have to use a Gauss theorem. It's a direct application of Gauss theorem. Get the electric field. Okay? Step number one. Get the electric field. And then evaluate the expression. Okay? Let me solve it now. Suppose this question says that a system consists of a uniformly charged sphere of radius capital R. Okay? This is that sphere of radius capital R. And let's say it has charged capital Q. Okay? Now the surrounding medium is filled with a volume density of this much. Okay? So everywhere outside, outside of this, okay? Has a volume charged density of rho is equal to alpha by R. Alpha is a positive constant, and R is a distance from the center of the charge. The charge of the sphere for which electric field outside the sphere is independent of R is what? Okay? Now let's first try to get the value of electric field itself. See the thing is it is a spherically symmetrical case. So electric field has to be radial. Okay? So if electric field has to be radial, let's assume a Gaussian surface which is spherical in nature. So this white circle represents a sphere which is Gaussian surface of radius small r. Fine? So what does Gauss law says? Gauss law says that integral of E.ds is equal to Q enclosed by epsilon naught, right? Now electric field has to be radial because of symmetry. So since electric field is radial, electric field makes zero degrees with the area. Okay? Because of that this integral E.ds simply becomes E into ds. Okay? And also because of the spherical symmetry, the magnitude of electric field throughout this sphere is constant. So E comes out of integral. So it simply becomes E into ds integral. So that is E into surface area. So this left hand side becomes simply E into 4 pi r square. Okay? This is equal to Q enclosed by epsilon naught. Now I need to get Q enclosed. Okay? Now what are the charges that are enclosed? Charges enclosed are charged on the sphere. Okay? So there is uniformly charged sphere. Let's say Q is at charge plus you have now volume surface, volume charge density of alpha by r. So alpha by r into 4 pi r square dr. Okay? This you need to get, you need to integrate this from capital R to wherever you're finding till the distance of r. You are able to understand, right? Any doubts? Quickly type in. Okay. So you're able to hear me clearly, right? Power went off. Should be some time back. So this is 4 pi alpha integral of r dr. Excuse me once again. But you are able to hear me, right? Now we are back. This is Q plus when you integrate, it will become 2 pi alpha r square minus capital R square. Okay? So this is equal to Q plus 2 pi alpha r square minus capital r square, okay? So electric field will become Q divided by this multiplied by 1 by epsilon naught as well. So this is, yeah, option C is correct. Q divided by 4 pi r square plus you will get alpha by 2. 4 pi r square alpha by 2, yes. Minus 2 pi alpha r square divided by 4 pi r square. Okay? Now I need to find out the value of, I need to find the charge of the sphere for which electric field is independent of r. Okay? Now what are the terms that are depending on the r, small r, so 1 by r square if I take common, it will be Q divided by 4 pi, okay? Minus, so this will be alpha by 2. R square by 2, okay? So this plus alpha by 2. Now this bracket term, okay? This bracket term, if this bracket term becomes zero, electric field do not depend on r, right? Because this is electric field and only this term is depending on r, okay? So Q by 4 pi when this become equal to alpha r square by 2, then electric field do not depend on r. So this will be equal to 2 pi alpha r square, the value of charge Q. Should I move to the next question now? Solve this one? Okay, Amogh has answered others. See, whenever you get such kind of questions, always try to see whether you can find out some series of parallel connection, okay? So are you able to see some series of parallel connection immediately? This wire, this wire I can treat as if it is a point. I can take this and join there. I can remove this wire and put this point over here. Nothing will change, okay? So the same thing can be thought of as this capacitance C, parallel with C and this C is parallel to that C, okay? These two are in parallel and those two are in parallel, fine? So you can remove this wire and join this point to that point. Similarly, you can remove this and join point this to that point, okay? So now these two are in parallel, so it becomes 2C. Here also it becomes 2C, okay? So you will get this is C, this is C, this is 2C and this is 2C, okay? This point A comes down, so this is A. This point B goes there, so this is B, okay? So now 2C and C, they are in series, okay? So between these two, the equivalent capacitance will be 2C by 3, okay? Now this is 2C by 3, this is 2C by 3. They are connected parallel. So two times of 2C by 3, so that is 4C by 3, okay? Now tell me whether you guys know this. If let us say you have a situation like this, between these two points, between A and B, you need to find equivalent capacitance. This is C1, this is C2, this is C1, this is C2 and this is C3. What will be the equivalent capacitance? Correct, so Kirchhoff's law, Kirchhoff, yeah? So this is Wheatstone bridge, okay? And not only just a Wheatstone bridge, it's a balanced Wheatstone bridge, so you can eliminate C3, okay? So I think we have discussed this in the class as well. Now try doing this one, C1, C1, C2, C2. Try solving this one, okay? So clearly, I mean, don't solve it using brute force, okay? Here also something similar to Wheatstone bridge you should be using, okay? This is extended Wheatstone bridge, extended or expanded, okay? So this is C4 only. So you can get rid of these two, okay? These two you can get rid, all right? So I can remove this, now it becomes simpler, all right? So remember this, not only you can balance in this kind of scenario, you can also balance the bridge in this kind of scenario, okay? So now you can solve it, right? C1, C4 and C1, they are in series and that is in parallel with C2, C4 and C2. C5 is not there at mess. C5 would complicate the situation. This one. Okay, so you have three concentric metallic spherical shells. Ready A, B and C, so this is A, B and that is C, okay? So surface charge densities are given for A, B and C, this is A, B and that is C, okay? We need to evaluate these options, you know? So first let us see what is potential at A, okay? So before starting to find the potential, let us try to see how we can write potential in terms of the potential of A, B and C. So we have the potential of surface charge density sigma, okay? So when you take a spherical shell, all right? You have radius R and you have given it a charge Q, okay? Inside potential will be equal to Q divided by four pi, epsilon not R, fine? Q is what? Sigma into four pi R square, so this will be nothing but sigma into R divided by epsilon not, okay? This is inside, okay? There is an assumption that C is the largest radius, but it doesn't matter, no, it may matter actually, but then yeah, we are assuming that A is the smallest radii, B is the next one and C is the larger one, okay? But it should have been written, okay? And but then if it is not written and if you have to assume, then you assume the logical thing. Q is again sigma into four pi capital R square divided by four pi epsilon not R. Try that sign here after the class and let me know whether you're getting the same answer, okay? Sigma R square by epsilon not R. So this is outside potential. So we can quickly derive it, you know, the inside and outside potential is this and in terms of sigma and then go on to solve the question, okay? So potential at A is what? Inside all these spheres, okay? A is inside of all these spheres, so potential at A will be equal to potential due to charge on A plus potential due to charge on B plus potential due to charge on C, okay? Due to A, it is sigma A by epsilon not, okay? And due to B, it is minus sigma B by epsilon R because sigma is minus sigma, okay? And due to C, it is sigma C by epsilon not, okay? So had it been minus over there, then it would have been a correct option but right now it is not correct, all right? Now let us look at B. For option B, potential at B, this point is, you know, this point is outside the innermost sphere, okay? So if it is outside the innermost sphere, this formula will use for the innermost sphere because that point is outside the innermost. So V1 potential due to the innermost sphere will be sigma into B square by epsilon not A, okay? V2 is sigma into, for V2 I can use this or that, it doesn't matter, it is on the surface of the second one, okay? So sigma into, this is minus sigma into B divided by epsilon not, okay? And V3 is sigma C by epsilon not. So you can, this is, no, there is a little bit. So this will be sigma A square by B, all right? So that is the option B is correct. Any doubt, anyone of you just quickly revisit whatever we have done, is this thing clear? It should have been given that which radius is higher, which radius is lower, but it is not given. So I'm assuming the way it is usually in the question, the innermost radius is given first. These questions are difficult than those who came in your J mains, right? I'm assuming that, okay? So this VB is the potential on the surface of the second sphere. So that point lies on the surface. So you can take either of those two formulas, okay? Just substitute the values, you'll get it. And all of you use paper and pen, don't just watch it, okay? You also try it. Ramcharan is saying B. I just suppose you score 96 percentile in physics, 96 percentile in chemistry and 96 percentile in, let's say max, what could be your overall percentile? It will be 96 only or not? Okay, good that you guys are clear about it, okay? It will be closer to 98 higher, like 98.7 or something, but definitely more than 96, definitely more than, it can't be less than, understood? So you can see that some of you I've seen, it was really sad that some of you could score 97.6 percentile in one subject, 96 percentile in other subject, and the third subject you score somewhere around 85 percentile, okay? So right now your percentile could be hovering around 95, okay? But if this third subject, if this third subject also, if you would have got 96 percentile, you would have been 99 percentile above, okay? So if you work on this subject, in which you got already 97.6 percentile, okay? The scope of improvement is very less. So keep on working hard on the same subject, it not yield you, you know, there is not too, I mean there, although there is a scope of improvement, but not much. But if you work on this, the scope of improvement is tremendous. So this 85 could be, you know, if you're getting 97 and 96 in two subjects, you know, you are sure to get more than 92, 93 percentile in the third one also, because aptitude level is there, okay? This is your, if you get skewed score like this, in which two subjects you're getting very good and third one very bad, it is purely because of your carelessness, okay? So, you know, and at times you may get shocked that, okay, throughout the year I thought that my physics was strong and I got least marks in physics. Let's say throughout the year you thought my chemistry was the strongest and you got least percentile in chemistry, okay, it is because of your exam temperament, okay? It is not because of your knowledge. You need to understand these are two different things, okay? So if you have stopped taking mock test or you are like a person who doesn't feel good to take a lot of mock test, you just run away from taking mock test, then your exam temperament will not be there to, you know, translate whatever you know in marks. So don't, I mean, if you think that there's so many more marks you would have got, but you could not get because, and then you start blaming that on your silly errors and blaming it on, you find a reason that I could have spent lesser time on, you know, some questions I could not have done that, all that comes if you keep your temperament, you know, with respect to exam ready, okay? So make sure that doesn't happen in the up rail setting, okay? And if this 85% is your chemistry, if this is your chemistry, 85% is your chemistry, I mean, this is a crime, okay? You must, I mean, it's so easy to improve chemistry. You should, you must be like, you know, all of you should aim to get 98% above in chemistry. You know, there is no excuse. You should aim to get more than 98% tell in chemistry, okay? So make sure you, I mean, these are like, not a rocket science, you know, you might know it already. I'm stressing the same fact again, okay? And remember, no amount of discussion or no amount of introspection or research, you can't just, you know, take few documents and start researching and thinking that some miracle will happen and due to my research, my marks will go up by 50 marks or 60 marks. That will never happen, okay? So the path is very straightforward and obvious. So you need to take that path. Okay, attempt this, all of you, should I solve? Four identical charges are placed at the four vertices of a square line in a YZ plane. A fifth charge is moved along X axis. So X axis comes out of your screen. Your screen is your YZ axis. Potential energy along the X axis is represented by what? The variation of potential energy. Now, here again, the assumption is that the fifth charge has a same sign as the other four charges, okay? Now, when X is zero, the potential energy will be maximum, right? And when X tends to infinity or when X tends to minus infinity, will potential energy go towards zero? Answer me this. Will potential energy will go towards zero when the X tends to infinity? See here, there will be two types of potential energy. Potential energy of system, A, B, C, D. This remains constant throughout. This never becomes zero, all right? Potential energy between this and this, this and that, this and this, this and this, this and that. So it will remain, it will be there, okay? The potential energy will never become zero, even though this charge goes to infinity, but distance between this charge and that charge, that is not becoming infinity, right? So again, this guy has ignored that fact. In fact, here we should have a situation like this. B option should have been like this, you know, your X axis should be like this and it will go like that, fine? This much is this potential energy of the system, okay? Otherwise, in this particular question, it should be written potential energy between the fifth charge and the system, okay? Because potential energy will not become zero just because one of the charges goes to infinity, okay? So option B is correct if it is like this. Otherwise, this is not a correct question. Understood? I'll go to the next question, not this. Somehow the time just ran away, it's one and a half hours already. This one, why not D? Okay, I did not see the D and this, okay? See potential energy, see here, it is going towards infinity. When X is equal to zero, okay? You need to understand that when you move the position of this charge, potential energy doesn't change suddenly. It is not very sensitive. You can slightly move up and down, potential energy more or less remains same, okay? So you can evaluate, you know, many a times we think that we think that physics is mathematics. So we try to get the expression and somehow get the exact function of the graph and then see which graph is correct. But what is expected is in physics, that you visualize, you feel what is happening and using that you answer these kinds of questions because if you start getting the expression for the graph, you know, it will take hell lot of time and you may not get it also, okay? So start using visualization in physics and more so in graphical question. So if there's a graph, it is expected that you will visualize a scenario rather than mathematically solving it. Even if it stopped there, another point is that it can't suddenly change. It has to have a zero slope, potential energy has to have a zero slope at X equal to zero. There again, you need to atomize balance assume electric field is E, then charge into electric field is equal to M omega square R. So you get electric field in terms of R and then integral E dot dr, minus of integral E dot dr is change in potential. So just take one electron and balance the forces. One more thing, guys, that start looking for a variety in questions. Don't count that, okay, today I've solved 100 questions, 150 questions. Just on a look out of some good varieties of questions, you can, you don't need to solve entire book. Right now the stage is towards the end. So you just browse through in the internet or look at whatever books you have. Even Etsy Verma for that matter, if you look at the exercises, towards the, I mean in between or towards the end, you can find very, very interesting questions. And one more thing, I have created video solutions of entire Etsy Verma on rigid body motion and full optics. So if you want, you can come to the center and watch it. You know, it is around eight to 10 hours videos there. So you can watch it in a stretch of three hours per day. And while solving, I realize that some of the questions are very nice. Should I solve this one? Yes, simply it is work, work energy theorem, okay. Work done is change in counting energy plus change in potential energy. Fine, I'll quickly solve this. Four identical particles, each of mass m and q are kept at the four corners of a square of length L. Final velocity of this particle after setting them three is what? So work done is u2 plus k2 minus u1 plus k1, okay. There is no other force doing work other than electrostatics for which you are considering the potential energy. k1 is zero, u2 is zero, okay. I want what? I want the final velocity. Final velocity will happen when the particles will separate from each other and go to infinity. So that is why u2 is zero. u1 is four times k q square by a plus two times k q square by a root two, okay. So there, this is the potential energy initially, fine. k2 will be equal to half into mv square into four. Now you can say that there'll be gravitational potential energy between the two masses. That is very, very less. So we are ignoring. We are just taking electrostatics potential energy, okay. So just solve this expression and you will get the option, okay. So many of you saying option is correct, but whatever it is, okay. You have to solve this one and you'll get the answer for velocity. All right, let's go to the next one. This one. Option A, how did you approach this question? Initially, the electron stick to it and charge it. Charge it, sorry for that. Yes. Others you still stick with a, w is zero, k1 is what? Half m into v0 square, okay. Now these electrons are coming from infinity. So u1 is zero, okay. And when at the steady state, what does it mean steady state is that the charge on the, charge on the sphere doesn't change with time. So k2 is zero because if electron speed becomes zero just before reaching the sphere, the charge will not get into the sphere. So basically what is happening is that the charge on the sphere remains constant. So k2 has to be zero and u2 is, the charge just reaches the surface, I mean about to touch. So u2 is k, q1, q2, so that is, let's say q was the final charge on the sphere, k, q1, q2, which is electron now. So qe divided by r, okay. So this is u2. Now one more thing you need to understand that whatever charge comes on the sphere, it's a negative charge, okay. While solving, you may get q as positive, but actually it is negative, it is electron. So that is why all the options have minus sign in it. So k is one by four pi epsilon naught. So let us expand that. So four pi epsilon naught. So we'll substitute and equate that to zero, so you'll get mv naught square, fine. So q will be equal to, is two, so two pi m epsilon naught r, v naught square divided by e, okay. So it is option A, there'll be minus sign also. What happened Amog? Oh, changed again. All right, any doubts on this question? Anyone of you? This one. So let us keep the level of questions at high because when you solve higher level questions, the easier one you'll anyway start feeling comfortable with. So if you prepare easier ones, even easier ones, you may not be very comfortable with. So let's keep it at advanced level going forward, even though we'll solving less number of questions, but the amount of concepts involved will be more than, if we solve double the number of questions, easier ones. Okay. Both A and C. Others, it is just X component that is given, okay. Y component electric field is not there. Like C is this point, C is not that point. Potential difference between A and B is point related. But when you talk about C, there's a potential difference between B and C also. Vertically as well, there can be electric field. Sorry, not between B and C. What I'm trying to say you understand, right? We are just taking care of X component of electric field. So when you integrate from A to B, you're keeping Y constant. So electric field will not change along Y direction. But if you go from A to C, there's a potential difference for that. You need to account for electric field along Y axis as well. Fine. So C cannot be the option, okay? Only A is the option, okay? Between A and B, oh wait, it's a metallic body, right? E axis is given. So V A minus V B. If it's a metallic body, then potential at A should be equal to potential at any point over here. And potential at B should be equal to any potential over here. So potential integral ex dx, so 3x square plus 0.4 dx. And my x goes from zero to 20 centimeter, which is 0.2. Getting 0.088 volts. This is a potential difference between A and B, okay? And VB should be equal to VC. Component of field is, this is potential difference and the equivalent opposite charges are given, which is 0.88. So capacitance is Q by V A minus V B. So 0.88 microcouplem divided by 0.088. So you will get 10 microfarad, that's correct. But even option C should be correct because they're metallic bodies, right? So potential here should be same as potential there. Yes, both ANC are correct. It's a metal object, okay? Fine, let's go to the next question now. Zero, no, it cannot be zero. You remember Vendigraf generator? Use the basics to solve. J will never test you on remembering direct formulas or something like that. You need to apply the basics. So potential at the common center will be due to the, will be because of the, because of the both the spheres, right? So just add a potential due to both the spheres at the center. That will be the total potential of the center. Okay, let us see. Q is distributed over two concentric hollow spheres such that surface charge densities are equal. So let's say Q1 and Q2 are the charges. So Q1 plus Q2 should be equal to capital Q, okay? And surface charge densities are equal. So Q1 divided by four pi epsilon naught R square equals to Q2 by four pi epsilon naught capital R square. So this is Q1 by, let's say, which one is smaller R? So this is equal to, this is I'm equating surface charge density. So there will be four pi epsilon naught term will also come in, but that I'm ignoring, okay? So from here, you'll get the value of Q1 and Q2 using solving these two equations, okay? And potential will be potential due to the sphere one which is Q1 divided by four pi epsilon naught R plus due to sphere two that is Q2 divided by four pi epsilon naught capital R, okay? So when you simplify this, these two get the value of Q1, Q2 and substitute there you'll get the total potential, fine? And the answer is A, okay, fine? So continuous two hours like this. So right now it is 630, okay? We will, I mean, I will come back after 15 minutes, okay? So this is my break time, not yours. So you should solve this question, okay? Solve this question once I come back if you happen to solve this question quickly take a break and come back before 6.45 we'll start 6.45, all right? Okay, now I must be audible. Sorry, I could not. That's why I was wondering why you guys are not replying. Okay, fine, so this is the situation, right? D minus A, from D minus A the separation goes to D plus A. So I need to find out the capacitance, all right? Anyone of you tried? No one? Okay, so you can utilize this entire situation like this, you know, that you can assume all this width is DX. Okay, and this one, this is Y, okay? So just give me one second, okay? So A is the edge length of the square plate, okay? I'm assuming that, okay? Can you tell me what will be the capacitance of this capacitor? This small capacitor that I have considered, what will be the capacitance for that? Quickly? Yes, it's not ADX by DY, fine? ADX is the area of this portion of the plate, okay? It goes A length inside and DX is the width, okay? Now the problem is that there are two variables, DX and DY. So DY, can you get in terms of X? Y, DY, yeah, it is just Y. Y is the separation between the plates, it's just Y, sorry. Can you get Y in terms of X? I have been trying to do that for last 15 minutes. Arey, Atmesh, why? We're not able to do that? Okay, you can use similar triangles, no? Draw a line like this and another line like that. These small, small tricks, you will understand when you solve a lot of questions, okay? So these are like routine tricks, lack of problem solving or lack of varieties of problem solving. So if I solve, let's say 20 questions with you, you should solve 100 questions your own, okay? The multiplying factor should be five, that is minimum. Okay, so this distance is X, okay? And this distance is A. Why I say A? Because D is very less compared to, so D is very large compared to A. Wait, what is D? Yeah, wait, wait, wait, wait, wait. This cannot be, let's take it as L, okay? Take it as L, now solve it. The question is incomplete actually. So let's assume it as L, now can you solve it? In question, it is not clearly specified whether A is the side length of the square plate or D is the side length of the square plate. So that is why there is a confusion. But anyways, the question is good, you know? So focus on the concepts. This distance is how much? This one, that is A, and this distance is also A, okay? Because from D minus A, it should become D plus A, okay? So we have A by L equals to, let's say this distance is B, it will be equal to B by X, okay? So B will be equal to A, X by L. Yes, actually that's what I thought later on, but then it is still a guess, Shweta, okay? So it should be clearly specified D is the edge length. But anyways, we are assuming A to be edge length and L is given. So let's continue like that only. I hope you are getting the concepts. I don't care about the final answer. Concepts should be focused here. You may not get any of the options because we are assuming something else probably, okay? So B is this. So the Y is what? Y is D minus A plus two times of B, right? There is B down also. Yes, on the, that's what I'm saying that the question is incomplete, but since question is nice, we are assuming something and solving it anyways by assuming something and our assumptions may not be correct as in what the author may intend. So two A by L, X, okay, so this is Y. The question is nice, so I don't want to skip it. So we will substitute the value of Y over here. That is in terms of X. So I get DC in terms of X now entirely and you can see that the capacitors, how they are connected, all these small, small capacitors, they have the same potential difference, isn't it? So it will be as if they are connected parallel to each other. So summation of this is equal to equivalent capacitance and when you add up all the continuous variables with tap biting differential elements, okay? It becomes integral. So when you integrate this expression, you'll get the answer for equivalent capacitance, okay? So like this, you have to approach these kind of questions, okay? The way we have assumed probably we'll get a log term also in our expression. All right, let's go to the next one, this one. See, we may not be able to cover all the varieties of questions in just today's class. So I hope you'll be solving a lot of questions your own, okay? So these kind of routine that I set in that this week we'll be doing capacitance and other topic will help you to focus your efforts. So you know this week you have to do electrostatics and some other topic. So like that, just follow the plan which we have set in, okay? Not, you know what Mesh, we have taken L, right? We have taken horizontal length itself we have assumed and that's how we have taken it, okay? We have used similar triangles. L is the, X is the horizontal and so A by L is B by X. So we got B, Y is equal to this plus two B. Theta doesn't come in picture, okay? If question would have been clear, you could have solved till the end. Solve this one, this is also nice. I hope you're able to see many different types of questions today. So that's what you have to do. Just look for the varieties. Yeah, Mesh, I understand your argument, but that'll be a very, very small length. Okay, D? No, that's not correct. That's not correct. Yes, Meshra got it correct. Okay, should I solve? Fine, let me solve this now. See this, let's, this switch be closed. So let's say the battery gives the charge, X amount of charge, okay? So this particular plate will have Q plus X and how much charge will be there on that plate on the left, so on the right plate, how much charge will be there? If on the left plate, it is Q plus X. I am talking about total charge, okay? I'm not talking about charge between the plates. If Q plus X on the left plate, what will be the charge on the right one? It'll be minus X. Yes, okay? So between the plates, what is the electric field? This is sigma Q plus X by A. This is sigma by epsilon naught, sigma by two epsilon naught. This is because of the left plate plus because of the right plates. That is X divided by two A epsilon naught. Fine, so this is the electric field between the plates. Understood, so this will be equal to one divided by, let's keep charge separately. So Q plus two X like this, okay? So potential difference between the plates will be what this electric field is uniform. So E into distance between the plate will be the potential difference, okay? So D divided by two A epsilon naught into Q plus two X will be the potential difference. All of you understood till now, any doubt? Okay, how much this V should be equal to? This should be equal to what? This should be equal to E, the source, Kirchhoff's loop rule, okay? Epsilon naught A by D is capacitance. So I can say that Q plus two X divided by two C is equal to E. So Q plus two X is equal to two C E, okay? So X is a charge, right? So X will be equal to Q by two minus, minus X will be this, Q by two minus C E, okay? In fact, on the right plate, minus X is the charge only, so minus X is this, okay? We'll go to the next question now. Use the conservation of energy. It's an important question. How did you do Sondarya? Not actually. Okay, should I solve or you guys are in between? Work done is change in potential energy, change in current energy, plus heat that is liberated. This is the hint. We ignored heat earlier because we only used to conserve mechanical energy, okay? But if you just use plain and simple energy conservation, then work done will be change in potential energy plus change in current energy plus some part of work will be gone as heat. Who's doing work? Battery is doing the work. How much work battery does? Potential difference into charge that is flowing through the battery. See, work done will be with the battery's potential which is E two, okay? Into the charge that is flowing through. Change in current energy has no meaning over here. Change in potential energy will be there in the capacitor only, fine? So change in potential energy will be equal to, let's say Q two is the charge on the capacitor finally. So Q two square by two C is a final potential energy minus Q one square by two C. So this is change in potential energy, fine? And one more thing you must notice over here that earlier it was negative polarity and it is like this. Now it is positive and negative is over here, okay? So earlier let's say it has minus Q one like this later on this minus Q one has become plus Q two, fine? So total charge that must have flown through the battery is Q one plus Q two like this. Understood? All of you able to get it? Q one is C times E one, okay? It was connected from here and Q two is C times E two. So just substitute these values, you'll get the answer. Quickly tell me what is the answer you're getting? Who does the work here battery and how much work battery does? The work done by the battery is the charge that is flowing through the battery, okay? Multiplied by the potential difference of the battery. So Q into V is the work done, right? If charge flowing through a potential difference of V the work done is Q into V, okay? So that is what it is. Earlier charge on the capacitor was minus Q one on the upper plate and now the charge has gone up to plus Q two. So the total changes first Q one get neutralized it becomes zero and then plus Q two. So Q one plus Q two must be flowing through the E two battery so that the charge on the upper plate of the capacitor becomes plus Q two from minus Q one. What is the answer? Silly, Reshweta. Yes, you get B as the answer, okay? So have this observation now when you solve questions related to heat liberated in a capacitor, you know have this observation that, okay I have done like this. Can I solve similarly each and every question? So you'll notice that when you do change in energy in the capacitor, when you say that is equal to heat, okay? There you'll see that work done by the battery will be zero. Okay, so you need to notice these small, small things. All right, Silly, I would rather take one more question because it is mathematics from here, all right? Just try it yourself once again. Okay, okay, let me do it. Amok, can you share the screenshot with me? I'll share that same screenshot with everybody. Have you solved, right? Amok, we'll take one more question. I'll share the screenshot of what Amok did with you guys. Simon, you can also share the screenshot with me. I'll share with Shweta and Sanjana, all right? Okay, we'll take one more question. We have some time. This one. Fine, Shweta, I'll share the screenshot. Shweta and Sanjana, if doubt persists, let me know. I'll solve and share with you, screenshot. Solve this question. Last question for today. Amok, that you can do after the class. I'll share it after the class. Solve this question first. See, oh, is it that simple? All of you getting see? Oh, it is straight forward. See, flux is nothing but integral of E dot DA. And luckily, this is also equal to Q enclosed by epsilon naught. So Q enclosed is epsilon naught, sorry. Q enclosed is equal to lambda times the length, this length. Now, this length, how will you find y is given? So this is radius, this is radius, this is 90 degree. So this distance, this distance is root over r square minus y square, fine? Double of that is the total length. This into lambda is the charge, okay? Divided by epsilon naught is the flux. Q enclosed by epsilon naught, so, yeah. So option C is correct for this. Fine, so straight forward question. So we have to do one more, not this, this, that you take as homework, this one. Equipotential surface should be perpendicular to electric field, right? You think that B is perpendicular to it? So field, what is the slope of the electric field? Slope of the electric field is minus one by root three, fine? So electric field is like this. So we need a line with slope of root three first of all. Multiplication between two slopes should be minus one. Slope is tan of theta. So tan of theta should be root three for the equipotential surface. So theta should be 60 degrees. So it cannot be B, it cannot be A. That is 30 degree, all right? Now let's see whether, you know, this is lower potential or this one is the lower potential, fine? So electric field is going in this direction like this, okay? And along the direction of electric field, the potential decreases, right? If you have any doubt, you assume a positive charge like this. So electric field lines are like that. So potential decreases along the field line, right? So that is why option C is correct. Any doubts? Any doubts on this question? Quickly tell me. Am I audible now? Okay, so what I was saying was that those who got 97 percentile or close to 98 percentile, they can straight away start focusing on J-advanced, okay? And those who got very less percentile in chemistry, it's a cause of big concern, okay? Because that is something which is not acceptable. You need to do pretty well in chemistry. It's an easy to score, all right? So do all that and, you know, I know that many of you will be concerned about your marks, but, you know, you have to show your concern. If you are very concerned, you are not very happy with your marks, please call me, okay? And discuss with me your strategy. I cannot call, let's say, 70 to 80 students, you know, one by one and address your concern. The help doesn't come like that. You need to ask for help, okay? So just let me know if you are confused about your preparation about your plan one by one. You can message me that you want to discuss your strategy with me and I'll be happy to help, okay? So that's it for today. We'll meet again. Bye-bye.