Hello everyone, I hope you are all fine and today I have brought you a one shot video of electrostatic potential class 12 physics. As always, we will finish this whole lesson in just one video and after watching this whole video, the concepts of electrostatic potential will be crystal clear. So I am Roshni from Learnohub, the free learning platform where you can study physics, chemistry, maths, biology, everything absolutely for free at learnohub.com. So, in this video we will finish electrostatic potential and in the next video we will study capacitance. So, I hope all of you are ready. Let's get started. So, in the last video we talked about electric charges and electric field. Today we are going to talk about what happens if a charge is in an electric field. It experiences a force and this force displaces the charge. When force is applied on any object, displacement moves from one point to another. Force or displacement when they come together, then what happens? Work done, exactly. And this work done, this gets stored as potential energy of the body. So this is our general story. Now we will see that our story of electrostatic potential is very similar to the story of gravitational potential. You will ask how? Remember what is gravitational force between any two objects? We had studied Gm1m2 by R square in gravitation. On the other hand, if I ask what is electrostatic force between two charges? Remember Coulomb's law which was read in the last lesson which says that electrostatic force is KQ1Q2 by R square. If you pay attention to both the expressions, then what will you see? that gravitational force as well as electrostatic force both are inversely proportional to R square. That means both of them follow the inverse square law. Not only that, both of these forces are conservative forces and because of this, the story of gravitational potential and the story of electrostatic potential is very similar. So, if we talk about gravitational potential, then what used to happen in that story? Something like this used to happen in that story, that if I have a ball at this point right now, if I want to take it from this point to this point, then what will I have to do? To take it from here to here, I have to apply some force on this ball. By applying force, displacement is happening in it. Some work is done by combining force and displacement. I am doing some work on this ball. And this work is stored in it like potential energy. So when it is reaching this point, the potential energy in it is basically the work we have done to bring it from here to here. Now when I will leave this ball from here, then the same potential energy will be converted into kinetic energy. This is how our story in gravitation used to be. Think in the same way here in the case of charges. When a charge is inside an electric field, if I want to move that charge from one point to another point, then what will I have to do? I have to do some work. What do you get without doing work? Nothing. So I have to work hard. So if I have to take that charge from one point to another point, So, I have to work against the gravitational force. Because I didn't want to use gravitational force. Similarly, if I want to take a charge from one point to another point against the electrostatic force, then I have to do some work. And this work gets stored as electrostatic potential energy. So, in simple words, this is the story. Do you get it? Did you understand? Let's move on. So, let's look at it in detail. Let us suppose we have a charge Q. Now I have to take this charge from point R to point P. I have to take it from point R to point P. Now you will ask me, ma'am, you just said that this is not possible without work. So, yes, I know that I have to do some work. But how much work will I have to do? Who will tell me? How much work will I have to do? So, kids, here is a very simple logic. See, I am taking from R to P. So, the potential difference between R and P, I have to work that much. It is like that I have 100 rupees in my pocket and I have to buy a book whose price is 150 rupees. So, how much extra money will I have to invest? 50 rupees. So, the difference between the amount in my pocket and the price of that book, I will have to invest that money. So, I will have to invest 50 rupees from my side. See, in the same way, here the two points, we have to go from R to P, then the difference of potential energy between these two will have to work if we want to take that charge from R to P. Did you understand the thing? You must have understood this in simple words. So now if we look at it mathematically, then what can we say that electrostatic potential energy difference between R and P because we are taking it from R to P. So, my final point is P and my initial point is R. So, let us suppose that the potential energy of final point is UP. Let us say that the potential energy of initial point is UR. So, UP minus UR is equal to WRP. WRP means work done to take the charge from R to P. So, with this, children, we can define electric potential energy difference. Right? Because, now we have talked about potential energy difference. So, electric potential energy difference is basically The work done by an external force to take the charge from one point to another point in an arbitrary charge configuration. So, to take any charge from one point to another point, the work done is equal to the potential energy difference between the two points. So, children, now we have understood what is the difference between electrostatic potential energy. Because we have talked about difference. But how will we define electrostatic potential energy? How will we define it? If we have to define only potential energy So think carefully What did we just say? That the work done to take any charge from one point to another point Like work done to take a charge from R to P is equal to the potential energy difference between these two points Now if I modify this a little bit and say that suppose I am taking this charge from infinity to some point P. Do you understand? What I was saying earlier, I am taking it from R to P. Now I am saying that I am taking it from infinity to P. Okay? Now how much is my potential energy in infinity? Zero. So that's what we assume that the value of R that is infinity and 1 upon R becomes 0. So that's why potential energy at infinity is always 0. Okay, so if I see now, then my work done to bring the charge from infinity to point P, that is W infinity P is equal to what? UP minus U infinity. Okay, so what is U infinity? Potential at infinity is 0. So, if we have to define the potential energy, electrostatic potential energy, then we can define it like this, that work done to bring a charge. from infinity to a particular point P in an arbitrary charge configuration without accelerating the charge. So that will be my electrostatic potential energy. Okay? Good. Now when it comes to electrostatic potential energy, it is important to know how we denote it. We denote it with capital U. That is why I am writing UP everywhere, that is potential energy at point P. U infinity means potential energy at infinity. U R means potential energy at point R. If we talk about units, SI unit is joule. In other units, electron volt is another unit. So that's about electrostatic potential energy. So okay, till now what all we have covered? We have covered electrostatic potential energy difference. We have covered electrostatic potential energy. Now it is time for electrostatic potential What is electrostatic potential? So slowly words are getting less First it was electrostatic potential energy difference After that electrostatic potential energy Now electrostatic potential So kids we define electrostatic potential very easily So what is potential? Ability We often say that this kid has a lot of potential He will do something very good in future What is potential? capacity, ability, capability so that's potential. So here also the word potential means that how much capacity is there in any object that How much already stored energy is there? So that's what is potential. So how do we define potential? So electrostatic potential is work done per unit charge. Ache se samajhna hai work done per unit charge in bringing the charge from infinity to a particular point. Thik hai, toh matlab jaise humne electrostatic potential energy ko define kya tha exactly waise lekin potential energy per unit charge. So that is potential. So electrostatic potential ko hum denote karte hai capital V se. We denote potential energy as U and potential as V. What is the difference between both? Electrostatic potential energy per unit charge becomes electrostatic potential. This is a scalar quantity. So, now the time has come that we will see electrostatic potential energy and electrostatic potential. We will see both of them mathematically. We will see the mathematical expression of both of them. So, first of all, let's understand the situation. Let us suppose we have a point charge, let's say, plus Q. Now at any point P which is situated at a distance of R from this plus Q, at that point P I have to get electrostatic potential. So how much will be my electrostatic potential? So understand the logic of electrostatic potential, I had just defined it. Electrostatic potential means work done per unit charge means work done in bringing a positive unit charge. from infinity to the point P. From infinity to that point P, you have to do as much work as you can to bring a unit charge. That is the electrostatic potential. So, mathematically, the electrostatic potential here is KQ by R. How do we denote electrostatic potential? From capital V. Here, capital V is the function of R. Because as R changes, V value changes. So, what will be V of R? KQ by R. K is 1 by 4 pi epsilon naught. We know this. We are learning this from Coulomb's law. Isn't it? So, this is my expression of electrostatic potential. If we have a charge given at any point, I have to get electrostatic potential, so it will be KQ by R. Okay? Now, let's talk about electrostatic potential energy. So when we talk about electrostatic potential, we assume that we are bringing a unit charge infinity to a particular point. But when we talk about electrostatic potential energy, then we say that we are bringing any charge q to a particular point from infinity. So here electrostatic potential energy will be K Q into small q divided by R where we are assuming that we are taking this small q charge from infinity to point P. So, that is why my electrostatic potential energy will be U. So, U of R is equal to KQ into Q divided by R. Clear? So, looking at both the mathematical expressions, what is the difference between them? If you do potential energy per unit charge, then what do you get? Electrostatic potential. And let me tell you children, these expressions are super duper important because we will solve many numericals based on them. Based on them, our future potential story is based. So, that's why you have to pay attention to them. So, we have discussed three things till now. One is electrostatic potential, another is electrostatic potential energy and the other is electrostatic potential energy difference. Now, when we talk about electrostatic potential and potential difference, then many times confusion comes in the mind. We are talking about potential and potential difference. Why? Why are we talking about potential difference so separately? Because the concept of potential difference is physically more significant. Because the story is that when there is a difference of potential between two points, then only the work is done. So that's when the story is happening when there is a difference of potential between two points. So that's why potential difference is... Physically it is more significant than potential. However, electrostatic potential and electrostatic potential difference both have same units. We measure both in volts. Not only this, if we talk about infinity. On infinity electrostatic potential is always taken as zero. Why? Mathematically I had told you the expression what is electrostatic potential? KQ by R. So when we go to infinity then R's value is infinity. One by infinity is zero and therefore potential is zero at infinity. So children these three things we have discussed potential, potential difference and potential energy in terms of electrostatics you don't have to get confused in them. So children now we will see that potential at a point. Due to a system of charges, what have we seen till now? We have seen that the potential will be on 1 point P due to a point charge. If there is a charge Q, then the potential on point P, which is at the distance r, will be KQ by r, where K is 1 by 4 by epsilon naught. The story was simple. Now I said that I have a system where there are many charges. Q1, Q2, Q3, Q4, Q5, Q6 and so on. There are so many charges and now you tell me that point How much will be the potential at point P? So we will apply very simple logic. Here we will apply our old known superposition principle. Which says that potential at point P due to all these charges will be equal to potential at P due to Q1. plus potential at P due to Q2 plus potential at P due to Q3 and so on. So, means all these individual potentials, if you sum them up, then it will be net potential at point P due to a system of charges. So, means our potential will look something like this, KQ1 by R1 plus KQ2 by R2 plus KQ3 by R3 and so on. and that's potential at a point due to a system of charges. So, let's see how much potential there will be at an axial point due to an electric dipole. Now, what is an axial point? By now, everyone knows that. It means any point which is lying on the axis of this dipole, like this point P. and this point P let us suppose is at x distance from dipole and we know from dipole that all distances we measure from center of the dipole so this means this distance is x and this is my dipole so this is let's say plus q and this is let's say minus q and the distance between both is length of the dipole that is 2a So this is my situation. So what I have to find is, I have to find potential at this point P. Now the potential at point P will be 1 because of this plus Q and 1 because of this minus Q. So let us suppose we call this point A and this point B. So the potential at P will be 1 because of A and 1 because of B. So because of A, How much will be the potential at point P due to this plus Q? KQ by R, R means distance from the charge. So distance from plus Q will be X plus this distance A. So this will be X plus A. Potential at P due to B will be KQ divided by R means this distance. This distance will be X. minus a because this distance is also a so minus a from this whole x so this will be x minus a ok now what i have to find out is how much net potential will be on p so net potential at p will be potential at p due to a plus potential at p due to b now here we have missed one thing potential at p due to b will have minus sign because this q is negative charge Ok, now we will add both, so what will come? KQ by X plus A minus KQ by X minus A. Now from here we can take minus KQ common. So what will be left inside? Inside will be minus 1 by X plus A plus 1 by X minus A. So let's calculate it further. If we take LCM of these, then it will be x plus a into x minus a. Above it will be minus x minus a plus x plus a. Ok, I am doing simple maths. So, let's do further calculation on this side. So, it will be minus k cube minus x plus a plus x plus a divided by x square minus a square. So here minus x plus x cancel will be done. So this will be minus kq into 2a divided by x square minus. a square. Okay. Now, if we pay a little attention here, then what we are seeing is that charge into distance between the charges or length of the dipole, what we call it dipole moment. What is it? Q into 2a. This is called dipole moment. So, here we can replace it with p. So, it will be minus kp divided by x square minus a square. So this will be net potential at an axial point. Now there can be a special case of this. If I talk about a special case. What kind of special case? Suppose this point P is very far from dipole. That means value of X is very very greater than A. That means point P is very far from the length of dipole. So in this case mathematically, the net potential at the axial point will be x square because it is very big compared to a square. So we can write x square minus a square as x square only. Right, so it is like if you are doing minus 1 out of 1 lakh, then it is a very small reduction, almost negligible reduction. So in that case you can write minus Kp by x square. So, let's see in this way with this logic that at an equatorial point, what will be the potential at an equatorial point of a dipole? Equatorial point is any point that lies on the perpendicular bisector of any dipole, like this point P. So, let's assume that this point P is also at x distance from the dipole, where this point is measured from the center of the dipole. Anyways, and this is plus Q. this is minus Q, this is A and this is B and length of the dipole is 2A so this distance will be A and this distance will be A So this is our given situation. Now what will happen? The net potential will be on P. One will be because of A, that means because of plus Q. Second potential will be on P because of B, that means because of minus Q. So let's first find out the values of both. What will happen? KQ by R. What will be R? The distance between A and P, which is this distance. So, what will be this distance? See, what is this? This is a right angled triangle. So, apply Pythagoras theorem. This is hypotenuse. So, what will be this? Root over x square plus a square. Can you see this clearly? Perpendicular is x and base is a. So, root over x square plus a square. So, write it here. This will be root over x square plus a square. Similarly, let's find because here minus Q charge is divided by root over this will be this distance Pb and Pb value will be same root over X square plus A square because this is also a right angle triangle so this will also be root over X square plus A square So now when we calculate net potential, it will be potential at P due to A plus potential at P due to B. And if you see both the values are same but opposite sign which means if you add both the values then the answer will be 0. So basically at equatorial point of any dipole, the potential of net is 0. Now here we notice a very interesting thing because remember when we were calculating electric field. for electric dipole. So we had noticed that at equatorial point the electric field net that comes out is non-zero. But here we saw that when we calculate potential, then potential is zero. Now many students will think that how electric field is non-zero. Now I will not repeat that because in the lesson of electric charges and fields, we calculated the electric field at the equatorial point of the dipole. Those who have seen that video and if they remember, then quickly practice once, practice will also be done and it will be fun. So this was an important point to note. So, children, we have seen two special cases, axial point and equatorial point. So, if we generalize it more, that at any point, that is, if any point is P, that is, it is not that it is equatorial point or axial point. There is any point P which is lying at a distance of R from the dipole in this way, which is making angle from dipole theta. Take care. So, let's assume that this is my dipole minus Q plus Q. So, in this case, how much value of potential can we write? So, in this case, we see that our net potential can be written as KP cos theta divided by R square. So this is a general case. Now see how the axial and equatorial points that we talked about in general case also get applied here. So when we were talking about the axial point, so what was happening on the axial point, my P was lying somewhere here. Correct? In the case of axial point, my point P was lying somewhere here. So in that case, what was the value of my theta? Because P was coming here, so the value of this theta was Zero, correct. So when we were talking about axial point, then theta is zero degree. So in this case, the net potential will be cos zero is one. So it will be Kp by R square. So if you go back and see, just 5 minutes ago, the derivation we took out, the value of V net was Kp by X square. So here we have considered this distance as R. So that's why Kp by R square. Okay, now let's talk about equatorial point. So when we talk about equatorial, then my point P will lie somewhere here. Right? So when it lies somewhere here, then what will be the value of theta? In that case, the value of theta will be 90 degrees and cos 90 is 0, so the net potential will be 0 and that is exactly what we have derived for equatorial point. So, in general, if there is any point P which is making theta angle with dipole and is an hour distance away from dipole, then you... Net potential at that point is Kp cos theta by R square. So, let's try a question. Two charges 3 into 10 to the power minus 8 Coulomb and minus 2 into 10 to the power minus 8 Coulomb. Two charges are located 15 cm apart. So, let us suppose that here is a charge at point A. Here is another charge at point B. So, here is plus Q1 and here is minus Q2. and the distance between them is R. Here Q1 is 3x10-8 and Q2 is 2x10-8. The distance between them is 15 cm. This 15 cm is the value of R. At what point on the line joining the two charges is the electric potential 0? means which point is above this line where net electric potential will be zero. Now for now I don't know, so let's assume, let us suppose, electric potential that is 0 at point P. So, we assume any point P. Like I assumed that this is my point P, which is at x distance from the charge of plus Q1. So, if this is x, then how much will be the remaining distance? R minus x. Okay. Now we have to find out the net potential at point P should be 0 So first thing we have to find out is the net potential at point P The potential at point P will be due to A and B So individually first we will find out the values of the potential at P due to A will be KQ1 by X square Potential at P due to B will be KQ2 minus KQ2 divided by R minus X square. Sorry, X square will not be there. Potential is KQ by R. Right? So KQ by R. I put square by mistake. Okay. So this will be potential. So now let's talk about net potential. So then at point P, the net potential will be, potential at P due to A plus potential at P due to B. Okay. Means this will be KQ1 by X minus KQ2 by R minus X. according to question, the potential on point P should be zero because we have assumed that P is the point where electric potential is zero so as per that, according to question or according to our assumption we can say that this potential has to be equal to zero Now we will solve this. Now we will take K as common. If we bring it here, then K will disappear. So what is left? Here Q1 by X minus Q2 by R minus X is equal to 0. So from here I have to get the value of X. So what will happen? Q1 by X is equal to Q2 by R minus X. Now let's try to put the terms of x on one side. So what will happen? Q1 into R minus x is equal to Q2 into x. So this will happen. Q1 R minus Q1 x is equal to Q2 x. So we can write this. Minus Q1 x minus Q2 x is equal to minus Q1 R. or if we take common minus from here then what will be this? Q1 plus Q2X is equal to minus Q1R. So, this minus minus cancel is done. So, from here X is equal to what? Q1R divided by Q1 plus Q2. Now, see we have all the values given. What is the value of Q1? 3 into 10 to the power minus 8. What is the value of R? 15 centimeter. That means 15 into 10 to the power minus 2 meters divided by Q1 plus Q2. So, what is the value of Q1? 3 into 10 to the power minus 8. Q2's value is 2 into 10 to the power minus 8. Means the magnitude basically. So when we add both of them, then what will come? 5 into 10 to the power minus 8. So if we calculate this, then it will come 9 into 10 to the power minus 2 meters. Means 9 centimeters. And this will be our answer. So, let's see problem number 2. A regular hexagon of size 10 cm has a charge 5 microcoulomb at each of its vertices. So, first of all, what will we make? Hexagon. Hexagon is a polygon which has 6 sides. So, 1, 2, 3, 4. 5, 6 and regular hexagon, regular means all its sides are of equal length. Ok, it is saying that on every vertex it has a charge of how many coulomb? 5 microcoulomb. So let us suppose on every vertex there is a Q charge. such that Q is equal to 5 microcoulomb. Okay. Calculate the potential at the center of the hexagon. We have to calculate the net potential at this point O. Okay. So, now think logically. So, the potential at this point O, what will be the reason for that potential? That potential will be there because of every vertex. Means because of every Q. Right? So here how many Q are there? 1,2,3,4,5,6. 6 Q are there. So let's talk about 1 Q. So let us say this point is A. This is A. We consider this point as A. So potential at O due to A. We can calculate this. What will be KQ by R? R will be distance of this point A from O. So, in the case of hexagon, how much distance will this be? Let's calculate and see. Now see, since this is regular hexagon, I will make it complete because we will need all these lengths when we will take it out for everyone. Since all the sides of this hexagon are equal, the middle angle of this will be 60 degree. Because there are 6 angles in total, all are equal angles, so the sum is 360 degree. So, how much will be each angle? 360 by 6 means 60 degree. Apart from this, since this is a regular hexagon, so the sides of it, the lengths of it will be equal. Okay, this is the second thing. Now what is happening, if we focus on one triangle, Let's say if we focus on this triangle only. Here we can see that all three sides are equal. If both sides are equal, then both angles will be equal. That means both angles will be 60 degree, 60 degree each. So basically this triangle is an equilateral triangle. Okay? Do you understand till here? Amazing! So, because it is an equilateral triangle, then how much will be the length of OA? The length of the side of hexagon is 10 cm. So, the length of AO will also be 10 cm. i.e. 10 x 10-2 m. This will be potential at O due to A. Now, in the same way, potential at O due to B, C, D, E, F will be there because of all. So, the net potential will be the sum of all. Now see, because of sub, the potential will be equal. Why? Because the value of R, KQ by R, is the same for all. Because this length is 10 cm for all. Right? So, net V will be 6 times of KQ divided by 10 into 10 to the power minus. 2 so here we can put value instead of k 9 into 10 to the power 9 instead of q 5 into 10 to the power minus 6 coulomb because this is given in micro coulomb and 1 micro coulomb is 10 to the power minus 6 coulomb so when we calculate this then answer will be 2.7 into 10 to the power 6 volts so this will be net potential at the center so you will see that in these types of numericals formula to the go ekhi laga normal potential ka formula but it was more about ki ye jo values hai na jishe RK value hai ye aap kaise nikhaloge which was from simple mathematics chalo aage bade toh bacho ab hum baat karne wale hai ek bahut hi important bahut hi interesting and trust me exam ke liye super important hai ek concept ki jiska naam hai equipotential surfaces Equipotential means equal potential. A surface whose potential is same at every point. Now you will think that is such a surface possible? Think think think, it is absolutely possible. Let's take the example of point charge. If we have a point charge Q, then the potential at the distance of R distance from that charge is KQ by R. We know this. Now, We will talk about this point charge and let's say we say that this point charge is lying at the center of this ring. I am giving the example of this ring so that you can understand the concept. So if there is a point charge at the center, then at this point, my potential will be KQ by R where R is the radius. At this point, what will be my potential? KQ by R, where R is the radius. At this point, what will be my potential? KQ by R, where R is the radius. So, kids, if we see this whole surface, then the potential is equal because of this point charge. Is it or not? So, this means that if we talk about a point charge, then we will get spherical surfaces. They are all equipotential surfaces. And what type of equipotential surfaces are these? These are spherical equipotential surfaces. On similar lines, if we talk about a line charge, like a long stick which has a charge. So, for this line charge, how will the equipotential surface be? Like a cylinder? Look at the picture on the screen. This is my line charge and if you make a cylinder on it, then you will see that the potential of every point that lies on the cylinder is the same. Why? Because the potential is KQ by R. Now this cylinder has the same distance on the surface. So these are cylindrical equipotential surfaces. Now if we talk about electric field, let's say there is a uniform electric field just like these. So how will be its equipotential surface? Its perpendicular. So it is like if our electric field is going like this. So these are the equipotential surfaces because the potential on these surfaces is just the same. On this surface, everyone's potential is same. Similarly, on another surface, everyone's potential is same and so on. So these are the equipotential surfaces. So now let's see the equipotential surfaces in some specific situations. For example, I have a situation on this side where There is a system which has a positive charge and a negative charge. So when there is a system of positive and negative charge, then see how my equipotential surface is there. See the pattern is made. Where we are seeing that the equipotential surfaces inside are very close to each other. But as we are coming out, the gap between them is increasing. On the other hand, if you see here, then here in my system, both have like charges, that is, both are positive positive charges or both are negative negative charges. In their case, if you see the pattern of the equipotential surface, then it is quite different. So, children, can you guess that why these two different patterns are visible to us in both cases? Why are these patterns different? Why are we seeing such a pattern? Think about it, the concept of electric potential surface is linked to electric field. When we are talking about two charges, then what will come in the picture? Electric field lines, we had read in the last lesson, right? If there is a positive and a negative charge, like here, then electric field lines will come out of the positive one and enter the negative one. Right? So, if you see the pattern of their electric field lines, then that pattern will look something like this. And in the perpendicular of these electric field lines, we will get the equi-potential surfaces. So, due to this distribution of electric field lines, we get to see the pattern of equi-potential surfaces. On the other hand, when we see the electric field lines here, since both are like charges, then what happens is that the electric field lines disappear. So basically here they don't meet. So that's why the pattern of equipotential surfaces is of this type. So I hope with electric field lines you must be able to understand that in the case of opposite charges and like charges, why our pattern of equipotential surfaces is different. In the context of equipotential surfaces, one more important thing is that work done to move a charge on an equipotential surface is always zero. Means work done to move any charge from one point to another point on an equipotential surface is always zero. Why? According to definition, What is potential? Potential is work done per unit charge. So what can we write as work done? Charge into potential. Simple mathematics? Absolutely. Now let's assume that we have an equipotential surface. Equipotential means that every point on that surface has equal potential. Now I have to take a charge from point A to point B. So how much work done will I get? Q into potential at B minus Q into potential at A. Right? Now, potential at B and potential at A are equal. So, if we subtract both, what will come? Zero. That means, work done is zero. Logically, if you think that if I am taking any charge from one point to another point, So work done is done only when there is a potential difference between the two. For example, I told you that I have 100 rupees in my pocket. The book I have to buy is of 150 rupees. In that case, I have to pay 50 rupees from my side. But suppose I have 100 rupees in my pocket and the book is also of 100 rupees. So in that case, how much money do I have to pay separately from my pocket? Zero. So that's the case here. Because potential is same on all points. then there is no need to do any work done separately and therefore work done on equipotential surfaces is always zero another special thing about equipotential surfaces electric field is always perpendicular to equipotential surfaces why? let's prove it The next important thing is that electric field is always perpendicular to equipotential surfaces. Now this is very important and interesting thing that electric field E is always perpendicular to equipotential surface. As we can see here electric field and equipotential surface are perpendicular to each other. But why you will assume this? So let us try to prove this mathematically. ok, to prove this we assume, let us assume that electric field is making any angle theta with the equipotential surface we have accepted this, we don't accept that it will always be perpendicular so we are saying that suppose this is a surface and here my electric field is making angle with this surface, it is making theta angle and this is a surface angle So, we have to prove that theta will always be 90 degree. So, that's what we have to prove. So, I have assumed that electric field E is theta angle. So, if I resolve it in horizontal and vertical components, then E cos theta will be on this side and E sin theta will be on this side. Basically, both its components are E cos theta. E cos theta is parallel to the surface. E sin theta is perpendicular to the surface. Now let's talk about a charge. The horizontal component of electric field is E cos theta. Let us suppose that this point is A and here is a charge dq. Now, in horizontal direction, E cos theta is electric field. So, under that influence, let's say that this DQ charge is moving in this direction from A to B. So when dq will move from A to B, then work done will be to take dq from A to B. This will be dq into dv, where dv is potential. Right? What is work done? Work done is done because of potential difference. Right? So dq will be dw dq into dv Ok? There is another way to calculate work done force into displacement So in that way we can write this small work done we can write it as f dot dx and we can write it as f dot dx Now, what can we write as F dot dx? Force and displacement are both vector quantities. What can we write as F dx cos theta? can you write it? so that means dW how much is it? instead of F we can write in terms of electric field because here we are more interested in electric field so what we write for force electric field EQ so here Q is the value dQ so write it as dQ so dW is here ok, is it clear till here? ok Now see, the work done we are taking out, the surface we are taking out, what kind of surface is that? The surface we are talking about here, basically, we are talking about an equipotential surface. Okay, if equipotential is a surface, that means the potential is equal in that entire surface. Now if the potential is equal, then how much work done should be done to take any object from one point to another point? ideally it should be zero. So, we have read that for equipotential surface, work done that should be zero. Now, if that is the case, then I can write zero instead of dw. So, I can write edq dx cos theta is equal to zero. Now E cannot be 0 because electric field is given. Now DQ charge is also not 0 because there is a charge. If there is no charge, then the story will end. So I am sure that DQ is also not 0. Now if we talk about DX. So displacement is happening here, A is going somewhere from A, that's why this story is going on, that is, DX is also not zero. So what is left? In the end, the remaining cos theta, that is, if this DW has to be zero, which it will be because it is an equipotential surface, then only one thing can be zero and that is cos theta. So if cos theta is zero, then what will be the value of theta? It will be 90 degrees. This means if, If we are talking about a surface which is equipotential, then theta has to be 90. So, dear children, now we will see about equipotential surfaces. Third interesting point and that is that two equipotential surfaces never intersect. Koi bhi do. Different equipotential surfaces cannot intersect with each other. Why? Logically, we have two equipotential surfaces, one is red and the other is blue. I said no, it is wrong, we can intersect them. Okay, you are forcing it so much, then let's do it. As soon as we intersect them, in between them, The intersection area will be the common area of both. Now, suppose the red surface had the potential of V1. The blue surface had the potential of V2. Now, these are equipotential surfaces. That means, the potential of V1 was at every point of the red surface. The potential of V2 was at every point of the blue surface. Now I have made both of them intersect. So, the common area of the two, children, tell me how much will be the potential of that area? How much will be its area? V1 will be there, V2 will be there, V1 plus V2 will be there, V1 minus V2 will be there, how much will be there? So, if we make them intersect, then this one potential surface will not remain. Because after intersecting, now there is no potential in this whole surface. Do you understand now? So, that is why we say that two equipotential surfaces never intersect. So, now we will see the relationship between electric field and electric potential. That is, we will see the relationship between E and V. And to see this, let's suppose we assume that we have a charge DQ which we are taking from point A to point B. That is, the displacement of this charge is happening in this direction. Now, since we are taking this charge from one place to another place, then we have to do some work. So, the small work that has to be done to take this small charge is denoted by dW and that will be dQ into dV. dV means the difference of the potential of both these surfaces. Assume that we will do the same thing. Now we are assuming that if there is some work done then it means there is some potential difference then only work done is done. Now see we can write work done in terms of force and displacement also. What can we write? F dot dr. What can we write? F dr cos theta. What can we write F? We can write F in terms of eq. E means electric field. Thank you. Okay, we can write this DR cos theta. Now if we talk about theta, then see this displacement is happening in this direction. Our applied electric field is in the upper direction. So basically theta is the value of theta here is 180 degree. So cos 180 degree will be minus 1. So this will be minus EDQDR. This will be DW. So earlier also we have written an expression for dw in terms of charge and potential. Now also we have written an expression for dw which is in terms of electric field and displacement. So we can say that from 1 and 2 we can equate the right hand side of both. dqdv will be equal to this. So we can write dqdv is equal to minus e dqdr. so this dq dq will be cancelled so E is equal to minus dv by dr so this will be the relationship between electric field and potential so what we get from this is that see the gradient of potential means along with distance potential will decrease as it is minus sign so that relation is of electric field okay so if we talk about the direction of electric field then we see that electric field is in the direction in which potential decreases steepest where potential is decreasing the most right because dv by dr means potential with respect to distance that how much it is decreasing and that is because of this minus sign mathematically now if we talk about magnitude of electric field then we see that its magnitude is given by change in magnitude of potential per unit displacement normal to the equipotential surface at that point ok so here both these surfaces are equipotential surface right means this first surface is equipotential if you move charge from one point to another point inside this surface then your work done will be zero but if you are moving from one surface to another surface which means let's say if this surface potential is V then the potential of this surface will be V plus DV V So that extra potential was the reason for this work. So electric field is minus dv by dr and this is a very important relationship. So after talking a lot about equipotential surfaces, now it is time for electrostatic potential energy. Because till now we were reading the story of potential. Now we will see mathematical expression of electrostatic potential energy. Let's see. So for a system of charges where there is more than one charge, how will we calculate the potential energy there? We will see this step by step. So in the first step we will see a system which has only two point charges. So this is the simplest case. So let us suppose there is a system where Q1 is lying on A, this is Q1 and Q2 is on B. Now kids, how did we define potential energy? How much work is done to bring any charge from infinity to a particular point? So for now let's assume that Q1 is already at A. And Q2 is brought from infinity to B. So we assume that Q1 is at A and Q2 is brought from infinity to B. Now if we talk about this point B. So, at this point B, the electric potential will exist because of point A and Q1. So, the potential at point B will be KQ1 by R. Let us suppose the distance between R and Q1 is given by R12. So, R12. This will be the potential at this point. Because the potential at this point will be due to the charge in the surrounding. That's why its potential will be there. We have studied this. Okay. So, now what I have to calculate is potential energy. And children, how do we define potential energy? Work done in bringing a charge from infinity to a point. So, here work done in bringing Q2 from infinity to B. What will happen? This will happen. Q2. into the potential difference means Q2 into from where we are taking it from infinity to B means my initial position is infinity and my final position is B so what will happen Q2 into V potential at final position minus potential at initial position means VB minus V infinity VB minus V infinity now what we take as potential at infinity 0 so it will be 0 so what will happen into VB. What is the value of VB written above? We have written Q2 into KQ1 divided by R12. So, what did we get in total? Potential energy of this system. What will happen? KQ1 Q2 by R12. So this will be potential energy of a system of 2 point charges. Now let's complicate it a little bit. Let's complicate it a little bit. So now let's talk about case 2 where we talk about potential energy of a system of 3 point charges. Now we have 3 charges. Like here we can see Q1, Q2 and Q3. If we have 3 charges, then how much potential energy of this system will be? Its potential energy will be... K which is 1 by 4 pi epsilon naught. So basically what is this? K is K. K Q1 Q2 by R12. Q1 Q3 by R13. Plus Q2 Q3 by R23. So when 2 was of system. So what was there? Interaction was only between those two charges. If there are 3 systems. So what will happen? First interaction will be between 2nd. Second will be between 3rd. First will be between 3rd. So basically here 3 terms will come. See format is same. K Q1 Q2 by R is the format. But. In first one, 1 or 2 are involved, in second one, 1 or 3 are involved, in third one, 2 or 3 are involved. Now let's talk about potential energy of a system of n point charges. What if we have 10 charges? What if we have 20 charges? So here the logic will be that we have Q1, Q2, Q3, Q4, there are many charges. So Q1 we are bringing from infinity to point A. So when we are bringing Q1, then there is no existing charge here. means there is nothing here, right? because no electric field is existing because till Q1 came there was no charge here right? so that's why Q1 let us suppose work done to bring Q1 from infinity to A is U1, so how much will be U1? 0, to bring this we don't have to do any work done because electric field doesn't exist here Now when we bring Q2 from infinity to a particular point, this point, then the potential will exist, will exist for sure. And for what reason? Because of Q1. So it is like a two system, two particle system. So how much will be U2? KQ1Q2 by R12 assuming the distance between these two is R1. Now suppose there is another charge Q3 in this system. So we are taking Q3 from infinity at this point. So now its potential will be due to this, Q1 and Q2. So now U3 will be KQ1Q3 by R13. Let's say this is R13. plus KQ2Q3 by R23. Okay. Now, in the same way, when we bring Q4, so when Q4 comes, then the potential will be because of Q1, Q2, Q3, with Q4. In the same way, we will also remove U4. So, it will be KQ1Q4 by R14 plus KQ3Q4. Q2 Q4 by R24 plus K Q3 Q4 by R34. So, this way, if we write the values of U1, U2, U3, U4 till UN, then finally, how much will be the net potential energy of this whole system? So, we will see that overall the net potential energy of this system will be some of these potential energies U1, U2, U3 up to UN. Okay, and when we will do the sum of all these, then we will see that the value of U will come out, summation KQIQJ divided by Rij. where ij represents all the pairs finally we have to pay attention to one thing we have already applied the summation but the value of i and j should be different because if you see these terms then they are R12, R13, R23 R11 is nothing r22 is nothing right, what will be 22? nothing will happen, there is no distance between q2 and q2, so r11, r22, r33 they don't exist, so finally our potential energy will be, summation of i is equal to 1 to n, summation of j is equal to 1 to n, assuming that total n charges are there, so i value will vary from 1 to n, j value will vary from 1 to n, such that i is not equal to j, this is very important, and this will be KQIQJ divided by Rij and this will be halved. Why did I halve it? This is very important children. Why did I halve it? I halved it because if you see for every pair, suppose if the value of i and j is Q1Q2 or Q2Q1, then it is the same thing. So we will take only one term right. So we cannot say that this will be summation for all pairs. We cannot say that in this system. So in this system not for all pairs but for one pair only one term will exist. For the interaction between Q1 and Q2 KQ1Q2 by R12 is enough. For this we will not consider KQ2Q1 by R21. So that we will not consider this term. and for J values we will consider only one term. So this is how we get potential energy of a system of n point charges. Here more important than derivation is to understand the logic from which we write this. So let's see one problem. If one of the two electrons of H2 molecule is removed, we get a hydrogen molecular ion H2+. Let's understand this first. A hydrogen molecule, let's say H2, how many electrons and how many protons will there be? It is H2, right? That means there will be 2 electrons and 2 protons. Now we have removed one electron from 2. Electron is gone, so it is positively charged because it has more protons and less electrons. So now how many electrons are there? One electron because we have removed one. And there are still two protons. Now what happened? In the ground state of an H2+, the two protons are separated by roughly 1.5 angstrom. So the situation is that these are two protons. And how far apart are these two protons? From 1.5 angstrom. And the electron is roughly 1 angstrom from each proton. So electron let's say is somewhere such that it is at roughly 1 angstrom distance from both protons. Determine the potential energy of the system. Now I have to calculate the potential energy of this system. So this is a 3 particle system in which 3 charges are involved. Which are these charges? Electron Qe that is 1-1.6 x 10-19 Coulomb Charge of proton P1, P1 and P2 are called both protons. So what is this? Plus 1.6 x 10-19 Coulomb Similarly, Qp2, the charge of the second proton is also 1.6 x 10 to the power minus 19 coulomb. Now, the distances between them are also given, 1 angstrom, 1.5 angstrom and 1 angstrom. Now, the potential energy of this whole system will be, So if we see the system of these three particles, then see what will be their potential energy? KQ1Q2 by R12, this and this. Then KQ2Q3 by R23, so this and this. In the same way, KQ1Q3 by R13, i.e. electron and proton 2. So this will be my net potential energy and now see there is nothing special, we can put all the values. K value 9 x 10 to the power 9, QE value minus 1.6 x 10 to the power minus 19, QP1 value plus 1.6 x 10 to the power minus 19 divided by REP1 means 1 angstrom. Remember to change the unit, 1 angstrom is 10 to the power minus 10 meters, so 1 x 10 to the power minus 10. plus QE will be again minus 1.6 into 10 to the power minus 19 into QP2 that is 1.6 into 10 to the power minus 19 divided by REP2. This will also be 1 angstrom which is 1 into 10 to the power minus 10 plus QP1 QP2. means it will be 1.6 x 10 to the power minus 19 x 1.6 x 10 to the power minus 19 divided by R P1 P2. How much distance between P1 P2? 1.5 angstrom that is 1.5 x 10 to the power minus 10. When you calculate it, the answer will be minus 30.78 x 10 to the power minus 19 joules. So this will be the net potential energy of the system. so now let's see one more problem calculate the electric potential energy of the system of charges so here how many charges are visible to us? we are seeing 4 charges and what is this? it is a square whose side lengths are equal so let us say this is my charge number 1 this is charge number 2 this is charge number 3 and this is charge number 4 why did we do this? just for our convenience and let's assume that each side of this square has a length that is a So now what we are seeing here is Q1, Q2, Q3 and Q4. These are 4 charges and their value is plus Q, plus Q, minus Q, minus Q. So the potential energy of this whole system will be KQ1Q2 by R12 plus KQ2Q3 by R23. plus KQ2Q4 by R24 plus KQ1Q3 by R13 plus KQ1Q4 by R14 plus KQ3Q4 by R34. Okay, this will be the net-net potential energy of this system. Okay, we have done 1 with 4, 2 with 4, 3 with 4, we have covered all. So here also what we will do, we can add their values, Q1, Q2 values are plus Q plus Q, so this will be Q2 divided by R12 is A, Q2, Q3 means minus Q2, this will be minus Q2 divided by R23. R23 will also be A. plus Q2 Q4, means minus Q square, divided by R24, R24 means, the distance between 2 and 4, so, diagonal of a square is, root over 2 into A, so, this will be root 2A, plus q1 q3 that is again minus q square divided by r13 r13 again root over 2 into a plus q1 q4 that is minus q square divided by r14 that is a plus q3 q4 that is q3 q4 plus q square divided by a So when we calculate this completely, we will get potential energy. Q square divided by, instead of K we will write 1 by 4 pi epsilon naught A into minus root over A. So this will be our answer. So kids, now we are going to talk about potential energy in an external field. Understand my point very carefully. We are talking about potential energy from the beginning. Potential energy, potential, we are discussing about all these things. But here I am putting more emphasis on in an external field. Why? Because here we will talk about something like this. That if I have a charge Q and I have to get potential energy of charge Q. In an electric field, which is not produced by Q. Which means, if I have to extract the potential energy of Q charge on an external electric field. Till now we were talking about a point charge and I have to find potential on a point. Basically the field that was being created was happening because of that point charge. So our story was something like that. But now our story is that I have a charge Q and it is in an electric field which it has not produced itself. We don't know who produced it. It doesn't matter whether we know or not. The main thing is that the electric field is not produced by the Q charge. So Q is an external electric field. So if I want to extract the potential energy of Q in that external field, then how do we do it? Okay, did you understand what we are studying now? So, potential energy of any charge Q in an external field is given by Q into V of R. Where Q is obviously that charge Q which we are talking about. And what is V R? V basically is the potential because of the external field at a particular point. So, in similar lines, we will see potential energy in an external field when we have a system of 2 charges. I mean, we just talked about 1 single charge. So, if we have 2 charges, then what will happen? Let's see. So, kids, here we are going to talk about system of 2 charges. That means, in this whole system, there are 2 charges. Let's say Q1 and Q2. So, first of all, what will we calculate? Potential energy of Q1. So we will calculate this first. This means that to bring Q1 from infinity to a particular point, we have to do so much work. So that is equal to U1. So Q1 we have just brought from infinity to this point. So that is work done to bring Q1 from infinity to P1. P1 let us suppose that this point is P1 so this point is P1 so this will be Q1 into potential at P1 minus potential at infinity Potential at infinity is 0 so this will be Q1 into V P1 means the potential at P1 point So let us suppose P1 point has potential V which is a function of R so this will be Q1 V R Now where did this R come from? Now see, now you will understand everything. Now in the same way, to bring Q2 from infinity to this particular point, let us say this particular point is P2, so to bring it to P2, how much work we had to do, that will be the potential energy of Q2, which we will denote with U2. So how much will this be? Now when this will come, then Q1 is already existing here, right? So what will happen here? This will be Q2. into potential at P1 due to P2. Correct? Because when P2 is coming, when Q2 is coming, then Q1 is already here. So, because of Q1, there will be a potential at this point, P2. So, that is potential at P2 due to P1. This will happen. Is it or not? Secondly, when we are talking about U2, then we have to do some work to bring this Q2 from infinity to here. That is Q2 into Vp2 minus V infinity. So, the sum of both will be there. So, when we are bringing Q2, then we have to do some work to bring it from infinity to here. Plus, due to the existing charge, there is a potential on it. So, how much will this be? This will be Q2 into potential on P2. Due to P1, how much will this be? The potential of two system charge. That is KQ1Q2 divided by R12. Where R12 is the distance between these two. Let us suppose if we make coordinates like this. So if the position vector of P1 is R1, if the position vector of Q2 is R2, then the distance between these two will be R12. Okay, so this will be R1 2 plus here Q2 into VP2 means potential at point P2. Okay, that means we can write this as V which is a function of R2. Okay, so now net net of this whole system's net potential energy will be U1 plus U2. So, that means we can write this as Q1 VR1. plus Q2 VR2 plus Q2 or plus KQ1 Q2 divided by R12. Okay, this will be done. Now, here we have written Q2 twice by mistake. What is the potential? KQ by R. Right, so this was the potential. Okay, so this will be our net potential energy for two system charges in an external field. So, let's move on to the next topic and now we are going to talk about the potential energy of a dipole in an external field. We have already discussed about dipole in previous videos. So, let's see the situation. Let's assume that we have a dipole with two charges separated by some distance. And we have placed this dipole in an electric field. Let's assume that the electric field is in this direction and let us say my dipole is in this direction. Okay? Let's assume for now So what will happen? There is a plus Q at one end of this dipole and a minus Q at the other end So the force applied on this end will be Qe in this direction The force applied on this end will be Qe in this direction So net force is zero on this because equal and opposite forces are applied But what will it experience? Rotation Because the force above this will push here The force below will push here So as a result what will it start doing? Rotate So as a result, what does my dipole experience in a uniform electric field? Because of rotation. Because of torque. And how do we calculate torque mathematically? Torque is equal to P cross E. Where P is the dipole moment of this dipole. And E is the electric field. So now let us calculate mathematically how we will calculate the potential energy of this dipole. So, we have just seen that any dipole will experience torque in a uniform electric field. And how do we denote that torque? P cross E where P is its dipole moment and E is its external field. So, what can we write? P E sin theta A cross B is equal to AB sin theta. We already know this. We already know this. Okay? Now we will see what is the work done by the external torque. Now how much work done is due to this torque. So that work done will be like normally when we talk about linear motion then work done is F dot ds force or displacement. So here what will come torque or angular displacement that means here in picture will come torque or d theta angular displacement. So what we can write this P e sine theta. d theta, from where to where, let us suppose that this dipole, initially if theta naught was its orientation, later it became theta, because torque is acting, so it is rotating, right, so assuming that this angle, the angle that it is making with the electric field, okay, so this angle was theta naught, now it is rotating, so now it has become theta, so from this we will integrate theta naught to theta, so work done will be equal to P E integration of sine theta will be cos theta, so this will be cos theta. 0 minus cos theta 1. okay, so we can write it like this, work done, now this work done from torque, this work done will be stored as potential energy, we know this, work done is stored as potential energy, correct, so what we can write as potential energy, potential energy will also be a function of angular displacement, that is theta, and that will be equal to pe cos theta naught minus cos theta 1, So this is my expression for potential energy. Now if we see some special cases here, special cases means special values of theta. So let us suppose initially theta0 is pi by 2, in that case cos theta0 will be equal to 0, means initially if it was perpendicular, then in that case cos theta0 will be 0, so in such case u will be minus pe. cos theta 1, which we can write as minus p dot e, because a b cos theta is a dot b, right, so in a special scenario, we can write u, that is, potential energy, as minus p dot e vector, okay, so this is the story of dipole when it is on an external field, So, Machapati, with this we have come to the end of this video and I hope that with this video, the concepts of electrostatic potential will be crystal clear and I will meet you soon with the next video in which we will have a very exciting discussion on capacitance and after that, as always, a special video of numericals will come. So, I hope that your concepts will be crystal clear after watching this video and if this has happened, then definitely write in the comments that the concept has been crystal clear. If you have any feedback, please do share. So, I will meet you soon with the next video. Till then, stay home, stay safe, take care. Bye-bye.