Understanding Electric Field of Charged Ring

Hello kids, so we have come to the lecture number 6 of electric charges and field. And today we are going to learn how to find electric field. On the axis of a charged ring. Charged ring. Ring means continuous charge distribution. Like we learned in lecture number 5. There we talked about questions like JEEV. But this topic. On the axis of a charged ring This is asked a lot in the board Directly ask its derivation of 3 number Or ask a formula-based question It comes in competition too but it is very important for the board That's why I have brought a different video of it See I am trying my best to make the board content in front of you. So let's see how to deal with it, what is its formula and what are the good levels of questions can be asked on it. So my name is Alok Pandey and this is your very cute channel physics wala. Let's start. Electric field heading will put electric field on the axis of a charged ring Now many people find my video slow, so they do this, that YouTube gives the option below to increase the speed. So go to speed and do x2, then you will see that I will teach you electric field, this is how it happens, I also see it many times. Come, so we will have a charged ring like this. We have a ring, okay. Beautiful ring. Wow sir, it's so beautiful sir. So beautiful sir. There is a beautiful ring. And why is the charge uniformly distributed on this ring? How? Like this. Uniformly. On the whole ring, Q charge I have distributed uniformly then we said on the axis of this ring on the axis of the charged ring. Let's say its center is O. Let's assume that we take O. Let's say at the distance of X, at this point, what is the electric field? So, the topic is electric field on the axis of a charged ring. How far from the center? X distance away. Here we have to tell the electric field. Look, if this point was charged, if this point was charged, assume its radius is R. Its radius is R. If the point was Q then it would have been easy to find the electric field. At the distance of X from Q, the electric field would be KQ by X square. The positive charge field would be away from it. And the negative charge field would be on its side. The magnitude would be KQ by X square. If it was plus then the field would be like this. If it was minus then the field would be like this. But this was true when Q was the point charge. What is the negative charge field? Why is there a point charge here? No, here Q is a charge distribution, an extended charge. Now don't even think that the center of this charge is here, so keep Q here. Q is far from the center, so X is far from the center. then KQ by X square is also wrong. You cannot apply this too because in the formula of electric field, X square is below. This concept that we used to do in the center of mass, that put the whole thing in the center, this would have been right if E is directly proportional to x. When it is linear, then this concept, we used to say that if we consider the charge at the center, then the existence is far away. kq by x square. Why this happens? If E is proportional to x. But if it is not, then this is wrong. So what can we do for such a system? We have a formula for point charge. There is no formula for such charges. So we will understand the whole story of integration in this. I am going to make its integration very easy. See. I will do one thing. I cut a small element here. Small, small, cute, pretty. Here small. And charge is dq on this small element. Charge dq on this small element. Let's assume this element. From this place, small r distance is at distance. Will this element be like point charge? It is a small charge. So this is point charge. That means dq can behave like point charge. That means that formula will be valid on dq. So, how will the electric field be here because of dq? See, this is a positive charge, how does the electric field happen because of this? Away from the charge, it will happen like this. Small electric field, dE. Electric field from dq is here, away from the positive. And instead of E, I said dE. Small electric field of small charge. Not electric, electric field. Okay, now tell me one thing, what will be the value of this dE? Now we will use the point charge formula. K, how much is the charge? dq and how much distance is it from it? r square. Because of this, electric field is formed here. If you look at this electric field carefully, then two components of it can be made. One is going in x and the other in y. This is called angle theta. So, kids, this angle is also theta. So, the component of DE comes here. DE cos theta. Where theta is there, cos theta. And the component of DE comes here. DE cos theta. Why all this has to be done? Because this formula of electric field is of point charge. We cannot apply this formula on distributed charge. Yes, if you see a small charge from distributed charge then this is like a point charge. formula is valid on this angle. This is like a point chart. So, for this, k dq by r square. Then I said, consider this angle as theta. So, this angle is theta. One component of d is d cos theta and one is d sin theta. That means, electric field will be in some x and some in y. Note this. It seems that electric field will be in x and y but the truth is that electric field will only come in x. Why not in y? Because, the reason for this is that symmetrical dq here, just opposite of it. Due to this, de will be like this. The cos theta of that de will come here and the sin theta of that de will come here. So, de sin theta and de sin theta will be cancelled. That is, of this element, of dq element, This component DE sine theta is being formed by a mirror of this one element here, another DQ the one in front of it, vertically opposite The DE of the opposite one will be like this its d cos theta will remain here but its d sin theta will remain like this d sin theta d sin theta cancel. I will make it and erase it. This will be d, this theta, this theta, this theta, d cos theta. and above it will be dE sin theta. So, what will happen to its dE sin theta? It will be cancelled. It will be removed. dE cos theta will be added. Its dE cos theta is with this. Okay. Similarly, children, for every component, when you will see dE, then the sin theta term of every component will be cancelled in front of it. Of every component. Suppose, there is a component here. Okay. Now, see, the whole story has come. See, I have told you three times. This is the ring and this is the axis. So, if you see a DQ here, then this DQ is like this and this DQ is like this. This is D cos theta and this is D cos theta. This is the same direction. This is the ring. This is DQ and this is DQ. Because of this DQ, DE is like this and because of this DQ, DE is like this. Now see, DE cos theta, DE cos theta is added. DE sin theta, DE sin theta is cancelled. That means there will be a DQ opposite to every DQ. Which will cancel DE sin theta. For example, see a DQ from here. From anywhere, from here. This is a DQ. There will be a DQ just opposite to it here. This is d, this is d, d cos theta is added, d sin theta is cancelled. Do you understand? Add, cancel. So, d sin theta components will be cancelled. So, we can write d e sin theta component of each element dq is cancelled. DQ. Did you understand the language? My English is a bit bad. Opposite element. The opposite of every DQ is D sine theta. That D sine theta is cancelled. Like this. D, D. D cos theta, D cos theta is added. D sine theta, D sine theta is cancelled. Okay, one thing, understand this too. Take any component. Any one from here. For everyone, the distance will be r. For everyone, the angle will be theta. Imagine this in 3D. Any component. I will make it like this and show it once. This is like this. I will do it in some books. See this is the story. Can it be understood like this? Yes or no? So you will take from here, you will take from here. Tell the distance from here. Small r. Tell the distance from here. Small r. Distance will come from everywhere. And let's talk about the angle. angle, then this is theta, this angle is theta. The angle of all components has to be theta. Why theta? This is what I am getting the feel and I will explain why theta. Look, understand it this way, for each dq, Let's do one thing, let's join it like this. What will happen? Radius R. For each dq R, R, X remains same. Yes, that's true. R, R, X all are same. See, for any dq, the radius will remain the same. Suppose we take the DQ here, for that also its radius is the same, for that also x distance is the same, and from here the distance will be small r. Small r, small r, this is like this. and this is an axis, so if the distance from here is 1 meter, then this is also 1 meter, this is also 1 meter I have seen a ring like this, and this is a point, this is a ring like this, and this is a point, I kept it like this So the distance from all the places will be the same, this R is like this, think like a cone This is the story of a cone. This distance is R, this distance is R, this distance is R, this distance is R. The distance is R from everywhere. X is also same for all the places. Do you understand? R, R, R. Take this like this. R, R, R. For all, x is the same. And for all, radius is the same. So, if R, R, x is fixed, then theta will also be the same. Triangularly fixed. Okay. For all, theta is the same. Now, what we have to do? We have to find out the value of every dq. of each dq, because each dq's d sin theta is being cancelled by its opposite. So, keep cutting each dq and keep cutting each dq and keep cutting each dq and connect all the dq's d cos theta component. So, net electric field will be sum of, that is, integration will have to be done, d e cos theta due to all components. And in this direction, the electric field will be in this direction. This is also shown. Let's remove this. De, k dq by r square, into cos theta. Now tell me, is k constant? Absolutely. Is r also constant? Absolutely. What is the value of r for dq, dq, dq, dq, dq? Same. So this also came out. Is theta also constant? What will theta be for all the components? Same. So, cos theta is also out. How easy is integration of DQ. Feel? What is theta for everyone? Constant. What is RRX? Constant. For any element, X value is same, R value is same, theta value is same, angle is same. So I said that K is constant, R square is constant, cos theta is also constant. Integration of DQ. DQ, DQ, integrate. 1 DQ, 1 DQ, 1 DQ, 1 DQ, 1 DQ. When you integrate all these DQs, then the total charge will come. capital Q This charge is broken in small dqs If we add all tqs, then Q is there Where is the end? Q is out of here KQ cos theta by R square is the answer Easy KQ cos theta by R square Note this much, but we don't write like this Now we are expanding it more KQ cos theta by R square Understood? K out R square out cos theta is also out because theta is same for everyone Dq is the indication of the end I have written all the things here. D, sin, theta, etc. are all cancelled. Clear? Ok. Now, let's move this derivation forward. Let's start keeping values. We are taking out the unit. k Q cos theta. Here, the value of cos theta will be x equal to r. That is right. Base upon hyperthin. Upon r square. So, this is kqx upon r cube. If you look at this triangle carefully, then can we say that the square of r is equal to capital R square plus x square? Here the angle is 90 degrees. R square is equal to R square plus x square. That means, R is equal to R square plus x square power 1 by 2. under root Means we can write R square plus x square power 1 by 2 instead of R So this is the final expression which is found in the books KQX upon R square plus x square power 1 by 2 and power 3 is also there, so it will be 3 by 2. If we replace it with power of r square plus square, we will do half into 3 power. This is the final expression for electric field due to a charged ring on the axis of ring at a distance x from the center of ring. Now if it is positive charge, then net electric field will come like this. If it is negative charge, then net electric field will come towards the ring. This is the final. Expression. This derivation can come in the exam. Is it easy? Is it easy? Did you understand this? R is equal to power half. R square plus square power half into 3. Note it. Pause and note it. Delete it. Pause and note it. Okay. Now we will play with this expression. Come. kQx upon r square plus x square power 3 by 2. Okay, so this comes in the board. I wish you remember this. Again, E knit, x k long, how much? kQx upon r square plus x square power 3 by 2. Let's do a little post-mortem of this. If I ask you how much electric field is there at the center of the ring, I have already taught you that in uniform charge distribution, if there is symmetric charge distribution, then the field at the center will always be zero. The force at the center of the charge will also be zero. Because if there is no field, then how will the force be formed? Tell me. So definitely zero should be at the center. Let's verify if it is correct or not. Will there be zero at the center? Because of this, it will be like this, because of this, it will be like this, because of this, it will be like this, and so on. Everything will be cancelled. At center, what will be the value of x? Zero. Brother, So, the value of x will be 0 as it is increasing. The value of x is 0 and the expression is correct. So, the value of x is 0. Now, let's talk about infinity. What will happen at infinity? If you go to infinity, then the value of x will tend to infinity. Brother, something is tending to infinity below, and something is tending to infinity above. Okay, you will have to solve this for the case of infinity. Do one thing first, listen. Let's take an easy case. If x is much much greater than a. First take this case, then you will understand the infinity one very well. x is much much greater than a. Listen, x. A, no, radius r is much much greater than radius r. The ring is like this and x is very far away. If you look from there, the ring will look like this. It will look like a point charge. If x is very big than radius, I have a ring here and I am looking at it from 1 km away. So will this ring be visible from there? If you have a bang in your hand from 1 km away, will it be visible? Sir, where did this bang come from? It will not be visible, right? Instead of a bang, a point will be visible. So this ring is not going to be visible from this distance. Now, x is good. If the point charge goes too far then the ring will not be visible. It will look like a point charge. Why won't it look like a distributed Q? Let's see if this is true or not. Let's try to stop this here. E NET is equal to K Q X upon Now if x is very much bigger than r, then r square will be zero in front of x square. Yes or no? If this is the condition, then x square plus r square will be nearly equal to x square only. r is very small. x is very far, 1 km, the value of r is 1 meter, this much big, 1 km is nothing in front. Below, this term will be 0, this will be left, x square power 3 by 2. So we will be left with x power 3. This will be equal to x square. into 3 by 2, 2, 2 cancel, x to the power 3. So, E net will come k, q by x square. What's the matter? Did it behave like a point charge? Did it behave like a point charge? So, this is what we said. If it looks like a point charge, then it looks like a point charge. So, it starts behaving like behaves like point charge. Feel is here. When will it happen? When x is greater than r. Now you will understand the case of infinity better. It has reached here. x was greater than r, so this formula. Now x is going to infinity. If it goes to infinity, then it will be greater than r. Can we use this case or not? Oh brother, if x goes to infinity, then it will be greater than xr. Use this case directly. If you keep x near infinity, then e net zero will come. Thank you. Did you understand? That means, even at the center, the electric field is zero. Even at an infinite distance, the electric field is zero. That means, the electric field will increase somewhere in the middle. The graph will be like this. It will rise and then fall. Zero in the beginning and zero in the end. I said that x is bigger than r so r square x square is 0 so x square into 3 by 2 to cancel x to the power 3. I said that this is k cube by x square. Then I said that if x is infinity then it will be bigger than r. Now if it is bigger than r then I used this formula and put infinity in it. So e is zero in the beginning and zero in the end. It will increase somewhere in the middle. Clear? Now let's make a graph of this. Note this much and then we will make a graph of this. Let's make a graph of this. Note it down. I don't know if it is constant on the way or if it is big somewhere. Let's see. Let's take a question. Can I erase it? Ok, so we understood that from a distance, a distributed charge looks like a point charge and behaves like a point charge. If we see this ring from a distance, we can see a point charge. What should be the formula for point charge? KQ by X square. Did you get it or not? It's a beauty of physics. Now let's take a question. Let's take this question through a question. There is only one question. For? a uniform charged ring charge the loss paper Q radius they look are fine the value of x such that E net is maximum. Means we have to find out that on the axis where is the electric field maximum. Also calculate E maximum. Also draw graph of E versus X. Come on. All these concepts will be completed in this question. First of all, it is saying that find the value of X such that E net is maximum. Come on. Let's solve. Brother, how much was the value of E net coming? The value of E net is coming. K Q X upon R square plus X square's power 3 by 2. Is it coming? Is it going like this? So, what is it doing? Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik dy by dx is 0. What is the variable here? It is x. The radius is constant, k is constant, charge is constant. So y is maximum when dy by dx is 0. To do Emax, we have to do 0 for de by dx. We have to differentiate the form of uv. uv Do you remember the differentiation of uv? How did we differentiate uv? How did we differentiate u by v? We used to write v square below. What did we do above? v du by dx minus We used to write v square below. We used to write v square below. We used to write v square above. This is the 11th thing. We used to write v du by dx minus u dv by dx. Earlier we used to take the lower one and differentiate the upper one. Minus. We used to write the upper one and differentiate the lower one. Come, let's put this thing here. Let's differentiate. Take the kq outside. Now we have to differentiate x and y. What will be the constant? First we will take the square below. R square plus x square power 3 by 2 square. This will be left. Below is v square. If we square this, this will be left. Then, v into du by dx. R square plus x square This is v into du by dx. Let's differentiate this. x's differentiation is 1 minus u dv by dx. We took u and we will differentiate this. If we differentiate this, 3 by 2 will come out. Power will be 1. This is the definition. nx minus x power n minus 1. If you subtract 1 from this, 3 by 2 minus 1. How much will be left? 1 by 2. Power 1 is less, 3 by 2 comes out. Then power 1 is less, 3 by 2 minus 1. 3 minus 2, 1 by 2. Then we have to differentiate it inside. R square's definition is 0. X square's 2x is equal to 0. This is the whole story. I took k and q out. Now see what we have in front of us. This is clear. u by v square, v du by dx minus u du by dx. First we write v, the lower one. We differentiate the upper one. Then write the upper one and write the lower one. First I wrote it down and then I differentiated the upper one. Then I wrote the upper one and then differentiated the lower one. Ok, let's try to solve this expression. So from here, this term goes there and becomes 0. This term becomes 0, so we are left with we are left with r square plus x square power 3 by 2 minus 2 say to cancel xxx square 3x square Into R square plus x square key power 1 by 2 is equals to 0 Yes, they call R square plus x square key power 1 by 2 common. Who Jaya What will be left? R square plus x square. What will be left? Power 1. 1. See here, 1 into 1 plus 2. Sorry, into, sorry, plus. See, when you take this out of here, half. This will be left. See, it will go inside. Power will be added. 3 by 2. Minus, this is completely gone out. 3x square is equal to 0. Now either this is 0 or this. This cannot be 0. How can root of square be 0? Under root r square plus x square is equal to 0. Not possible. This will be 0 only when x is also 0, r is also 0, r is 0. Then there is no ring. This will be 0 only when this is also 0, this is also 0. This is written. What does half mean? Under root. This will be 0 only when r is also 0, r is 0. Then there is no ring. R0 means nothing? Then this should be there. R square plus x square minus 3x square is equal to 0. That means R square minus 2x square is equal to 0. That means x square is equal to R square by 2. That means x is equal to plus... minus R by root 2. Yes, this is the answer. At this value of X, plus or minus, means in the front as well and in the back. Electric field is on both the sides. This side also at R by root 2 maximum. This side also at R by root 2 maximum. Is it clear? We had to differentiate. Are you erasing? Are you erasing? Okay. Next part is, calculate maximum value of E. When is E the maximum value? When X is the value of R by root 2. Keep any value, plus or minus. The value of both will be the same. KQ, instead of X, R by root 2. Below, R square plus X square. X square will be R square by 2. What will be x square? R square by 2 and here instead of x, we will solve it by R by root 2 Ok, we will solve it So, kq R by root 2 What is being made below? 2 R square plus R square 3 R square by 2 power 3 by 2 How to do this? Let's do one thing, first take the root of the whole, then this half story will end. So, kqr by root 2 And take the root of the whole, under root 3r square by 2 whole cube. Now it will be easy. kqr by root 2 upon, now r square will come out and cube will be applied. So 1 is r cube. From here r will come out and cube. Root 3 into root 3 into root 3. So root 3 into root 3 is 3 and 1 root 3 will be left. Below root 2 into root 2 into root 2. Now we have the cube of root 2. What did I do? I wrote power half in the root first. R square comes out and R is a cube. Root 3 into root 3 into root 3. Root 3 root 3. 3 becomes 1 root 3. Root 2 into root 2 into root 2. Root 2 into root 2. Root 2. Finished. Root 2 from root 2 cancels. From R, R square becomes. So we are left with KQ. This 2 will come up. 2 KQ upon 3 root 3 R square. This is the maximum value of A. E maximum is 2kq upon 3 root 3 R square. Value of such x such that E net is maximum. So, x's value is plus minus R by. Two answers are done. Also draw the graph of E versus X. Let's erase this. Graph of E versus X. Last part of the question. Let's make it. Let's assume this is a charged ring. I have taught you graph of E versus X in lecture number 4th. In electric field lecture. E and X. See, it is very easy to make on this side. I have told you that when X is 0, then electric field is 0 at the center. When x was becoming infinite, then also the electric field was zero. And when x is plus r by root 2, then E is maximum. That means the graph that will be formed, here E is zero, there also it is becoming zero, maximum in the middle. Will something like this happen? When will this zero be exactly? Going to infinity, okay, like this. parallel lines meet at infinity make it parallel so that it is zero at infinity ok This side is made. This side is made below in books. Which are good books. Some are making it above, some are making it wrong. Some people are not making it. First, where is this? This is the value of x plus r by root 2. And here, what is the value of e? Maximum. What is the value of e? 2kq upon 3 root 3 square. So, the graph that is being made here is below. Why? See, you are plotting the electric field in y. Understand, electric field is a vector quantity. What kind of quantity is it? Vector. So, children, this side, like this is the charge ring. When you see the electric field on this side of the charge ring, it will be like this. And if you see this side, it will be like this. Isn't it? Away from the charge. So, if you are considering this electric field as positive, then you will not consider this electric field as negative. This is what was taught in lecture 4 in E versus X graph. That the one going back is negative. And on which axis is E? First understand, we said that the electric field on the front is positive, so the one on the back will have to take negative. Electric field on this side will be positive and on this side will be negative. So, on which axis is E? Look at this side, it is made above the x axis because E was positive on this side. And what should E be on this side? E should be negative on this side, less than zero. So, how will we represent this negative thing here? By making it below. So, this graph. Now it is perfect. What is this value? This is minus r by root. Is it clear? So this is the graph. So here we stop this lecture. This was electric field on the axis of a ring. Keep studying, keep doing well, keep doing well, all the very best.

Hello kids, so we have come to the lecture number 6 of electric charges and field. And today we are going to learn how to find electric field. On the axis of a charged ring.

Charged ring. Ring means continuous charge distribution. Like we learned in lecture number 5. There we talked about questions like JEEV. But this topic. On the axis of a charged ring This is asked a lot in the board Directly ask its derivation of 3 number Or ask a formula-based question It comes in competition too but it is very important for the board That's why I have brought a different video of it See I am trying my best to make the board content in front of you.

So let's see how to deal with it, what is its formula and what are the good levels of questions can be asked on it. So my name is Alok Pandey and this is your very cute channel physics wala. Let's start.

Electric field heading will put electric field on the axis of a charged ring Now many people find my video slow, so they do this, that YouTube gives the option below to increase the speed. So go to speed and do x2, then you will see that I will teach you electric field, this is how it happens, I also see it many times. Come, so we will have a charged ring like this.

We have a ring, okay. Beautiful ring. Wow sir, it's so beautiful sir.

So beautiful sir. There is a beautiful ring. And why is the charge uniformly distributed on this ring?

How? Like this. Uniformly. On the whole ring, Q charge I have distributed uniformly then we said on the axis of this ring on the axis of the charged ring. Let's say its center is O.

Let's assume that we take O. Let's say at the distance of X, at this point, what is the electric field? So, the topic is electric field on the axis of a charged ring. How far from the center?

X distance away. Here we have to tell the electric field. Look, if this point was charged, if this point was charged, assume its radius is R.

Its radius is R. If the point was Q then it would have been easy to find the electric field. At the distance of X from Q, the electric field would be KQ by X square.

The positive charge field would be away from it. And the negative charge field would be on its side. The magnitude would be KQ by X square.

If it was plus then the field would be like this. If it was minus then the field would be like this. But this was true when Q was the point charge.

What is the negative charge field? Why is there a point charge here? No, here Q is a charge distribution, an extended charge. Now don't even think that the center of this charge is here, so keep Q here. Q is far from the center, so X is far from the center.

then KQ by X square is also wrong. You cannot apply this too because in the formula of electric field, X square is below. This concept that we used to do in the center of mass, that put the whole thing in the center, this would have been right if E is directly proportional to x. When it is linear, then this concept, we used to say that if we consider the charge at the center, then the existence is far away. kq by x square.

Why this happens? If E is proportional to x. But if it is not, then this is wrong. So what can we do for such a system?

We have a formula for point charge. There is no formula for such charges. So we will understand the whole story of integration in this. I am going to make its integration very easy.

See. I will do one thing. I cut a small element here.

Small, small, cute, pretty. Here small. And charge is dq on this small element. Charge dq on this small element.

Let's assume this element. From this place, small r distance is at distance. Will this element be like point charge?

It is a small charge. So this is point charge. That means dq can behave like point charge.

That means that formula will be valid on dq. So, how will the electric field be here because of dq? See, this is a positive charge, how does the electric field happen because of this? Away from the charge, it will happen like this. Small electric field, dE.

Electric field from dq is here, away from the positive. And instead of E, I said dE. Small electric field of small charge.

Not electric, electric field. Okay, now tell me one thing, what will be the value of this dE? Now we will use the point charge formula. K, how much is the charge? dq and how much distance is it from it?

r square. Because of this, electric field is formed here. If you look at this electric field carefully, then two components of it can be made. One is going in x and the other in y.

This is called angle theta. So, kids, this angle is also theta. So, the component of DE comes here. DE cos theta. Where theta is there, cos theta.

And the component of DE comes here. DE cos theta. Why all this has to be done?

Because this formula of electric field is of point charge. We cannot apply this formula on distributed charge. Yes, if you see a small charge from distributed charge then this is like a point charge.

formula is valid on this angle. This is like a point chart. So, for this, k dq by r square.

Then I said, consider this angle as theta. So, this angle is theta. One component of d is d cos theta and one is d sin theta.

That means, electric field will be in some x and some in y. Note this. It seems that electric field will be in x and y but the truth is that electric field will only come in x.

Why not in y? Because, the reason for this is that symmetrical dq here, just opposite of it. Due to this, de will be like this.

The cos theta of that de will come here and the sin theta of that de will come here. So, de sin theta and de sin theta will be cancelled. That is, of this element, of dq element, This component DE sine theta is being formed by a mirror of this one element here, another DQ the one in front of it, vertically opposite The DE of the opposite one will be like this its d cos theta will remain here but its d sin theta will remain like this d sin theta d sin theta cancel.

I will make it and erase it. This will be d, this theta, this theta, this theta, d cos theta. and above it will be dE sin theta. So, what will happen to its dE sin theta?

It will be cancelled. It will be removed. dE cos theta will be added.

Its dE cos theta is with this. Okay. Similarly, children, for every component, when you will see dE, then the sin theta term of every component will be cancelled in front of it.

Of every component. Suppose, there is a component here. Okay.

Now, see, the whole story has come. See, I have told you three times. This is the ring and this is the axis. So, if you see a DQ here, then this DQ is like this and this DQ is like this.

This is D cos theta and this is D cos theta. This is the same direction. This is the ring. This is DQ and this is DQ.

Because of this DQ, DE is like this and because of this DQ, DE is like this. Now see, DE cos theta, DE cos theta is added. DE sin theta, DE sin theta is cancelled.

That means there will be a DQ opposite to every DQ. Which will cancel DE sin theta. For example, see a DQ from here.

From anywhere, from here. This is a DQ. There will be a DQ just opposite to it here. This is d, this is d, d cos theta is added, d sin theta is cancelled. Do you understand?

Add, cancel. So, d sin theta components will be cancelled. So, we can write d e sin theta component of each element dq is cancelled. DQ.

Did you understand the language? My English is a bit bad. Opposite element. The opposite of every DQ is D sine theta.

That D sine theta is cancelled. Like this. D, D. D cos theta, D cos theta is added. D sine theta, D sine theta is cancelled.

Okay, one thing, understand this too. Take any component. Any one from here. For everyone, the distance will be r. For everyone, the angle will be theta.

Imagine this in 3D. Any component. I will make it like this and show it once. This is like this. I will do it in some books.

See this is the story. Can it be understood like this? Yes or no? So you will take from here, you will take from here.

Tell the distance from here. Small r. Tell the distance from here.

Small r. Distance will come from everywhere. And let's talk about the angle.

angle, then this is theta, this angle is theta. The angle of all components has to be theta. Why theta?

This is what I am getting the feel and I will explain why theta. Look, understand it this way, for each dq, Let's do one thing, let's join it like this. What will happen? Radius R. For each dq R, R, X remains same. Yes, that's true.

R, R, X all are same. See, for any dq, the radius will remain the same. Suppose we take the DQ here, for that also its radius is the same, for that also x distance is the same, and from here the distance will be small r.

Small r, small r, this is like this. and this is an axis, so if the distance from here is 1 meter, then this is also 1 meter, this is also 1 meter I have seen a ring like this, and this is a point, this is a ring like this, and this is a point, I kept it like this So the distance from all the places will be the same, this R is like this, think like a cone This is the story of a cone. This distance is R, this distance is R, this distance is R, this distance is R. The distance is R from everywhere.

X is also same for all the places. Do you understand? R, R, R. Take this like this. R, R, R. For all, x is the same.

And for all, radius is the same. So, if R, R, x is fixed, then theta will also be the same. Triangularly fixed.

Okay. For all, theta is the same. Now, what we have to do? We have to find out the value of every dq. of each dq, because each dq's d sin theta is being cancelled by its opposite.

So, keep cutting each dq and keep cutting each dq and keep cutting each dq and connect all the dq's d cos theta component. So, net electric field will be sum of, that is, integration will have to be done, d e cos theta due to all components. And in this direction, the electric field will be in this direction. This is also shown. Let's remove this.

De, k dq by r square, into cos theta. Now tell me, is k constant? Absolutely. Is r also constant? Absolutely.

What is the value of r for dq, dq, dq, dq, dq? Same. So this also came out. Is theta also constant? What will theta be for all the components?

Same. So, cos theta is also out. How easy is integration of DQ.

Feel? What is theta for everyone? Constant. What is RRX? Constant.

For any element, X value is same, R value is same, theta value is same, angle is same. So I said that K is constant, R square is constant, cos theta is also constant. Integration of DQ. DQ, DQ, integrate.

1 DQ, 1 DQ, 1 DQ, 1 DQ, 1 DQ. When you integrate all these DQs, then the total charge will come. capital Q This charge is broken in small dqs If we add all tqs, then Q is there Where is the end? Q is out of here KQ cos theta by R square is the answer Easy KQ cos theta by R square Note this much, but we don't write like this Now we are expanding it more KQ cos theta by R square Understood?

K out R square out cos theta is also out because theta is same for everyone Dq is the indication of the end I have written all the things here. D, sin, theta, etc. are all cancelled. Clear? Ok. Now, let's move this derivation forward.

Let's start keeping values. We are taking out the unit. k Q cos theta.

Here, the value of cos theta will be x equal to r. That is right. Base upon hyperthin.

Upon r square. So, this is kqx upon r cube. If you look at this triangle carefully, then can we say that the square of r is equal to capital R square plus x square?

Here the angle is 90 degrees. R square is equal to R square plus x square. That means, R is equal to R square plus x square power 1 by 2. under root Means we can write R square plus x square power 1 by 2 instead of R So this is the final expression which is found in the books KQX upon R square plus x square power 1 by 2 and power 3 is also there, so it will be 3 by 2. If we replace it with power of r square plus square, we will do half into 3 power. This is the final expression for electric field due to a charged ring on the axis of ring at a distance x from the center of ring.

Now if it is positive charge, then net electric field will come like this. If it is negative charge, then net electric field will come towards the ring. This is the final. Expression. This derivation can come in the exam.

Is it easy? Is it easy? Did you understand this? R is equal to power half. R square plus square power half into 3. Note it.

Pause and note it. Delete it. Pause and note it.

Okay. Now we will play with this expression. Come.

kQx upon r square plus x square power 3 by 2. Okay, so this comes in the board. I wish you remember this. Again, E knit, x k long, how much?

kQx upon r square plus x square power 3 by 2. Let's do a little post-mortem of this. If I ask you how much electric field is there at the center of the ring, I have already taught you that in uniform charge distribution, if there is symmetric charge distribution, then the field at the center will always be zero. The force at the center of the charge will also be zero. Because if there is no field, then how will the force be formed? Tell me.

So definitely zero should be at the center. Let's verify if it is correct or not. Will there be zero at the center? Because of this, it will be like this, because of this, it will be like this, because of this, it will be like this, and so on.

Everything will be cancelled. At center, what will be the value of x? Zero.

Brother, So, the value of x will be 0 as it is increasing. The value of x is 0 and the expression is correct. So, the value of x is 0. Now, let's talk about infinity. What will happen at infinity? If you go to infinity, then the value of x will tend to infinity.

Brother, something is tending to infinity below, and something is tending to infinity above. Okay, you will have to solve this for the case of infinity. Do one thing first, listen. Let's take an easy case.

If x is much much greater than a. First take this case, then you will understand the infinity one very well. x is much much greater than a.

Listen, x. A, no, radius r is much much greater than radius r. The ring is like this and x is very far away.

If you look from there, the ring will look like this. It will look like a point charge. If x is very big than radius, I have a ring here and I am looking at it from 1 km away. So will this ring be visible from there?

If you have a bang in your hand from 1 km away, will it be visible? Sir, where did this bang come from? It will not be visible, right?

Instead of a bang, a point will be visible. So this ring is not going to be visible from this distance. Now, x is good. If the point charge goes too far then the ring will not be visible. It will look like a point charge.

Why won't it look like a distributed Q? Let's see if this is true or not. Let's try to stop this here.

E NET is equal to K Q X upon Now if x is very much bigger than r, then r square will be zero in front of x square. Yes or no? If this is the condition, then x square plus r square will be nearly equal to x square only. r is very small. x is very far, 1 km, the value of r is 1 meter, this much big, 1 km is nothing in front.

Below, this term will be 0, this will be left, x square power 3 by 2. So we will be left with x power 3. This will be equal to x square. into 3 by 2, 2, 2 cancel, x to the power 3. So, E net will come k, q by x square. What's the matter?

Did it behave like a point charge? Did it behave like a point charge? So, this is what we said. If it looks like a point charge, then it looks like a point charge. So, it starts behaving like behaves like point charge.

Feel is here. When will it happen? When x is greater than r.

Now you will understand the case of infinity better. It has reached here. x was greater than r, so this formula. Now x is going to infinity.

If it goes to infinity, then it will be greater than r. Can we use this case or not? Oh brother, if x goes to infinity, then it will be greater than xr.

Use this case directly. If you keep x near infinity, then e net zero will come. Thank you.

Did you understand? That means, even at the center, the electric field is zero. Even at an infinite distance, the electric field is zero. That means, the electric field will increase somewhere in the middle. The graph will be like this.

It will rise and then fall. Zero in the beginning and zero in the end. I said that x is bigger than r so r square x square is 0 so x square into 3 by 2 to cancel x to the power 3. I said that this is k cube by x square.

Then I said that if x is infinity then it will be bigger than r. Now if it is bigger than r then I used this formula and put infinity in it. So e is zero in the beginning and zero in the end. It will increase somewhere in the middle.

Clear? Now let's make a graph of this. Note this much and then we will make a graph of this.

Let's make a graph of this. Note it down. I don't know if it is constant on the way or if it is big somewhere. Let's see.

Let's take a question. Can I erase it? Ok, so we understood that from a distance, a distributed charge looks like a point charge and behaves like a point charge.

If we see this ring from a distance, we can see a point charge. What should be the formula for point charge? KQ by X square. Did you get it or not?

It's a beauty of physics. Now let's take a question. Let's take this question through a question. There is only one question. For?

a uniform charged ring charge the loss paper Q radius they look are fine the value of x such that E net is maximum. Means we have to find out that on the axis where is the electric field maximum. Also calculate E maximum.

Also draw graph of E versus X. Come on. All these concepts will be completed in this question.

First of all, it is saying that find the value of X such that E net is maximum. Come on. Let's solve. Brother, how much was the value of E net coming? The value of E net is coming.

K Q X upon R square plus X square's power 3 by 2. Is it coming? Is it going like this? So, what is it doing? Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik Pichik dy by dx is 0. What is the variable here? It is x.

The radius is constant, k is constant, charge is constant. So y is maximum when dy by dx is 0. To do Emax, we have to do 0 for de by dx. We have to differentiate the form of uv.

uv Do you remember the differentiation of uv? How did we differentiate uv? How did we differentiate u by v? We used to write v square below. What did we do above?

v du by dx minus We used to write v square below. We used to write v square below. We used to write v square above. This is the 11th thing.

We used to write v du by dx minus u dv by dx. Earlier we used to take the lower one and differentiate the upper one. Minus. We used to write the upper one and differentiate the lower one. Come, let's put this thing here.

Let's differentiate. Take the kq outside. Now we have to differentiate x and y. What will be the constant?

First we will take the square below. R square plus x square power 3 by 2 square. This will be left.

Below is v square. If we square this, this will be left. Then, v into du by dx. R square plus x square This is v into du by dx.

Let's differentiate this. x's differentiation is 1 minus u dv by dx. We took u and we will differentiate this.

If we differentiate this, 3 by 2 will come out. Power will be 1. This is the definition. nx minus x power n minus 1. If you subtract 1 from this, 3 by 2 minus 1. How much will be left?

1 by 2. Power 1 is less, 3 by 2 comes out. Then power 1 is less, 3 by 2 minus 1. 3 minus 2, 1 by 2. Then we have to differentiate it inside. R square's definition is 0. X square's 2x is equal to 0. This is the whole story. I took k and q out.

Now see what we have in front of us. This is clear. u by v square, v du by dx minus u du by dx. First we write v, the lower one. We differentiate the upper one.

Then write the upper one and write the lower one. First I wrote it down and then I differentiated the upper one. Then I wrote the upper one and then differentiated the lower one. Ok, let's try to solve this expression. So from here, this term goes there and becomes 0. This term becomes 0, so we are left with we are left with r square plus x square power 3 by 2 minus 2 say to cancel xxx square 3x square Into R square plus x square key power 1 by 2 is equals to 0 Yes, they call R square plus x square key power 1 by 2 common.

Who Jaya What will be left? R square plus x square. What will be left? Power 1. 1. See here, 1 into 1 plus 2. Sorry, into, sorry, plus.

See, when you take this out of here, half. This will be left. See, it will go inside.

Power will be added. 3 by 2. Minus, this is completely gone out. 3x square is equal to 0. Now either this is 0 or this.

This cannot be 0. How can root of square be 0? Under root r square plus x square is equal to 0. Not possible. This will be 0 only when x is also 0, r is also 0, r is 0. Then there is no ring. This will be 0 only when this is also 0, this is also 0. This is written. What does half mean?

Under root. This will be 0 only when r is also 0, r is 0. Then there is no ring. R0 means nothing?

Then this should be there. R square plus x square minus 3x square is equal to 0. That means R square minus 2x square is equal to 0. That means x square is equal to R square by 2. That means x is equal to plus... minus R by root 2. Yes, this is the answer.

At this value of X, plus or minus, means in the front as well and in the back. Electric field is on both the sides. This side also at R by root 2 maximum.

This side also at R by root 2 maximum. Is it clear? We had to differentiate.

Are you erasing? Are you erasing? Okay.

Next part is, calculate maximum value of E. When is E the maximum value? When X is the value of R by root 2. Keep any value, plus or minus.

The value of both will be the same. KQ, instead of X, R by root 2. Below, R square plus X square. X square will be R square by 2. What will be x square? R square by 2 and here instead of x, we will solve it by R by root 2 Ok, we will solve it So, kq R by root 2 What is being made below?

2 R square plus R square 3 R square by 2 power 3 by 2 How to do this? Let's do one thing, first take the root of the whole, then this half story will end. So, kqr by root 2 And take the root of the whole, under root 3r square by 2 whole cube.

Now it will be easy. kqr by root 2 upon, now r square will come out and cube will be applied. So 1 is r cube. From here r will come out and cube. Root 3 into root 3 into root 3. So root 3 into root 3 is 3 and 1 root 3 will be left.

Below root 2 into root 2 into root 2. Now we have the cube of root 2. What did I do? I wrote power half in the root first. R square comes out and R is a cube. Root 3 into root 3 into root 3. Root 3 root 3. 3 becomes 1 root 3. Root 2 into root 2 into root 2. Root 2 into root 2. Root 2. Finished.

Root 2 from root 2 cancels. From R, R square becomes. So we are left with KQ. This 2 will come up.

2 KQ upon 3 root 3 R square. This is the maximum value of A. E maximum is 2kq upon 3 root 3 R square.

Value of such x such that E net is maximum. So, x's value is plus minus R by. Two answers are done.

Also draw the graph of E versus X. Let's erase this. Graph of E versus X.

Last part of the question. Let's make it. Let's assume this is a charged ring.

I have taught you graph of E versus X in lecture number 4th. In electric field lecture. E and X.

See, it is very easy to make on this side. I have told you that when X is 0, then electric field is 0 at the center. When x was becoming infinite, then also the electric field was zero.

And when x is plus r by root 2, then E is maximum. That means the graph that will be formed, here E is zero, there also it is becoming zero, maximum in the middle. Will something like this happen?

When will this zero be exactly? Going to infinity, okay, like this. parallel lines meet at infinity make it parallel so that it is zero at infinity ok This side is made. This side is made below in books.

Which are good books. Some are making it above, some are making it wrong. Some people are not making it. First, where is this? This is the value of x plus r by root 2. And here, what is the value of e?

Maximum. What is the value of e? 2kq upon 3 root 3 square.

So, the graph that is being made here is below. Why? See, you are plotting the electric field in y. Understand, electric field is a vector quantity.

What kind of quantity is it? Vector. So, children, this side, like this is the charge ring. When you see the electric field on this side of the charge ring, it will be like this.

And if you see this side, it will be like this. Isn't it? Away from the charge. So, if you are considering this electric field as positive, then you will not consider this electric field as negative. This is what was taught in lecture 4 in E versus X graph.

That the one going back is negative. And on which axis is E? First understand, we said that the electric field on the front is positive, so the one on the back will have to take negative. Electric field on this side will be positive and on this side will be negative. So, on which axis is E?

Look at this side, it is made above the x axis because E was positive on this side. And what should E be on this side? E should be negative on this side, less than zero.

So, how will we represent this negative thing here? By making it below. So, this graph.

Now it is perfect. What is this value? This is minus r by root.

Is it clear? So this is the graph. So here we stop this lecture.

This was electric field on the axis of a ring. Keep studying, keep doing well, keep doing well, all the very best.

Transcript for:Understanding Electric Field of Charged Ring

Transcript for:
Understanding Electric Field of Charged Ring