LECTURE 7 | Electric Field Due to Continuous Charge Distribution | Electrostatics | Class 12 Physics

[00:00:00]Hello everyone, welcome back to lecture 7 of electrostatics. In this lecture I will be teaching you electric field due to continuous charge distribution. But I will prefer before this you just go and watch lecture number six in which I have taught you what is electric field. I gave proper feel of electric field and we have done some questions based on that. In that particular lecture I have taught you what is electric field due to a point charge. If you will see that lecture this will be easier for you.

[00:00:26]Okay. I just want to revise one thing from the previous lecture that is due to a point charge at R distance electric field is KQ by R² and from positive charge this electric field is away from positive charge this electric field is away if it's a negative charge electric field will be towards this is what we have done and we have done multiple questions based on this is it electric field due to a point charge which is KQ by R² in magnitude and direction. From positive charge, electric field is away.

[00:01:02]Negative charge, electric field is 2.

[00:01:05]Clear? I hope this part is clear. Okay.

[00:01:07]Now we are going to deal with electric field due to continuous charge distribution. By seeing the term continuous charge distribution, I hope most of you understood that here we have to play with integration. Okay. Now I want to make one thing very very clear.

[00:01:21]Whenever you are integrating a scalar quantity, no problem. You can directly integrate without thinking but whenever you are integrating vectors they should be in same direction. You can integrate vectors directly when they are in same direction. If the vectors are not in same direction then what you have to do then you have to take component of vectors. Take component of all the vector vertical horizontal vertical horizontal and then you integrate integrate all the vertical component because they're in same direction integrate all the horizontal component because they're in same direction get the vertical field whatever you got get the horizontal field and then you can do what vector summation of that whatever I told just keep in mind in question I will say you what I want to tell from this clear so keep in mind scalers can be integrated ated directly.

[00:02:15]Vectors if they are in same direction integrate without thinking integrate but if the vectors are in different direction you need to take component horizontal vertical integrate all the horizontal integrate all the vertical then you can find the resultant don't worry I will show you okay so let's start electric field due to continuous charge distribution again one more thing I want to tell you before this what that is different type of charge distribution.

[00:02:46]First is lambda which is linear charge distribution.

[00:02:51]Linear charge distribution symbol is lambda.

[00:03:01]Whenever we have distribution of charge along the length we call it linear charge distribution. I'm not saying 1D you might be confused when the charge is distributed in one dimension. No, when the charge is distributed along the length for example charge on a rod rod I have given total charge Q length is L. So how this lambda is defined?

[00:03:28]Lambda is defined as charge per unit length. You all can write lambda is defined as charge per unit length.

[00:03:35]Clear? So this was a 1D body. Fine. Now if I draw a semic-ircular ring, this is 2D. But how is the charge distribution? The charge distribution on this semic-ircular ring is one dimension. Not one dimension basically along the length. The charge distribution is what? Along the length.

[00:04:00]This question might come you can you can do mistake in this that charge distribution on 1D body is known as linear charge distribution. No charge distribution along the length along the length is known as linear charge density. Symbol is lambda charge upon length. Here what you will get lambda is equal to charge. Let's say radius is r.

[00:04:21]So length will be semicircle length will be pi r. Full circle 2 pi r semicircle p<unk> r. Clear? Second is sigma surface charge density surface charge density or you can also say aerial charge density when the charge is distributed across the area again you will think sir it means it's a 2D body no again I'm saying it can be 3D body also the charge distribution has to be along the area across the area for example if I show Let's say this is a square plate.

[00:05:01]If I give this charge, if I give charge on this plate, area is A. Now the charge is across the area, total charge is Q.

[00:05:12]So sigma will be defined as charge per unit area. Lambda is charge per unit length. Sigma is charge per unit area. I can show you a 3D body.

[00:05:24]Let's say sphere. A sphere having charge on the surface. A 3D body. A sphere having charge distributed on its surface. So sphere is a 3D body.

[00:05:38]But as the charge is present only on the surface, what you will define for this surface charge density sigma. And that sigma will be let's say total charge is Q. Radius is R. So sigma will be charge upon surface area. The charge is on the surface area. Surface area of a sphere is 4 pi r² so 3D body but the charge distribution is what sigma clear third is volume charge density I'm rubbing this third is volume charge density row this is volume charge density when the charge is distributed throughout the volume again I have a sphere but in this case the charge is present in the entire volume. Let's say total charge is Q, radius is R. So what is row? Row is defined as charge per unit volume. Row is defined as charge per unit volume which you can write here as in this particular case Q upon 4x3 pi rq volume of a sphere. Is it clear to everyone?

[00:06:51]Understood? Same I can take a cube. I can fill the charge inside the cube.

[00:06:55]Then row will be what? Charge upon volume of a cube. So Q upon A cube if A is the side. Clear? Or if you just pause and write. Pause and write. So sigma can be defined for a 3D body when the charge is distributed on the surface and for the same sphere when the charge was in the bulk we called it what? Charge upon volume. Q upon 4x3 pi rq. You can pause and write. Clear? What is lambda sigma row? Fine. because we are going to deal with continuous charge distribution. So these things you should know. Fine.

[00:07:27]Okay. Now just for your understanding let's say this is a body on which charge is distributed uniformly or non-uniform doesn't matter. Okay charge is there from this body total charge let's say total charge is Q. Okay. From this body at a point P, I want to find out electric field.

[00:07:55]From this body at a point at a distance here at a point P, I want to find out electric field. Now what you will say?

[00:08:05]Sir, I know the formula for a point charge. For a point charge, the formula is K Q by R². But this is not a point charge. It's a continuous body having charge everywhere. Having charge everywhere. So what we will do not only here in every continuous body what you will do? We know formula for point charge. You will find a point on the body. What you will find a point a point on the body. This point will be very very small. I will take a very small element on the body. That a small element will behave like a point. So I have taken a very small element. A very small element will have what? Very small area DA.

[00:08:49]This very small element will have very small charge dq.

[00:08:54]Because of dq charge, I will find electric field. So it's a positive charge. So from positive charge, electric field is away. A away small electric field D. What I did? I know the formula for point charge. So on the continuous body I have taken a very small point a very small element which has very small charge dq. This this this will behave like what? A point charge. A small element behave like what? A small element behave like what? Point charge.

[00:09:24]Because of this point charge I will find electric field here. Then again I will take one more small element here. A small element will have a small charge and because of this element I will find electric field de again I will take one more small element one more small element because of this small element I will find electric field here it means what I'm going to do I am going to break this body continuous body into infinite number of small elements and because of all small elements I will find electric field here then I will apply principle of superposition Clear? It means I will add all the electric field vectorally.

[00:10:09]Is it clear to everyone? Is it clear?

[00:10:11]Which is known as integration. Is it?

[00:10:14]Summation of infinite infinite terms. We have to do what? Summation of infinite terms. Clear? How to integrate? What we have to do? You will learn. No need to worry. Understood? It was a continuous body. I have taken a small element.

[00:10:27]Because of that, I I calculated electric field. One more element, electric field.

[00:10:31]One more element, electric field. One more element, electric field. One more element. In this way, I will find electric field because of all the element. All the element, all the element at this given point. Then we will apply maths integration in a proper way to find out net electric field. Is it clear to everyone? Understood? Clear? What will be the value of this small electric field? If I ask you, you will say sir, K. This small electric field is due to this small charge. This is small charge is dq divide by distance between the charge and the point let's say r / r² in the same way I will find electric field because of all the elements then I will add them okay clear pause and write just for your understanding okay so we are going to start with our first element which already I have done when I was teaching you colum's law there I have shown you force Now electric field on the axis of a linear charge linear charged charge rod.

[00:11:48]Okay. So we are having a rod which is uniformly charged. I can make it non-uniform also. Don't worry. I will tell what to do in that. So this rod is having total charge Q. The total charge is Q which is uniformly distributed.

[00:12:03]Length is L. At a distance on the axis of this rod at a distance I'm making a point O. At this point O we have to find electric field. Now don't do one mistake. I have told before center of charge. There is nothing like center of charge. Electric field varies as 1x r².

[00:12:22]It's not a linear variation. You all right on the top there is nothing like center of charge. Keep in mind now I need to find out electric field at O clear. I want you all to go through all the questions which I'm doing today.

[00:12:37]After the class do it by yourself once again because you know these things can come in multiple chapters. In gravitation in place of charge I will I I will bring mass and I will ask you what gravitational field same we will do in magnetism and center of mass there also we need integrations and not this but I will take some semi hemisphere solid sphere. So all those things requires what integration. So from here I want you all to build what the fundamentals of integration. Clear? See how at this point O I want to find out electric field because of this rod.

[00:13:14]First thing is this is a linear charge.

[00:13:17]So for linear charge here what is defined? Lambda. What is lambda? Charge upon length. And it's a uniform distribution of charge. If you cut out 1 cm 1 cm both will have equal charge.

[00:13:31]Density is same. Clear? Now what I will see here okay I want to find electric field at O. I know the formula of electric field due to a point charge KQ by R² but this is not a point charge it's a linear charge so what I will do I will same thing cut a very small element from O I will go to a random distance X whatever words I'm using I will use in all the integration in all the chapters while solving any problem go to a random distance X go to a random distance X at random distance X cut a very small element having thickness dx this dx is very very small tending to zero 0 0 1 cm 1 m like that 01 okay this is very very small element so I went to x distance and took a very small element this small element will behave like a point charge and the small element will have very small charge very very small charge dq now you will say sir we know the formula of electric field because of a point charge. You made a point charge. Yes or no? By taking a very small element, what I did? I made this point as a what?

[00:14:46]Point charge. Very small element having what charge? DQ.

[00:14:51]What is the distance of O? X. Don't worry about DX. DX is very small. X plus DX is nearly X. No need to worry about that. Here we have the point O. So because of the small element, I know electric field. So what will be electric field? You will say sir it is a positive charge from positive charge electric field is away. So at the point O this small element will give a small electric field D.

[00:15:18]If I ask you what will be the value of DE you will say sir K a point charge DQ is it divide by distance between the charge and the point which is X which is X² clear? So I got a very small electric field due to a point element. Now you will say sir let's see what other points are doing. What about this point?

[00:15:44]Small element small charge. This is also giving electric field away. Positive charge give field away. Okay. So this element will also give electric field away.

[00:15:54]What about this element? This will also give away. What about this element? This will also give away. Okay. So we understood all the electric field are in same direction. When the vectors are in same direction we can add them like a scalar. Yes or no? When they are in same direction. So here all the vectors are in same direction. It means we can directly integrate without any difficulty. We can directly what?

[00:16:19]Integrate because all the field due to all the elements are in same direction.

[00:16:25]It means total field will be what? Total field will be DE plus D E1 D2 D3 D4 electric field because of all the elements will be in this direction and I will add all the electric field that will be the net field d1 D2 D3 D4 I will add all of them which is known as integration which is known as integration.

[00:16:48]Clear? So what will be the integration of DE? Integration of DE will be E which is equal to K is a constant. it will come out of integration. Now one problem is there. If I'm integrating with distance, operator should be dx. If variable is x, operator should be dx. But what we are having dq. Okay. It means I need to eliminate dq and bring what? dx. For that we will take help of what? Lambda.

[00:17:17]What is lambda? Charge upon length.

[00:17:19]Linear charge density. Lambda is charge upon length. Look at the element. The small element. This element has what length dx. This element has what charge?

[00:17:30]dq. So lambda will be charge upon length dq by dx. Therefore dq will be what?

[00:17:37]lambda dx. dq will be lambda dx which you can substitute here. So what you will get? K.

[00:17:46]Okay. Then in place of dq what you will write? lambda dx by x² k constant lambda constant integrate clear just a second lambda integrate okay now we have to put the limits of integration so see where will be the first element from O we are integrating what electric field due to element where will be the first element from O this will be the first element at What distance? A. So lower limit will be A. And if I go at the last, if I reach here, all the elements are covered. This last A element is at what distance from O? A + L. So upper limit will be A + L. Clear?

[00:18:36]Which is equal to K lambda. Integration of dx by X². This is X^ minus 2. So when we integrate X^ -2 + 1ide by -2 + 1 is it? So this will be x^ -2 + 1. So x^ -1 divide by -2 + 1 - 1. So -1 by x and limits from a to a + l is it. If I remove the minus limits gets interchange. So k lambda remove minus limits will interchange this will be a this will be a plus l. Put the limits.

[00:19:09]So 1x a minus 1 upon a + l which we can write as net field as k lambda lcm will be a into a + l and you will get a + l minus a a cancel out electric field is equal to k put the value of lambda what is lambda what is lambda charge upon length this is lambda charge upon length charge upon length into here we have length upon a into a + l. Length length cancel out electric field is kq upon a into a + l. This is the electric field on the axis. Kq upon a into a + l.

[00:19:57]For neat kits, I I want you all to remember one thing. You know electric field because of a point charge is KQ by R². You remember like this K Q by R² means R into R distance into distance first distance is A total distance is A + L. So distance into other distance A into A + L like this you can remember because this question can directly come in what neat examination.

[00:20:23]I hope you all understood clear. Now what maximum they can do in this? They can say that this lambda is variable.

[00:20:32]It means charge density is changing.

[00:20:34]They may say in this rod, the rod I have kept on this rod, the charge is not uniformly distributed. They will say lambda is equal to lambda not x where lambda kn is a constant and x is what this distance from origin.

[00:20:54]So as x will increase lambda will increase means as you can just pause and write this part. It means when x will increase charge density will increase.

[00:21:06]So here x is zero no charge as you will go right x will be increasing. So charge density also increasing less charge more more more here maximum charge you have maximum charge at the edge. So nothing to do everything will be same. You will take same type of element just here you took lambda out of integration. See focus here you took lambda out of integration because lambda was constant uniform. If lambda is variable nothing to do just this lambda equal to lambda x you put here lambda equal to lambda x.

[00:21:39]So lambda kn will come out but you will be having integration of x ds x dx which will be x² x dx by x² 1x 1x cancel it will be integration of dx by x ln x clear. So if the change vary lambda in place of lambda just put the function as the lambda is changing and then simply integrate. I hope it is clear to everyone.

[00:22:03]Great. Okay, you can just write if they change lambda in place of this lambda we will put the function and then we will integrate maths will be same. Clear? I hope you understood good.

[00:22:17]Second, second is electric field due to semicircular semic-ircular ring.

[00:22:40]Okay. So now we have a semic-ircular ring. This is the ring having uniform distribution of charge and the total charge is Q and the radius is R.

[00:22:57]Okay, total charge is Q. So what will be lambda? I told you here also we have charge along the length. It's a 2D body but the charge along the length. So lambda will be charge upon length and length of semicircle will be pi r.

[00:23:12]Clear? Now we have to find out electric field at the center O. Electric field at O. We have to find out that clear. Let's find out what you will do. You can see one thing. I want to find electric field at O. In the previous case, what happened? In the previous case, this was point O and the charge was changing with distance. See first element, second element, third element, fourth. Every element was at different distance from O. But in this case, in this case, every element is at R distance. See if I draw a ring here. Every element is at what? R distance. R distance. R distance.

[00:23:48]Distance is same for all the element. So what is changing here? Angle is changing. Here we have to play with angle. If you stand here, you have to rotate your head like this to see all the element. Is it? So here how we will find out electric field at O. How we will find out electric field at all? You will see. So do one thing. If the things are changing with angle from here you go to a random angle theta. There I went to random distance x. Here you go to a random angle theta.

[00:24:19]Fine. I went to a random angle theta.

[00:24:21]And you take a very small element very small element. This very small arc and this angle will be very very small which is d theta. So I went to a random angle theta and I cut a arc which makes what angle at the center? d theta. There I went to a random distance x and took a element dx. Clear? Now this is a arc very small arc. The length of the arc is let's say dl and as it's a very small element it will have very small charge dq.

[00:24:54]Again same thing I will find electric field due to a element. Then I will go for integration. Now what electric field will this element will give? You will say sir it's like a point charge yes or no it's like a point charge dq distance is what r positive charge electric field will be away d value will be k charge is dq point charge by r² yes or no so because of this small point charge I will find electric field which will be away yes or no let's say de which and you will say s you can write the to K DQ by R². Understood? Okay. Now if this is theta vertically opposite, this is also what theta. Now you know what is the problem here? The problem here is every element will give electric field in different direction. Every element will give electric field in different direction. Yes or no? If you look at this ring, if I take a point here, this positive charge will give field away. This positive charge away.

[00:26:02]This positive charge away. This positive charge away. This positive charge away.

[00:26:08]All the field are in different direction. We cannot integrate directly.

[00:26:11]Yes or no? Fine. So what we will do?

[00:26:14]What we will do? We will take component of DE for each element. I know electric field direction. I will take component of each electric field one horizontally and one vertically and I will add all the horizontal component and I will add all the vertical component yes or no.

[00:26:33]But you know here one more very good thing is there that is what if you see about this line the distribution is symmetrical. So whatever element we have here, here also we will have one element mirror element. Here also we will have one element and this element will also give electric field like this which will be de angle will be theta. Symmetrically this angle will be theta. So this angle also theta. Now this electric field I will take component what I will get? I will be getting de cos theta. Yes or no? Uh let me highlight this. I will be getting de cos theta and for the same I will be getting one down de sin theta.

[00:27:25]Same for this D. For this D de, this is theta. DE cos theta. This side D cos theta.

[00:27:36]And same uh just a second D cos theta and same this will be D sin theta D sin theta. Is it? You will say sir these two components are opposite.

[00:27:55]So these two components will get what?

[00:27:57]These two components will get cancel out. Horizontal component will get cancel out. Vertical component will get added up. In the same way this element and this element electric field electric field horizontal horizontal cancel out vertical vertical add up for every element on the right half we will have a element on the left half the mirror element which will cancel its horizontal field but their vertical field will add up. It means for every element here we have a element here which will cancel its horizontal field and their vertical vertical field will add up. It means for all the element de sin theta is getting add up for all the element which is vertically down. So for each element you can write for each element the vertical component gets add up. Which component?

[00:28:53]Vertical component. It means I can write I can write net electric field will be integration of all the vertical component for each element the vertical component is de sin theta.

[00:29:07]Is it clear to everyone?

[00:29:09]Now some of you will say sir I can see two element 2d e sin theta. You can take two but when you take two your integration will go only on the half this half part from here to here because you are taking two element at a time. If you take two then also same answer will come. Just be careful with your limits.

[00:29:28]If you take two it means you are taking two element at a time means you have taken this element automatically this element is covered. This element this is covered. This element this is covered.

[00:29:37]This element this is covered. This element this is covered. This element is covered. This element. So you will integrate only on the 1/4 part. But if you take d sin theta, it means you're taking only one component of a element.

[00:29:49]So you have to go on the entire half cycle to get net electric field. So what will be net field you will say? So net field will be integration of d. What is d? d is the small electric field because of the small element and that will be k.

[00:30:07]It's a point charge K dq by R². I wrote there K DQ by R². So you will say net field will be integration of K dQ by R² into sin theta. Oh, again a problem came. We are dealing with angle. Is it?

[00:30:27]We are integrating with angle. But the problem is with sin theta I should have d theta. Operator should be d theta. But what we are having dq. Okay. So again same thing we will do. You will say sir we know lambda. Lambda is charge upon length. Look at our element. Element has what charge? DQ. Element has what length? DL. Yeah. But I want d theta. If you all remember if I draw arc this angle is theta. Radius is r. Length of the ark is L. What is the length of ark? R into theta. So for our element what will be the length? What will be the length of the ark? What is the radius? R. And what is the angle arc is making d theta. Very small angle d theta. So dl will be r d theta.

[00:31:17]Therefore from here dq will be lambda r d theta. This dq I can substitute here.

[00:31:27]All of you pause and write till here.

[00:31:29]Pause and write. Fine. Great. I'm rubbing this part.

[00:31:34]Now net field will be look there integration of k sin theta what is dq dq just now we got lambda r d theta divide by r² fine so from here 1 r 1 r cancel out net electric field will be k is a constant lambda is a constant radius all All element has what distance? Radius which is constant integration of sin theta d theta. Now I am integrating with sin theta with theta. So my limits will be in terms of theta. So I from here I went to a random angle theta. If I want to cover all the elements what I will do if you look at this ring from here I have taken a random angle theta. Yes or no? So where will be the first element? So this will be the first element. this one. So for this angle will be 0°. Yes or no? Then I have to reach to the other end. So first I'm looking at the first element. Angle will be zero. Then finally I have to look at the last element. Last element angle will be 180°. Last element from here angle will be 180°. Yes or no? So limits will go from where to where?

[00:32:54]First element 0° last element 180°. Yes or no? So what you will get? K lambda by r integration of sin theta will be minus cos theta from where to where 0 to 180°.

[00:33:10]Put the upper limit or do one thing if you remove minus the limits get interchanged. So upper limit will be cos 0 minus lower limit will be cos 180. Yes or no? Is it clear? So what you will get k lambda by r cos 0 will be 1 minus cos 180 will be minus 1 this will be 2k lambda by r this will be the electric field and what will be direction of electric field you can pause and write you can see the cost horizontal component cancel out vertical add up so net field will be in what direction minus j direction is it clear to everyone I hope you all understood you can pause and write, pause and write.

[00:34:00]You know I have told vertical component get cancel out. If anyone have doubt in that what you can do if you see vertical component what we got de cos theta. So to find out electric field vertical component electric field vertical component you can do integration of de what cos theta and when you integrate you will get zero. Mathematically also we can prove.

[00:34:29]I hope it is clear to everyone what I have done. Listen, I went to a random angle theta. I took a element which is a arc. Angle is d theta. Because of the small element I got electric field d is it now every element was giving direction electric field in different direction. So for this element I got electric field like this. Symmetrically I took one element here. This give electric field in this direction. So I got two electric fields like this. I took their component their horizontal component horizontal component cancel out their vertical component add up and vertical component was d sin theta. It means for every element d cos theta is cancel out by the symmetrical element and d sin theta gets added up. So all d sin theta will be in what direction?

[00:35:16]Vertical down. So all the components will be in same direction. Now I can do very easily integration. I integrated and I got that answer. Is it clear to everyone? You can pause and write. Pause and write fast. Okay. Next one question. I'm giving you homework. I want you all to try and please write the answer in the comment section. Whatever answer you got. So this is a quadrant.

[00:35:43]Total charge is Q which is uniformly distributed and the radius is R. I want you all to find out electric field at O. Clear? I I will give you a hint. If you want you can skip the hint. Go to a random angle theta. Again take a element. This will be d theta. A small element will have a small charge dq. A small length dl.

[00:36:08]Because of the small element electric field will be like this. D. All the elements will give electric field in different direction. This will give like this. This will like this. This will like this. This is like this. Like this.

[00:36:18]All will give field in different direction. So what you will do now? If the directions are different we cannot integrate. So I will take component of all electric field. This is theta this also theta. I will take component of all electric field. I will get component in this direction de cos theta and I will get component in this direction de sin theta. Now any element you take for all the element one component sin theta will come down and d cos theta will come left. So you will get filled in two direction. One will be the field in y direction that will be integration of de sin theta and one will be electric field in x direction negativex this is negative y integration of de x will be d cos theta and this in this case this electric field in x will not be zero because there is no element to cancel out that. So you will get electric field in Y, you will get electric field in X. H you will get net field in Y and net field in X. Now to find out the resultant field you will take the resultant of both the fields.

[00:37:26]So resultant field will be under root of electric field in X whatever you got square plus electric field in Y whatever you got square this will be net field.

[00:37:36]So whenever the field is in different direction you take component horizontally vertically you add all the horizontal component with the integration you add all the vertical component with the integration. So now you got filled in this direction because of all the element you got filled in this direction because of all the element. So what will be net field? Net field will be vector sum of these two field. Is it clear to everyone?

[00:37:59]Understood? Try this as a homework.

[00:38:01]Pause and write. Fine.

[00:38:04]Let's generalize this. Now r can come 60° 90° 180 90 180 I I have shown you they can give 60° 120°. So let's generalize this because this comes a lot especially for neat kits I'm saying remember the result for neat and j means I want you all to remember the result direct question can come from here. Now it's a arc having charge Q radius is R and its substance at an angle theta at the center. I want to find out electric field at O. Concept is same. I want you all to pause and try. I want you all to pause and try. Just find the magnitude direction. You can find out positive charge. So away field will be down this much. I think you can guess. So what I will do? So I will draw a line here.

[00:38:58]Okay. And what I will be doing or let's do one thing. Okay. Fine. So let's make this angle beta.

[00:39:08]Let's make this angle beta. Okay. This total angle is beta. I will go from this vertical line. From this vertical line I will go to a random angle theta. From the vertical line I will go to a random angle theta. See whenever things will change angularly. This is what we will do. Taking element you will learn by doing a lot of practice. So go to a random angle theta and cut make a arc very small arc very small length very small charge dq which makes what angle? Very small angle d theta. Now because of very small element it will behave like a point charge. So this will give electric field away. DE yes or no. If this is theta, this also theta. Again I will take component of this. This will be D E cos theta. This will be DE sin theta. Yes or no? Symmetrically if I say this is a mirror these two are what? Mirror image.

[00:40:10]So symmetrically like here one element will be here which will also give electric field in the same way. Yes or no?

[00:40:19]Again whose horizontal this will be vertical. Again horizontal will get will get cancel out because of symmetry. For this element this element is there for this element this element is there. So symmetrically elements are there which will cancel like the previous case which will cancel the horizontal component and vertical component will get add up. So for each element on the right half a element is there on the left half which is cancelelling out its horizontal component. Like for this vertical horizontal one element is here which will give vertical horizontal horizontal vertical their vertical will get add up horizontal will get cancel out. It means for all the elements only which component will add up d cos theta.

[00:41:04]So what will be net electric field? You will say sir for all the element d cos theta d cos theta d cos theta d cos theta is getting added up. So add all the d cos theta component which is known as integration. So net field will be what? Integration of d e cos theta. What is d? d is a small electric field. What is d? A small electric field due to a point charge. Due to point charge what is the formula of electric field? K charge is dq distance is this is radius.

[00:41:34]For all the element for all the element distance is same r² into cos theta. Is it clear? Again same problem I am dealing with angle. So I have I should have d theta but I am having dq. So what I will take again I will take help of lambda. lambda is charge upon length. The element has what charge? Element has charge dq. Element has length dl. It's a ark. What is ark length? Ark length is radius into the angle substanted. So radius is r, angle is d theta. So this length will be r d theta r d theta. So dq will be what?

[00:42:15]lambda r d theta. So in place of dq I will put lambda r d theta. So integration of k lambda r d theta cos theta by r². You can see r r cancel out. So what I can do here is k is a constant, lambda is a constant, r is a constant integration of cos theta d theta. Yes or no? Now if you see limits now I have taken this was the vertical from vertical I have taken a element at an angle theta. Is it this is what I have done? So I have to cover all the element. To cover all the element what I will do? Look this is the vertical line.

[00:43:00]So from the vertical line if I go right from here to here. So from here to here.

[00:43:07]So total angle is beta. So from here to here this half angle will be beta by2.

[00:43:13]And if I go back this is clockwise this is anticlockwise. If I go back this angle is also what? Beta 2. So limits will be from where to where? Limits will be from from this vertical. This is beta x2 anticlockwise. This is beta x2 clockwise. So if I take clockwise positive, anticlockwise will be negative. Be careful. So limits will go from minus beta by 2 to beta by2. You want you can do from here also you can go from here at an angle theta. Take a element find electric field then integrate from 0 to theta beta.

[00:43:50]Integrate from 0 to beta. You can do like this also like I have like I have done for that semic-ircular ring. You can do answer will come same thing only is it? So this will be k lambda by r integration of cos theta will be sin theta limits beta by 22 to minus beta by 2 which is equal to k lambda by r upper limit sin beta 2 minus lower limit sin minus beta by 2 sin of minus theta is minus sin theta. So k lambda by r sin beta by 2 and sin of minus beta by 2 is minus sin beta x2. So minus and one more minus clear you can write sin of minus theta is minus sin theta. So 1 minus will come out minus minus will become plus so sin beta by 2 plus sin beta x2 sin beta x2 sin beta by 2. This is the formula of net electric field at this point O which is 2 k lambda by r sin beta by2 where beta is the angle that the ark makes at the center. One more thing you might be confused sin beta by 2 or sin beta x2. So sin beta by 2 clear this is the electric field at the point O and direction you can find by common sense is it you can just pause and write how to find out direction all are positive net field will be down anyway you are getting vertical component only but you can use common sense these are all the charges positive they all will give resultant down this is direction of electric field is it clear to everyone you can pause and Right? You can check also for the ring.

[00:45:42]If you look for the semic-ircular ring, what is the angle? 180°. Put 180. 180 by 2. 90. Sin 90 is 1. You get 2k lambda by r. If you check, same result came 2k lambda by r. 2k lambda by r. direction you can find with the help of common sense.

[00:46:02]Pause and write. Pause and write. Done.

[00:46:04]Clear. Fine.

[00:46:07]Next.

[00:46:08]Now we'll now third is very very important. This third or fourth whatever this is electric field at a random point.

[00:46:21]Random point at a perpendicular distance at a perpendicular distance d from a finite line charge. finite line charge. So we are having a finite line charge. The total charge let's say is Q.

[00:46:51]Length is L. This is not required. Just lambda is required. Lambda is defined as charge upon length. So for this rod lambda is given lambda kum per meter.

[00:47:02]And what is the unit? Kum per meter. So lambda unit is kum per meter. What is sigma unit? Charge upon area kum per meter squared. What is row unit? Charge upon volume. So kum per meter cube.

[00:47:14]Lambda kum per meter sigma kum per me²ared row kum per meter cube. Okay. From this finite line charge at a perpendicular distance d perpendicular distance d at this point let's say o we have to find electric field at O.

[00:47:33]Clear? Pause and write this. I want to go to next page. Fine.

[00:47:47]So what we will do?

[00:47:49]This is a line charge having linear charge density lambda.

[00:47:55]Charge per unit length lambda. From here at a distance I want or at d distance I want to find out what electric field again this is a continuous charge distribution I know because of a point charge. So what I will do I will go to a random distance y.

[00:48:17]I will go to a random distance y and I will take a very small element of what thickness? A very small element of thickness dy and this small element will have very small charge dq and this I will connect from here.

[00:48:36]Clear? This angle is theta. You can do with theta also. You can say so from here I will go to a random angle theta and I will take a very small element which makes what angle? D theta. Then also this length will come dy. Both you can do from here you can go to a random angle theta. You can take a small element which makes what angle d theta and this element will have what length dy whatever you want you can take no problem or you can go to a random distance y and take element dy. Now this element very small element will have very small charge dq and this is very small charge will behave like what point charge is it? And let's say this distance this distance is R. This distance is R from the point charge to the point. So basically what we got we got a point charge. This is our point charge. And from here R distance we have the point O. So a positive point charge a positive point charge will give electric field away. This will give electric field away. Is it electric field away? Which will be DE. Okay. And what will be the value of DE? You will say sir D will be K. Charge is very small DQ by distance between the point and charge is R². You can just leave it like this this part. Okay. You can just write for your understanding. Kisser has made a point charge because we know the formula for a point charge. Fine clear understood. Good. Now see what I will do. Because of the small element I will find I will draw electric field. I have shown you there but here also I will draw. So from the positive charge electric field will be away and very small electric field D is it?

[00:50:24]Now if this angle is theta this also theta vertically opposite angle. Now the problem is every element will give electric field in different direction different direction different direction different direction different direction.

[00:50:40]I will draw here. Is it? See multiple problems are there in this problem? Yes.

[00:50:46]What are the multiple problems? This element will give electric field in this direction. This element will give electric field in this direction. This will give in this direction. This will give in this direction. All the elements are giving electric field in different direction. Problem number one. Second problem is if you take this element, this is your y.

[00:51:10]Uh let me change the color not pink.

[00:51:13]This is your y.

[00:51:16]This is your r. Yes or no? And this is your theta.

[00:51:22]If you change the element, then what will happen? Okay. If I change the element, my y will change. Yes or no? My r will change. Yes or no? as well as my angle will change. Yes or no? It means we are having three variables.

[00:51:39]Clear? See basically concept is same.

[00:51:42]Maths will increase nothing else. Fine.

[00:51:44]So first thing is all the electric field because of all the elements are in different direction. All the electric field because of all the elements are in different direction. Second problem is we are having three variables. We are having three variables. So what to do?

[00:51:59]What to do? Again same. First I will resolve the problem of different direction of electric field. So I will take component for each element. For each element I will take the component of the electric field. One will be horizontal, one will be vertical. Clear?

[00:52:16]So I will take component of this DE. So this side what I will get D E cos theta and here I will get DE sin theta.

[00:52:28]And I will call this component as the perpendicular component of electric field. We'll write later. So for each element for all the electric field for all the electric field I will do component horizontal vertical horizontal vertical horizontal vertical horizontal vertical and I will add all the horizontal component to get perpendicular electric field. This is known as perpendicular. If I will add all the horizontal component I will get perpendicular electric field. And if I add all the vertical component I will get vertical electric field. So here I will be giving you two formula. One for perpendicular electric field and one for parall electric field. Clear?

[00:53:10]Understood? It means you will say you are saying you can write here we will take element of all the electric field.

[00:53:17]Clear? Clear? And then if I will add all the horizontal component the net electric field horizontally which I will name as perpendicular electric field because this is perpendicular to line charge is integration of d cos theta all the element has electric field all the elements electric field I have broken into horizontal vertical component. I will add all the horizontal component integration of d cos theta and I will add all the vertical component which will be net electric field what vertically which I will call as parallel electric field is equal to integration of d e sin theta any problem till here all of you just pause and check whatever I have wrote clear so I'm going to do integration you can draw this part at the corner to get the better feel. Pause and draw. Fine. And right there we are having three variables and also all the field are in different direction. So this problem I resolved by taking component. Clear? Done. Good.

[00:54:30]Next I am going to do integration. But before integration let's find out the value of D. I told three variables. Let's see. So D will be if you see what is D? electric field because of that element. So what will be that K what is charge there? dq divide by what is this distance small r² clear? Now the problem is I want to integrate with angle. You know we have three variables angle r and y. Y can change if I change the element Y will change. If I change the element R will also change. I have shown you as well as angle will also change on changing the element. Three variables we are having is it. So what I will do I want to integrate with theta because in the books and all the question which comes the formula is in terms of theta. So what I will do if you look here look at this small element. Can I write for this element uh lambda is charge upon length. What is charge on the element? The charge on the element is dq. What is the length of the element? The length of the element is dy.

[00:55:43]Therefore, dq will be what? dq will be lambda dy.

[00:55:50]Clear? So, in place of dq what I can put? You will say so you can write d is equal to k dq is lambda dy by r². I want to integrate with theta. So, let's do one thing. Let's put this d value. So, what will be e perpendicular? E perpendicular will be integration of K dQ by R² I remove DQ in place of DQ what I wrote lambda dy very simple that's what we are doing so k lambda dy by r² e perpendicular contains what cos theta so I want to bring everything in terms of theta if I have theta my operator should be d theta integration d theta should be the operator I want d theta But I'm having dy problem and r is also a variable. I want to write r in terms of theta again. So what I will do for that? I want everyone to focus on this triangle.

[00:56:51]Focus on this triangle.

[00:56:55]If you look at this triangle, you will find out that what will be the value of tan theta in that triangle. You will say so tan theta is perpendicular. This is theta. This is theta. You will say sir perpendicular y by base. What will be base? This is our base d is it. So from here y will be d tan theta.

[00:57:21]I want you all to differentiate this with respect to theta. So dy by d theta is equal to d constant. Differentiation of tan theta will be se² theta. So from here I will get from here I will get dy is equal to d se² theta d theta hey this d is the perpendicular distance perpendicular distance keep in mind don't think it's differentiation okay so in place of dy I can write e perpendicular as k lambda what is dy dy is d uh se² theta into cos this is cos theta. So what is dy d se² theta d theta d se² theta d theta this is our what this is our dy divide by r r is also a problem. So in the same triangle again if you observe in the same triangle you will get to know this is the solution of the first problem. Second problem in the same triangle what will be cos theta? You will say sir cos theta will be base by hypotenus. What is base? D. What is hypotenus?

[00:58:39]R is it? So r will be d by cos theta.

[00:58:45]So you can put r as d by cos theta. So you will write d by cos theta and square of that. So d² by cos² theta. Cos square theta will go on the top. So cos square theta se square theta will get cancel out. So cos² theta se square theta gets cancel out and what we will be getting you can just pause and write that part done. So what you will be getting this was the second problem solution.

[00:59:11]So this will be k constant lambda constant 1d 1d cancel this d is the perpendicular distance by d integration of what is pending cos theta d theta cos theta d theta clear now I am doing integration with theta so what will be my limits everyone pay attention I want to find electric field because of all the element and from here I have taken a random angle theta see From here I have taken a random angle theta. So from here if I go like this here see I have covered all the element.

[00:59:50]If I go like this I have covered all the element. And same if I go below I have covered all the element. Yes or no? So so if I just in uh uh which color red?

[01:00:03]So from here if I go and connect the top end this angle theta_1 if I say and same if I go bottom from here and this I write as theta_2 you can see that from here theta_1 from here theta_2 I have covered all the elements from here if I go clockwise theta_1 all the elements are covered. See all the elements and below if I go theta_2 all the elements are covered. So my integration will go from where to where you will say. So integration will go from from baseline if I take clockwise positive. So till theta_1 and below anticlock means minus theta_2 like the previous case. So our limit will go from where to where limit will go from minus theta_2 to theta_1.

[01:00:54]So this will be k lambda by d.

[01:00:58]Integration of cos theta will be sin theta upper limit theta_1 so sin theta_1 minus sin of minus theta 2 clear clear and just now I told you in the previous case that sin minus theta is what minus sin theta so I can just write it as what plus sin theta_2 one more thing keep in mind very very important derivation if you forgot for once it's fine but you have to keep in Find D is the perpendicular distance.

[01:01:31]Very very important. D is the perpendicular distance and theta_1 is the angle taken from perpendicular from perpendicular clockwise. Theta_2 is the angle from perpendicular anticlockwise. Yes or no? Look at the diagram. Because the way we are doing derivation in the same way we have to put the variables from the perpendicular one side you connect the end theta1 is the angle below if you connect the end theta_2 is the angle. So that distance D I should not take D you know because differentiation D is also there I should take some other A or R but fine no issue you can remember that this D is what perpendicular distance and from the perpendicular one angle is clock one is anticlock so in this also one should be clock one should be anticlock and I have put with sign and all so don't put with sign just put the values clear understood pause and write pause and write I will zoom that one second Yeah, pause and write this part. Pause and write. Done. So, right now we got which which electric field? Perpendicular electric field. So, perpendicular electric field is what? K lambda by d sin theta_1 plus sin theta 2. Now, one is remaining.

[01:02:57]We have two electric fields here. Yes or no? We got perpendicular. Next is parallel. So let's do on the next page.

[01:03:04]What is E parallel? D E sin theta. See D you got. Now this is D. This is D. D you got. Just put this value. Here we have cos theta. Here we have sin theta. And let's integrate. So this only will come again. Let me show you on the next page.

[01:03:21]So E parallel will be integration of DE.

[01:03:26]What we had? Integration of D sin theta.

[01:03:30]Integration of D sin theta. What was D?

[01:03:33]d was K. Charge was DQ. Distance was R² sin theta. Is it? So K, what was DQ? dQ was lambda DY by R² into sin theta. I'm doing again for you. Okay. dq was lambda dy. Yes or no? on this element small element length was dy charge was dq and lambda was given is it and then I told again we have to bring in terms of theta so in that triangle what I did k lambda in place of dy what we got you all can see d se² theta d theta we got d se² theta d theta / r² what is r² square if you check d² by cos² theta and limit went from minus theta_2 to theta_1 sec square theta okay I missed sin theta I missed sin theta sorry cos² theta cancel out dd cancel out and e paral will be k constant lambda constant d will come out integration of sin theta d theta from minus theta_2 to theta_1 and this will be giving U K lambda by D integration of sin theta will be minus cos theta minus cos theta upper limit theta_1 minus cos lower limit minus theta_2 is it correct upper limit oh we have minus now so one second so This will be minus cos theta limit from minus theta_2 to theta_1. If I remove the minus the limits will get interchanged. So minus theta_2 to theta 1. So the parallel electric field component you will get k lambda by d cos of minus theta 2 minus cos of theta 1. Is it clear? And cos of minus theta will be what? cos theta. So you can just write this as cos theta_2. This is the electric field parallel to the finite line charge. Clear? Understood? Pause and write. Pause and write fast. Okay.

[01:06:04]So what we got? Because of the line charge, this line charge with linear charge density lambda at a perpendicular distance d. I got electric field two component one perpendicular which came how much which came k lambda by d sin theta_1 + sin theta_2 d is the perpendicular distance theta_1 angle clockwise from the perpendicular theta_2 angle anticlockwise and this is e paral which is k lambda by d cos theta_2 minus cos theta 1 now you will say sir How you got to know? This will be perpendicular. This will be parall. It can be parall above also. For this you apply common sense. What common sense? You know basically here we have less charge. Here we have more charge.

[01:06:59]If if if you look about this perpendicular here we have more charge here we have less charge. So basically what is happening? This is giving electric field in this direction. This more is giving electric field in this direction. Who is dominating? This is dominating because more charge is giving yes or no. So if this is dominating so in these two who will win this will win.

[01:07:22]So net field will be what? Somewhere here. So its component will be like what? This and this. So this is E perpendicular. This is E parall. This is how you can judge. Let's say you forgot which was theta_2 which was theta 1. No need to worry. You write this theta 1 this theta_2.

[01:07:40]Just put a mod. Whatever answer you get put a mod and direction decide like this which segment is more dominating the segment which is more dominating according to that segment I will put my E parallel and E perpendicular direction of E parallel and E perpendicular I hope you understood clear like if let's say this point I'm finding then then what will happen you will say so this is less this is more so net field will be in this direction so this will be E perpendicular this This will be E paral.

[01:08:13]Is it clear? If someone asked what is the net field? What is the net field?

[01:08:17]You got E perpendicular, you got E parallel. So this will be your net field. And what will be the value of net field? You will say some net field will be vector sum of both the field under root E parallel² plus E perpendicular square. Yes or no? H I got two vectors E parallel E perpendicular. So to find out the net field I will do the vector sum of both the field. Is it clear to everyone?

[01:08:44]Fine pause and write. We'll do some problems.

[01:08:48]Okay. So for today some good problems are there. Don't worry this is not the end. Next lecture I will take some more elements. Some good elements. I will take disk, half disk, hemisphere and know for those we will find out electric field. Very interesting. So let's do some problem.

[01:09:06]lambda is given. This distance is A.

[01:09:11]This angle is 30°.

[01:09:14]This angle is 60°. Find E parallel and E perpendicular.

[01:09:21]Question number one. Question number two. Again lambda is given.

[01:09:28]This distance is given as A.

[01:09:35]This angle is 30°. This angle is 20°.

[01:09:39]This point is O. Or don't take 20. This also you take as 30°.

[01:09:46]Find electric field at O.

[01:09:50]And this distance is A. All of you pause and try done. So first case E parallel E parallel or let's find out first E perpendicular K lambda by perpendicular distance A sign from the perpendicular join the top end one angle should be clock okay it's clock plus other angle should be anticlock yes it's anticlock that's all E parallel K lambda by A cos theta_2 2 theta_2 means second angle first. So cos 60 minus cos first angle will be 30. Is it clear? Then how to find out direction?

[01:10:35]No need to worry. Just keep in mind and you know if you forgot which one theta 2 theta 1 no worry you write cos 30 minus cos 60 direction you have to find out.

[01:10:44]Take the mod. Always take the mod direction. How you will find out? You will say sir in this case this region is more dominating. If this region is more dominating, this region is more dominating. This will give filled like this. So net field will be in this direction. So you will get like this.

[01:11:00]This will be your E perpendicular. This will be your E paral. This will be your E perpendicular. This will be your E parallel. No need to find just you can just put a mod. This is the mod direction. I will find with common sense. Okay. So no need to worry about angle and all. Clear? Come to this problem. You know this type of derivations I will do in magnetism also.

[01:11:24]Even that part part is a bit easier because there we have only one direction of magnetic field not like E parall E perpendicular not like two component.

[01:11:32]Okay this question has came from magnetism a lot but they have not given till now from electrostatics they can give. See how very very important question. So you will say sir E parallel any anyway this is the charge. So you know the net field will be what? in this direction. So it will have two component E perpendicular E paral direction I got I need to just find the magnitude. What is E parallel? You will say no first let's find out perpendicular K lambda by A or D. K lambda by D. D is the perpendicular distance. Is it the perpendicular distance? Is A the perpendicular distance? Yes or no? No. A is not the perpendicular distance. You need to take perpendicular distance.

[01:12:17]What will be perpendicular distance?

[01:12:19]This will be the perpendicular distance.

[01:12:23]H just extend it. This is perpendicular.

[01:12:27]So this is a angle is 30. A cos 30. So perpendicular is a cos 30. So perpendicular distance is a cos 30. Then sin theta 1 + sin theta_2. From the perpendicular we take the first angle.

[01:12:45]So from the perpendicular you join the top end. So from the perpendicular first angle should be what? Clock. Clock. What is that angle?

[01:12:55]60°. Yes or no? So first angle is clock 60°. And we take the angle from the perpendicular. That's why I remember I told you to remember derivation you forgot. Fine. But keep in mind theta 1 clockwise from the perpendicular theta_2 anticlockwise from the perpendicular.

[01:13:13]And the distance we take that D should be always the perpendicular distance because the options will be confusing.

[01:13:19]They will keep the options confusing.

[01:13:21]Keep this in mind. Okay. They will not give you that directly you can get the answer. They will keep option in terms of A, in terms of D, in terms of angle they will try to confuse you. Plus sin theta_2 sine. Now theta_2 should be anticlockwise. So from here where you are finding the field, you join the lower end. So till now what was happening? This was upper end theta_1.

[01:13:45]This was lower end theta_2. Is it 1:00 one anticlock? But here what is happening? If I go from the perpendicular anticlock, this is the anticlock. From here, this is the anticlock. So if this is 30, this is 330°.

[01:14:02]So what you will write? Sin 330°.

[01:14:06]Yes or no? Instead of this, you know what you do? Whenever one angle is coming clock and other also comes clock.

[01:14:14]I think first of all what you thought sir in the in this case one angle is clock other is also clock. So whenever you get both clock put a minus sign because sin 330 is what you will say k lambda by a cos 30 sin 60 minus sin 30. This is the same thing which you will get. So either you take one angle clock and from here one angle to the other edge anticlock which is 330°. But to save your time concept you understood just keep in mind if both are in the same sense one you take positive other you take negative. That's one you take positive one you take negative.

[01:14:59]Problem solved. Clear? In the same way what will be E perpendicular? E perpendicular we got. Now what will be E parallel? You will say sir E parallel will be E paral will be K lambda by perpendicular distance is A cos theta and now we have cos theta_2 so cos what is theta_2 what is the other angle other angle is 330 330 so cos 330 or you can take what cos 30 is it minus cos what is the first angle First angle is 60.

[01:15:39]Is it clear to everyone?

[01:15:42]Understood? Fine.

[01:15:44]Okay. Can we move to next part? Pause and write. Keep in mind we have to take perpendicular distance. From the perpendicular distance you join both the end. From the perpendicular distance you join both the end. The top end. So one angle should be clock. The bottom end the other angle should be anticlock. But to save your time as I told you you go 330 you write sin 330 or you go in the same sense 30 and you write what minus sin 30 both will give the same value. So you should be good in trigonometry and this values as well. I hope you understood. Pause and write. If there's net field, you can find the net field.

[01:16:22]Not a problem, is it? Now, electric field due to infinite line charge.

[01:16:29]Infinite. This I will do with the help of Goss law also. You will learn Goss law. But here also I want to show you.

[01:16:36]This is infinite. Infinite.

[01:16:39]See practically infinite does not means that line charge will come from Venus and go till Mars. It's a relative term.

[01:16:46]Infinite point these are what relative term. For example, if I say out of this which is point you will say sir first one is point but no if someone is seen from far distance all three are point and if I bring a insect here insect will say that so this is very big disc none of them are point. So point and infinite both are what relative term. If I say infinite line charge if I say infinite line charge you know what does that mean? That means you are very close to line charge. When you are very close and look up and down, it will be like what?

[01:17:22]Infinitely long. It will be like what?

[01:17:24]Infinitely long. Let's say this is 100 m and you are 1 cm. So for 1 cm, I can say what? 100 m is nearly what? Infinite.

[01:17:34]Yes or no? This is the meaning of infinite. So from this infinite line charge at what distance? At d distance we have to find out electric field.

[01:17:44]Clear? Same thing E parallel E perpendicular. What will be E parallel and what will be E perpendicular? So you will say sir K lambda linear charge density is lambda by A. From the perpendicular you have to join both the ends. Now this is going till infinity.

[01:18:03]So if you want to draw a line which meet this line at infinity, we know I have to make it parallel. Parallel lines meet at infinity.

[01:18:13]Basically this is not actually infinity.

[01:18:16]You know basically what is happening?

[01:18:18]Let me explain you first mathematically.

[01:18:21]This is let's say 2 m and your point is at let's say 0.2 cm.

[01:18:29]So which will be what 0.002 m. You will say sir join the upper end.

[01:18:37]This angle is theta_1. In this triangle, what is tan theta? Perpendicular by base 2 m by 0.002.

[01:18:48]So tan theta_1 will be what? This will be nearly what? So perpendicular which is 1. Now sorry perpendicular is 1 and base is what? 0.00.2 2 into 10^ - 3.

[01:19:03]This is 10^ 3x 2 which is 500. So tan theta is 500. So theta is nearly what?

[01:19:11]90°.

[01:19:13]Yes or no? If you are very close and you want to join the upper end, you have to go nearly what? Parallel. This angle will be nearly what? 90°. Yes or no? So to reach to infinity, this line has to go parallel. So this angle will be nearly 90. 89.99999.

[01:19:32]And if here I want to go and meet, this will also go what? nearly at 90°.

[01:19:38]So what you will write? So you can just pause this just for understanding you can pause. Okay.

[01:19:46]So I got both of my angles. The angles are sin theta_1 sin 90 plus sin theta_2 sin 90 sin 90 sin 92 and I got k lambda by a 2 2k lambda by a in fact 80% 85% question comes from here only okay second is e parallel if you check e parallel will come zero why zero k lambda by a cos 90 - cos 90 cos 90 is 0 you will get zero and it's obvious now why because about this line upper half whatever charge we have lower half we have the same charge so this entire charge will give electric field in this direction this entire charge will give in this direction and as the charges are equal so for this electric field I will have two component for this field I will have two component there vertical will get cancel out horizontal will add up that's why parall is zero and we have only perpendicular I hope you all understood you know one more thing I can do here I can ask you that again I have a line charge for which linear charge density is lambda and this is semi-infinite wire find electric field at O what is the meaning of semi-infinite here it's going till infinity at this point you have to find electric field again you will say sir E parallel perpendicular distance is D E parallel E perpendicular.

[01:21:20]So first you write K lambda by perpendicular distance is D. K lambda by perpendicular distance is D. So I told you from the perpendicular you have to join both the ends. So one end will go till infinity. So to go to infinity you have to go nearly parallel. So this angle will be 90° and one ang one one end you will touch directly without rotating to any of the angle. So this angle will be 0°. So for semi- infinite one angle is 0. So sin theta 1 is 0 sin 0 plus sin 90 one angle is 90 and here also cos theta_2 so cos 90 minus cos theta_1 cos 0 this is what you will get I'm saying again if you forgot this theta 1 theta 2 no issue you write cos 0 minus cos 90 just get the magnitude direction we will find with common sense so with common sense direction will be entire charge is here so fill this away in this direction so this will be E perpendicular this will be E parallel direction. Is it clear to everyone? Understood? Fine.

[01:22:21]Let's do some more problems.

[01:22:27]Okay, the video is becoming a bit big.

[01:22:29]No issue. Infinity infinity linear charge density lambda lambda at this point a distance from here a distance from here at this point. O find electric field at O find electric field linear charge density is same pause and try pause and try done okay good so I will find individually first I will find because of this wire one let's name this as wire one so if you look at the wire one this is the point and if I join perpendicular distance is a if I join this and I join this is infinity now this will go straight so this angle will be angle one will be 90 and as it is a square a A it's a square. So this angle will be what? 45. Yes or no? So what I will get for the wire one? I will get two electric field. How? And one more thing. Here we have less charge. Here we have more charge. So net field will be in this direction because of the first wire. Is it? Are you understanding? So first wire will give field in this direction. So for this what will be the component? This will be perpendicular.

[01:23:39]This will be parallel. So this will be perpendicular. This will be parall. What will be perpendicular? K lambda by perpendicular distance is a first angle is 90 sin 90 is it minus plus sin other angle is 45.

[01:23:59]Can I write the value 1 + 1x <unk>2? So 1 + 1 by <unk>2. In the same way, what will be E parallel? K lambda by A cos theta_2 mean cos 45.

[01:24:14]So this will be cos 45 which is 1 by <unk>2 and cos 90 minus cos 90 which is zero. I got the electric field because of the first wire. Now I will check for the second wire. This is the second wire and this is the point. This is the perpendicular A. So I will join the ends. I will join the ends. This will go 90°.

[01:24:37]And again you can see from here this is 45°. It's a square other is 90 is it? So this angle will be what? 45°.

[01:24:46]I hope you can understand. Now in this case I know one will go parall. This is our line. This is our line charge.

[01:24:53]So in this case what you will be getting at this point which region is dominating here we have less charge. Here we have more charge. This is more. So net field will be in this direction because this is more. So net field will be in this direction. It will again have one e perpendicular component and this will be what? E parallel component. So let's write here. So for the second wire, for the second wire, what will be E parallel? In this direction, what is this E parallel?

[01:25:22]K lambda by perpendicular distance A.

[01:25:25]One angle clock, one angle anticlock. So this will be what? This will be what?

[01:25:30]cos just write the magnitude now I got the direction just write the magnitude no issue so one is cos 45 1x <unk>2 and cos 90 will be zero and one will be e parallel e parallel will be k lambda by a which will be then sin 45 + sin 90 which will be equal to what sin 45 1x <unk>2 sin 91 so 1 + 1x <unk>2 so these two directions right and Down I got because of wire one and left and up I got because of wire two. So I got four vectors. How many vectors? Four. Let's plot them. Let's plot them. So what I will be getting? This is right. How much? K lambda by a 1 + 1 by <unk>2.

[01:26:19]Then we have one down. How much k lambda by a 1x <unk>2 means by <unk>2 a is it?

[01:26:28]Is it here? up up how much? Uh, open the bracket. K lambda by A plus K lambda by <unk>2. Open this bracket also. K lambda by A plus K lambda by<unk> 2. And we have one left. What is that left? What is this left? This is K lambda by<unk> 2A. So down is K lambda by<unk> 2 A. Up is K lambda by <unk>2 A plus K lambda by A. So you can see this this and this will get cancel out and here this and this will get cancel out. Here we have a a I forgot to write a sorry is it? So what I will I will get finally. So see these two are canceling out opposite this is pending k lambda by a. These two are canceling out right and left this is pending k lambda by a. So net electric field will be what? Net electric field will be under root a square + b square + 2 a b cos 90 is it under root a square + b square + 2 a b cos 90 cos 90 is zero. So <unk>2 k lambda by a this is what you will get under root a square + b square both are equal now when two vectors are equal result comes root two times when they are at 90° and if two vectors are equal equal force if someone will pull me I will come exactly in between they that two are equal so net electric field will be exactly in between and value will be this pause and write understood so what I have done nothing right nothing I Just calculated electric field because of this parallel and perpendicular. Because of this parallel and perpendicular for this wire this segment was dominating.

[01:28:12]So net field was in this direction. So component was right and down.

[01:28:17]And for this this segment was dominating. So net field is in this direction. This direction up and left. I have just plotted all the four together.

[01:28:26]That got cancel out and I was left with k lambda by a up k lambda by a right. I just took the resultant. Is it clear to everyone? Is it clear? You can just pause and write. Pause and write. Fine.

[01:28:40]Next.

[01:28:42]Okay. I think you wrote. You wrote.

[01:28:45]Clear. Clear this part. Let me zoom it.

[01:28:49]Clear. Fine. Okay.

[01:28:52]Next is let's do one more problem. One more problem. I have a infinite line charge.

[01:28:58]Infinite. Let me connect with mechanics.

[01:29:00]infinite line charge at R1 distance I release a charge Q1 release Q1 at R1 distance I release Q1 find velocity of charge Q let's say Q when it is at distance when it is at distance at distance R2 see this is positive charge line charge infinite line charge So infinite line charge means what? There will be no par electric field. Parall field will be zero. It will give only what? Perpendicular field is it? So this charge positive charge positive line charge it will repel and this positive will move away. You have to find the velocity of this positive charge when it will be at what distance?

[01:29:51]R2 distance. Again this question I can do in one line when I will teach you potential. But right now let's say we don't know what is potential and all how to solve. So you can see here we will get only which electric field perpendicular. How much? 2 K lambda by D. If you check turn your pages check 2K lambda by D. D is the perpendicular distance. E parallel was zero. There was no E parallel. Now this electric field depends upon perpendicular distance. As the charge will go from here to here.

[01:30:26]The perpendicular distance will keep on changing as the charge will go.

[01:30:30]Perpendicular distance will keep on changing. If the perpendicular distance will keep on changing, it means electric field will keep on changing.

[01:30:38]And how much force electric field apply?

[01:30:40]If you all remember the second lecture, force because of electric field is QE.

[01:30:46]Charge multiply by the field in which charge is present. Charge multiply by the field in which charge is present.

[01:30:52]That is the force. Now if distance will change perpendicular distance will change as the charge will move electric field will change which means force will change. So what I will do again same I understood I have to integrate because something is continuously changing. I will go to a random distance X. At random distance X what will be the force? You will say sir force will be Q into E. What will be E there? E will be 2 K lambda. Random distance is X. What they're asking velocity and they are giving us distance. I need to connect velocity and distance. I know force from force what I can calculate from force I can calculate acceleration so we know force is mass into acceleration let's say mass is m so force is mass into acceleration so acceleration will be q 2 k lambda by m into x okay I got acceleration I want velocity acceleration is dv by dt but time is not given okay what I can do what I can do I have to relate with distance so I know if you all remember I told in one of the lecture acceleration is dv by dt. You multiply by dx and divide by dx. This dx by dt is what? Velocity. So acceleration is v. This is velocity dv by dx.

[01:32:11]Okay. So I can relate velocity and dx.

[01:32:14]Velocity and x.

[01:32:17]I can relate acceleration with v and x.

[01:32:19]So one is dv by gt. When time is given I will use dv by gt. Here velocity and x relation they are asking. So I will write acceleration as what? V DV by DX I will send DX there. So what we have Q 2 K lambda by MX and this DX I will send there. Clear? Clear. Now integrate.

[01:32:42]Integrate. How to integrate? Look at the limits. Charge is released at what distance? R1 distance. Yes or no? So when veloc limit of velocity is zero because charge is released and here we are integrating with distance. So particle starts at what distance? R1. So lower limit is R1. Then finally velocity is V. That's what we have to find out.

[01:33:04]And finally distance will be what? From here distance I have to find the velocity at R2 distance. So final distance will be R2. That's all. Now you can integrate uh V DV will be V² by 2 upper limit V lower limit zero. So v ² by 2 is equal to 2 constant k constant charge constant mass constant dx by x integration will be ln x limit r1 to r2.

[01:33:31]So what you will get? So v² I will send this two here 4 k lambda q by m upper limit ln r2 minus ln r1. We know ln a minus ln b is ln a upon b. So ln r2 - ln r1 will be what? ln r2 by r1. If I remove this square, if I remove this square, this will come under root. This will be the velocity when the particle will be at r2 distance. I hope you understood. Just pause and write. So I think this is sufficient for this lecture. I will come with one more lecture in which I will take some more pro some more better problems some advanced level problems or beyond advance we will do some questions in the next lecture I hope you understood for J kids kits especially for advance it's very important to remember all derivation in whatever way I will I have done for need kits just feel the derivation and try to remember the result is it clear to everyone okay so bye take Here.