 You know you can treat it like a numerical rather than derivation because nobody is going to ask these derivation in your school exam so even though I am Teaching you like as if it is a derivation treat it like it is a numerical okay case number two is Circular charge at the center. How much? find out There's a circle Charge per unit length is lambda Okay, radius of the circle is Capital R you have to find out the elective field at the center Okay, all of you It's a direct thing. Yes, it is equal to zero Okay, so I think many of you have got it You have to understand you're dealing with vectors Okay, so suppose you take a small charge at the top Take a small DQ at the top. Which direction the field will be because of this What do you get downwards? Just look at field you to DQ now take a Charge here DQ Which direction field you do this will be upwards They these two will cancel out So I'm saying for every charge you take at a location. They'll be diametrically opposite another charge Which will cancel out the electric field. So like this you will get vectors Elective field vectors like this They will all cancel out. So electric field is zero for uniformly distributed charge New formula all of you understood this Clear. Okay. Now do this one. Suppose. This is a scenario circular arc Okay, center was somewhere here, right? This is a center charge per unit length is lambda okay, and The scenario given to you is this radius is R This is theta This is also theta Try finding out electric field Due to this circular arc You also have to use the symmetry Can anyone tell me which direction the net electric field will be? Towards left towards the arc along the right, okay Okay Along the dotted line by sector correct. Okay. All right So I just demonstrate to you the direction of the electric field Suppose you take a small DQ here Okay, which direction the field will be because of that It will be like this now Take a symmetrical DQ over here as well Take a DQ here now The electric field will be like this So now between these two when you add these two vectors, which direction the field will be because of these two These two field will be in which direction their components Will get cancelled along this line The component will remain over here Yes or no Along the x axis right Okay, so electric field You can take any charge you will always get a charge Symmetrical you can take here and another charge here one Field will be like this other field will be like that So their component along this line will get cancelled and along that line will get added up Fine. So because of that, I already know that my net electric field will be in this direction only Electric field will be in this direction. So I will Integrate along that direction only Electric field is a vector quantity you take component and integrate Okay So can you do it now or should I solve it? Okay, I'll solve it now Take a DQ over here. I have a DQ Okay, electric field will be naturally in this direction. This is My DE Okay, this angle is a variable. Let's take it as Phi This angle the small angle is D Phi the small angle this element makes This element is DQ Okay Which is equal to lambda into DL DL is this small arc this much Okay, which is Rd Phi Rd Phi so lambda into Rd Phi This is my DQ Okay, so all of you understood till now This angle is Phi So I know electric field is a oriented direction So I will integrate the electric field along x axis which is DE Cos of Phi Okay, DE is what? K into DQ By R square cos of Phi All of you type in is it clear till now Everyone what is a doubt if there is Feel free to ask doubts, okay It is very easy to ignore What is I where is I? L where is L as a DL DL you mean this is DL The small arc length You have to consider small length for DQ right DQ will occupy small length because it is distributed on the length only Okay, so the charge on that DL small length DL is Charge per unit length lambda into DL Okay, and DL can be written as R into angle D Phi So DQ is lambda Rd Phi Understood this now DQ is lambda Rd Phi and Electric field is like this Okay, so I cannot integrate DE Why I cannot integrate DE because DE Direction will be different for every DQ I take suppose I take DQ over here My DE will be in that direction. How can I add this and that like a number? They are like this They are vectors If they are in the same direction, then I can add all the DE and integrate DE Like for example the previous case when it was a rod, but now All the DE's are at different different directions So what I have done is I have found out that electric field Will exist only horizontally Okay, so I have found out the component of DE in the horizontal direction and I am integrating that only Now is it clear DE is DQ by R K DQ by R square into cos of Phi is DE X see the When when I'm teaching this I am assuming that you guys have done basic calculus Okay, I am assuming that you have basic understanding of integral All right, so if you have not a basic understanding of integral it is all right, you can learn Okay, so message me. I'll send you a basic for this thing which is required for you to watch Otherwise, this will create a problem again and again throughout the class 12. It will be a problem It is very easy for me to not to teach this I can you know easily ignore this that you know this integral part People face a lot of trouble. So let me teach only simple simple things all of you will be happy that you are able to solve questions But when exam comes You'll get shock of your life Okay, so better you learn those difficult difficult things as well Then only you will be able to score well okay So again DX is K DQ by R square into cos of Phi DQ is Like we can we have discussed it is lambda R D Phi Cos of Phi. This is DEX Okay One R get cancelled away So when I integrate the electric field along x-axis my total electric field is that only because electric field along y-axis is zero So K lambda by R integral of cos of Phi D Phi What is a limit for Phi how my Phi is changing upper limit and lower limit? I have to cover the entire circular arc My Phi will change from what to what to cover the entire circular arc Phi should cover like this and Then like that So first I have to go anti-clockwise or first I have to go clockwise then anti-clockwise whatever So my limits will be minus theta to plus theta Okay Integral of cos Phi is how much everyone cos of Phi integral is what? minus of sine Phi Derivative of cos Phi is minus sine Phi integral of cos Phi is sine Phi Okay. All right. So the Shouldn't what shouldn't it be from theta to minus theta not the doesn't matter Okay, you are finding out the magnitude of electric field whether it is minus theta to plus theta theta to minus theta at the End you are anyway going to take mod Direction you already know electric field is K lambda by R Sine of Phi Minus theta to plus theta Understood and rug This would be to K lambda Sine of theta you're assuming anti-clockwise to be positive clockwise to be negative, right? So you can assume anti-clockwise to be negative clockwise positive. It's not a rule K lambda by R This is the expression for the electric field due to a circular arc Okay, and if the charge is positive electric field is away from the circular arc along the symmetry line If it is a negative circular arc charge Electric field will be towards the circular arc along the symmetry line Okay Fine for a full circle. What is the value of theta? For a full circle theta is what? 360 degree or 180 degree Look how theta is defined theta will be 180 Okay, from symmetry line, how much angle it makes, right? So when this theta becomes 180 the upper also becomes 180 So for full circle theta is 180. So electric field becomes zero and that is exactly what it should be Okay, for half circle, what is the value of theta? For half circle theta is what? 90 degree. So you can put theta is 90 you'll get electric field due to the half circle Fine. I hope things are clear Okay, some of you might be thinking Sir told us that class 12 physics is extremely easy. Now he's showing all kinds of integrals and headache is there Class 12 is easy why class 12 is easy because The variety in numericals are very less. If you focus really hard or well Concentrate well in learning the theory all these integrals somehow you have understood and everything is done Then the number of numericals you have to solve is very less in number Okay, that is why I told class 12 is easy Just understand the theory Numerical part when you practice the varieties are very less Just let's say a 15 to 20 varieties of questions per chapter, but in 11th Chapter like rigid body motion Millions of varieties. It is never ending. Okay, so that is why 12 is easy All right case number three Electric field due to the Circular charge Can you guess what it is? We have done it in gravitation. What it should be circular charge Correct along the Along the excess Oshek and hurryer and got it others haven't paid attention in 11th Okay Anyways, here is your second and last chance This is your circle half of it is Inside the page or inside your screen Half of it is inside half of it is coming out. Understand the circle is like this Okay So This is the axis fine Along the axis at a distance a away You have to find electric field charge per unit length is lambda and radius is capital R find out Remember electric field is a vector quantity Don't treat it like a scalar Very easy to get fooled Direction sense is there But before finding out the magnitude of electric field, can you tell me the direction of electric field net direction which one? along the axis all of you All of you along the excess So let us talk about how the electric field is along the axis Suppose I take these two Charges one at the top and one at the bottom the electric field Is like this So these two electric field What will happen? along this direction Net electric field will be what because of these two? Zero so net electric field will be in that direction because of these two so You can always find Dometrically opposite points. It's like a cone. It'll create like a cone One from here one from there one from here other one from there every time One electric field like this other one will be like that. So their components perpendicular to the axis get cancelled So when you visualize it like this It is a half a sphere No Why? How you're visualizing it to be half a sphere. I told it Three four times at the start how the figure is you're not listening. So you have a circle Half the circle is inside the screen half the circle is outside the screen. This is a circuit chart. It is written also here How can you say semi? You know hemisphere hemisphere? There is a sphere written here Okay, so this is the thing the net electric field is along the axis That all of you understood any doubt in that No doubts So now electric field, you know is along the axis you have to find component of electric field along the axis and integrate it Okay, try doing it yourself everyone If let's say I take a small DQ from here. This is DQ. This is theta Electric field will be in this direction D e How should I write D e? D e will be what? K D Q divided by this distance. What should I write this distance as? Everyone that distance what should I write? correct R square plus a square This is D e Now I have to integrate what can I integrate this directly D e? What is the problem in integrating this directly? What is the problem? What is the problem? D Q No, only one variable is there D Q is only variable. Is there any other variable? There is no other variable right only DQ is there So if you integrate D e it becomes KQ by R square plus a square and what is the problem? Why I cannot integrate D e? Correct. It is like scalar addition for a vector D e for this one is in one direction D e due to that one will be in other one other direction due to this one in some other direction How can you add vectors like a scalars? That is why you cannot integrate D e just like that All the vectors are not parallel So if you take components of it along the x-axis, so if this is theta You have to integrate D e cos theta To get electric field along x-axis, which is what is the net electric field? Okay, so E x is K into D Q divided by a square plus R square into cos theta integral Now how many variables you have here? Everyone is theta variable is theta variable all of you is theta variable No, it is not Okay, wherever you go all the DQs will make the same angle Theta only it's like a cone so imagine a circle and at the top All the points on the circle will make same angle theta. So everything is a constant You can take everything outside K cos theta divided by a square plus R square integral of D Q Indicul of D Q is what? Q total charge Okay, so electric field can be written as K Q Cos theta is what? From here you can write cos theta as a divided by root over a square plus R square divided by a Square plus R square. So this will become K Q a divided by a Square plus R squared to the power 3 by 2 All of you understood this type in is this clear? Clear, okay, but Q is not given to you lambda is given So Q can be written as lambda into 2 pi R That you can substitute here All right, so You can ask doubt from anywhere. Whatever we have done today today was little heavy okay, so That is how it is, you know, you have to learn these things Otherwise some things will be ignored and in the exam you will see these things It will be like that you will have never learned this Okay