Hello, welcome to lecture number 02. In lecture number 01, I have discussed about completely basic electric charge, field, coulombs, flux, formulas. I thought that I will finish the lecture in 20 minutes, but it really is 38 minutes. But don't worry, I will explain the basics so that you understand the concept. That is more important. We will use flux.
That is very important. So, in lecture number 02. ಈ ಈ engineering first day with my own money so and then that's the reason why it's very special and the other thing you don't have to worry when you go to your engineering and all you can still work and when you go to degree so after second piece no issues but when you go to degree and all you have to work on yourself and you order so that's the reason why it's quite special for me okay and now so yeah what is electric dipole we will discuss what is electric dipole so I will erase martini so you have to know the you know flow so first lecture united second lecture united ಈા ಈા ಈા ಈા ಈા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ��ા ಯಾಕೆ ಸರ್ಟ್ 2 ಯರೆ ತವಾಂದಿವಿ ಆ ತವಾಂಬುದಲ್ಲವಾದರೆ, ನೋ ಪ್ರಬ್ಬಲನ್ ಯು ಟೇಕ್ ಯಾಯಾಲ್ಸವು ಅದರ ಕಲಿಕಲಿಯಿಸಿಂಸಲ್ಲಿ ನು ಎಲ್ಲಾಕ್ರಡೆ ಆ ಬಾಯ್ ಟೂವಾಕ್ಪೆಕವತ್ತೆ. ನಾನು ಗೊತ್ತಿರೆ ನನ್ನು ಗೊತ್ತಿರೆ ಪ್ರಕಾರ್ ಚಿಕ್ಮಕ್ಲಿದ್ದಿವಿದ್ದಿವಿದ್ದಿವು ನು ಫ್ರಾಕ್ಷಿನ್ತನ್ನು ಇಸ್ತಾಗಲ್ಲಾ. ಬಟ್ ನಮ್ಬರ್ಸನ್ನು ಇಸ್� ಈ 2 equal and opposite charges separated by distance of 2a and then dipole moment p is given by q into 2a and direction is given from minus q to plus q. So, we need to find electric field due to dipole.
Why electric field due to dipole? Electric field is zero, minus q plus q is not there, plus q minus q is not there, cancel it and it is zero. No, it is not like that. Yes, there are some points where it can be zero. ಈ.
it is very important to understand the dipole. If you ask for the electric dipole, they can ask to find electric dipole. Find electric field due to dipole at this point, at this point, at this point. You can find it at any point. You don't have to find any point.
If your PUC board is very easy, you don't have to find all the points. You just have to find two points, that is on the axial plane and on the equatorial plane. Equatorial plane is like this. ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲಿ ಇಲ್ಲ� ಈ.
ಅದು ನಿಮ್ಮ ಪಿವುಸಿಯಬೋಡಲ್ಲು ಇಲ್ಲಡೆ ಕೇಳದು ಒಂದು ಅಗ್ಜಿಯಲ್ ಪ್ಲೈನ್ ಅಂತಾ ಅಗ್ಜಿಯಲ್ ಅಲಾಂಗ್ ಅಗ್ಜಿಯಲ್ ಅಂತಾ ಇನ್ನು ದೇಕ್ಕೊಟೂರಿಲ್ ಅಂತಾ ಇಲ್ಲು ಒಂದು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲು ಇಲ್ಲ� I will focus more on making you understand. After understanding that, you have to add all the derivations to the derivation. So, for that you have to add salt, pepper, and spices.
What do you mean by that? So, you have to explain. I am considered a dipole of length 2a. You have to write the definitions.
When you ask the question, solving equation 1 and 2. You have to write the statements. You will get the statement in any book. So, when you see any derivation, you will feel it is very lengthy.
ಈા ಈ ಈ. I told you that force and electric field are the same. Electric field is force per unit charge.
I have already explained all this in the first lecture. Now, here, the electric field due to, I will write red line, due to plus. This plus and plus are repulsion.
The direction is, this is the direction, electric field due to plus Q. Plus and minus are attracted. That is attracted electric field due to. ಈ.
ಈ Q into, actually this is Q term actually, electric field source charts actually, QQ divided by distance between them. ಈ, ಈ ಈ 2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2 we get the directions by considering the sign positive or negative. Otherwise, I don't have to consider and directly put negative.
Otherwise, this minus this will give electric field at this point. After that, I don't use the formula. Otherwise, electric field at point P is 1 by 4 pi epsilon naught, Q divided by R minus A the whole square.
Then, minus this minus this, 1 by 4 pi epsilon naught. q divided by r plus c the whole square. So, here, if you take plus in some books, here it is minus, so plus into minus becomes minus. Here, it becomes minus. If I am taking minus directly, I am not considering sign here.
Because I got the direction. What I am thinking is, because this is helpful numerical. Numerical is very helpful. This side is this much, this side is less.
So, obviously, the net force here is 10. This is 5. So, net force here is 10 minus 5. So, I am just taking like that. 10 minus 5. So, here the force is that minus this. I am just taking magnitude and then subtracting it. That is all. So, this is the formula.
Don't think of this as a math. So, electric field at this point P or axial. You can take 1 by 4 epsilon naught common and Q also.
So, 1 by R minus a the whole square minus 1 by R plus a the whole square. Next, simplify this and Q divided by 4 epsilon naught. 2a2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2a2b2 ಈિંડે સે રાતાગળં દે કા કિ સે રાતાગળં દે. સે રાતાગળં દે.
સે સે સે સે સે સે સે સે સે સે સે. સ્કયર તે મિનસ તુ એ ર બರತે. ಇದ್ದಿವಾಂದಾ ಕುಯಾ ಸಾರ್ತಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿ� qx4a, qx2a, qx4a, 2p, you can note it down, I will erase that.
what this becomes now? i am erasing this. coming from here now, p is 2p. so i can have it like 1 by 4pi epsilon naught. 2p divided by r square minus whole square.
here also i can do it here itself. you have to write all the assumptions. now if we change it, dipole will be very heavy. you can see in next chapter. ಈ.
ಅಗಿಯಿಂದ ಮಾಟ್ಸಲ್ ಯೋಚಿದುವಾಟಿಯಿಲ್. ಇದ್ದು ಬಿಗ್ ಕ್ವಾನಿಟಿಟಿಯಿಂದು ದೊಡ್ಡು ಕ್ವಾಯರ್ಯಿಂದು ಸ್ಕೈರ್ ಮಾಡ್ತಾಯಿದ್ದರಾ ಮತ್ತೆ ಸಬ್ಟ್ರಾಕ್ ಮಾಡ್ತಾಯಿದ್ದರಾ. ಸಮಾಲ್ ಮಾಡ್ತಾಯಿಂದು ಮಾಡ್ತಾಯಿಂದು ಮಾಡ್ತಾಯಿಂದು ಮಾಡ್ತಾಯಿಂದು ಮಾಡ್ತಾಯಿಂದು ಮಾಡ್ತಾಯಿಂದು ಮಾಡ್ತಾಯಿ� almost negligible. Remove this a square and then you can remove an additional subtraction not in multiplication.
20 x 0.1 is just 2. So you can't do assumptions in the multiplication division. Only in addition and subtraction. If there is a very small quantity then you can forget about it.
Next, R square is the whole square. Next, R cube is the denominator. Qx4a as 2p, p is Qx2a but 4 is 2p.
Here in the denominator we have only R cube and R is here. R cube and R are cancelled. I can show one more step.
R square. I will explain once again. 2p, a2 will forget about that a2 because a is very small and r2 and square will become r4, above r is r by r4 that becomes r3, sorry, oh sorry, r3 it is, I am sorry, r by r4 that becomes r3.
So again this is the expression for electric field due to and dipole along the axial expression actually, e is equal to 1. ಇದು ಅಗ್ಸಿಲ್ ಪೋಯಿಂಟಾವಿಂದಗಾ, ಇದು ಅಗ್ಸಿಲ್ ಪೋಯಿಂಟಾವಿಂದಗಾ, ಇದು ಅಗ್ಸಿಲ್ ಪೋಯಿಂಟಾವಿಂದಗಾ, ಇದು ಅಗ್ಸಿಲ್ ಪೋಯಿಂಟಾವಿಂದಗಾ, ಇದು ಅಗ್ಸಿಲ್ ಪೋಯಿಂಟಾವಿಂದಗ� ಅರ್ತಾಯಿದಲ್ಲವಾ, ಸುಬ್ಬಾಟ್ ಯೆವ್ರಾಯಿದ ಸ್ಟೇಟ್ಮಿಂಸ್ ಇರ್ರಿಂಡ್ ದೇರ್. ಸುಬ್ಬಾಟ್ ಇರ್ರಿಂಡ್ ದೇರ್. now due to axial plane, it is very simple, it is an equatorial plane, electric dipole, electric field due to dipole, this derivation is electric field due to a dipole on equator, on a point on an equatorial plane, I don't know the spelling of equatorial because equatorial plane, I have also studied in Kannada.
i have understood signs well but some where some spelling mistakes i have done so this is the equatorial plane what happens is, let's go with this one now so i am assuming, so the diagram should be good again it is very simple so this is a dipole i have to find a point here now so this is minus q here again same thing i mean this diagram will be good ಈ. because i'm taking it test positive again here also we're taking a test positive chart so positive negative attract out then positive positive ripple again that's all again same thing you have to and the dipole direction with the p negative in the positive it will be in this direction okay next one so again same thing you have to do it you just have this is a normal vector okay I have to add this and just see if, again you have to take two components, right? If there is an angle here, you have to take two components. If the angle is theta, I am drawing a line here, a line parallel to the base.
This too becomes theta, ok? Again, here is theta, here is theta. This too becomes theta because in the end it is an isosceles triangle.
We can prove all those things, do not worry, we can do maths. Theta, theta, this becomes plus q, it is very simple. ಈ ಈ ಈ ಈ so plus two minus two is same but obviously magnitude will be different obviously what happens? this is plus sign and this is minus sign but considering plus sign and minus only I got the sign this is left and this is right you can clearly see this is right and this is left that means if this is pulling down this is pushing up so the vertical components will get cancelled because the magnitude is same but they are in opposite direction it is magnitude is same but they are in opposite direction teta is same, both the theta is same, only horizontal components are left, you have to write this in the exam properly, so e plus q, e minus q, they are add up, so I can see add now, so what happened, so I got the sign, I got the direction considering the sign, derivation model I don't consider any sign, ok, so electric field at point P is given by an equilateral plane below, ಈ you have some root again square for that. so root cancela hota iga.
again hota iga. just r square plus a square. ade niti plus.
1 by 4 epsilon naught cross theta divided by, here also, same r square plus a square. So, if you write negative here, you get minus q here, sorry, I can write, ok, 1 by minus q. You can ask why you can't write minus q here, it will be subtracted, it will be zero. But both are adding up, no? So, you have to understand that.
So, here I am not considering the sign because I have already considered the sign and got this. Both are adding up. ಈ. R divided by, that is, A divided by, again, root of R square plus A square. Now, like this, so you know that already, so you have something without root, with root.
And then, x into root x is, that is, x to the power of 3 by 2, because this is 1, and I can write this as half, half plus 1 is 3 by 2, loss of exponents. ಈ. 2 q a 1 by 2, so always maintain uniform, you know, uniform, uniform notations everywhere, that's very important in your board examination, but when you do a cognitive exam, it's not, it's fine, you just have to remember, you know, how it's coming, that's more important, ok, here, 2 p comes, sorry, p just, because 2 into q into a, that is 2 a into q, that is just p, that's it, so p comes, so 1 by 4 pi epsilon naught will be nothing but p divided by, R2 plus A2 to the power of 3 by 2 We are writing this 2QASB R2 plus A2 equals 3 by 2 Again same notation we can do A is much much smaller Or you can just say R is much much greater than A Obviously I can neglect the R But the additional subtraction we can neglect it Ok R2 is 3x2. If you cancel it, you get the electric field at point P.
That means the axial plane, the equatorial plane. That will be 1x4.0. P divided by R2 is 3x2.
Something square, again 3x2. So, if you cancel 2x2, you will get Rq. So, that is the axial plane.
So, electric field due to and dipole on the axial plane is given by. 1 by 4 by epsilon 0 divided by R cube, it is very important. Here, I am going to give, usually, this is only derivation in your exam, but, in that case, you have to understand carefully, this is equal to real plane.
So, here, you have p divided by R cube. So, in the previous derivation, I checked, here, 1 by 4 by epsilon 0, 2p divided by R cube. See, there it was 2p, it was p, and then, you know, the electric field, if you take same distance R, the electric field on the axial plane is double compared to same distance equal to the plane i'll just give a note at the end okay it's barkowli first in the so write down this so take a screenshot and write it later or write it down pause the video and write down everything but if you have to add the statements here and there that's right but if you see this is very simple it is point down tv and resolve martini so resolve but then you don't want to add up so both are equal so add them so two times of one by four times not q 2 x 1 x 4 x 0, q x r square plus a square, cost e to the place of a x root of r square plus a square. so that's all about this, just add up everything and get this answer.
and once you return this, i'll just erase this and give a note for you. because this is very easy to ask in case-eater. usually when i do this, i prepare the computer exam simultaneously. so p-o-c and case-eater, but many people first p-o-c is derivation and derivation is problem solving.
It will be difficult to score in the competitive exams. Okay? So, here we understood that on the point R, same point if we take here also r, you can see electric field at this point on equatorial eq, now if we discuss 1 by 4 epsilon 0 p divided by r cube, but in this point only same distance, axial point we have 1 by 4 epsilon 0 2 p divided by r cube, you can clearly see that, so electric field along equatorial if you want to find it is half, electric field in the axial plane, e-axial.
And also, I forgot to show you the diagram. If you see the diagram, p comes here. So if you see the diagram there, the net electric field was in this direction.
So electric field in the axial plane was in this direction. So they both have the same direction. But if you see in this, it is the opposite direction, in both actually.
These are the usual p-versions of KCDL. it is half of this and it is in the opposite direction e is equal to minus half of e axiol so they are in opposite direction this is not important for POC but it is important for KCT this is a strong theory but I just got this fine now I will tell you ok so let us come to the next one now after this last topic number 2 we will discuss now so it is already 30 minutes when electric dipole is kept in the uniform magnetic field so it experiences torque so derive the expression for torque uniform electric field this is also heard a lot in the case in the case it is non-uniform so what to do with that? uniform electric field so what is the expression for torque?
when electric dipoles are kept in a uniform electric field, so it derives the expression for torque. I mean it experiences torque. First, we will discuss why torque is given.
It is a very simple relation. We will do it fast. First, we will talk about electric field.
So, if I take this as my direction of electric field, I will give a dipole here, that is plus Q and then ಈ. this is positive and this is negative. so obviously whenever there is a charge is kept in electric field it experiences a force. this is uniform. uniform is what is the force experienced by this charge?
f is equal to e into q. e is equal to f by q. in electric field the formula is f by q.
force is f into eq. so inside it is eq. so this is positive repulsion. therefore the force is in this direction.
ಈ, f is equal to minus eq. ಈ ಈ ಈ ಈ ಈ ಈ ಈ ಈ ಈ �ೋં ಈ �ঋ �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા �ই �ા ಈ ಕೇಸಿಡಿಯಾ ಕೇಳ್ಟಾಗರತ್ತರೆ ಅತ್ತರೆ ನಾವರು ಮಾಟ್ಟನು ಎಂದು ಪಿ� this cap is opening upwards. because when you apply two forces you are using two forces but the third force act coming upwards that means when we rotate two vectors when we rotate two forces we get another force that force is perpendicular to both the forces that is only called as vector algebra i can't show perpendicular here.
what is the meaning of this is meaning is when we rotate two forces ಈ ಈિ ಈિ ಈિ ಈિ ಈિ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ ��િ 2a cos theta 2a sin theta f x 2a sin theta 2 forces are acting and 3rd force is perpendicular this is also a vector, torque is also a vector, force is also a vector even the 2a sign is also a vector basically we will get force is 2a eq, force is what? eq, eq into 2a sin theta. so basically this is a vector coordinate. you are rotating two forces, you are getting one more force which is perpendicular to these two. this is left hand screw, right hand screw, this is fleming's left hand screw, all these come under this only.
ಈ. ಈ. ಅವರಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದಿದ�