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this is election No. 13 of a V232

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electromagnetic field theory

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today

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we will study a relationship between electric field

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and electro potential

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and also the electric typo

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this will cover section 4.8 and 4.9 of your textbook

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this is academic week No. 7 of your semester

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the learning objectives of today's lecture

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would be to find out

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a relationship between electric field

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and electric potential

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to derive Maxwell's second equation

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and to understand the working principle of electric

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typo

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first of all uh

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let's uh drive

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a relationship between electric field

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and electric potential and then uh

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we will drive a Maxwell's second equation

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so if we start up with this potential difference

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between two points that is between a and B

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so it is the difference between potential at point B

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minus the potential at point a

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if you're taking the reference point of infinity

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okay and then we studied that

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if we want to find out this potential

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difference between these two points

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then it can be easily found out by using this

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integral equation so

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the integration is of this electric field

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dotted along with it

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along with this part of travelling

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along which we want to find out the electric field

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interception

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so uh uh Indian uh Indian medical technique

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you can find out uh

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like this one that it is minus a uh

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in integral of a to B e dot B L

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so

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you have to take the line integral of electric wheel

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along the given part

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so if this part is from a to B

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so it can be integrated using this line integral

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so if you want to find out the potential difference

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between this B to a

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then so it will be

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it will be the same but in the minus direction right

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so the potential difference same

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but in the minus direction

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because in that case you're working

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you're working against the

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the original electrical difference so there

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there is a next gain endure uh

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endure uh endure uh that uh energy or endure potential

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potential difference so if you sum up these two things

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then you can find out that it is a closed integral

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of electric wheel

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around this closed bar that is starting from your

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from your this point a and terminating on point B

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and then starting from point B and then terminate

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terminating to B

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originating point

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so it is making up a close path in this case

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so for this a close path integral

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we can find out that this close integral is

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equal to zero so

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this means that no network is being done

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in moving a charge along a closed part

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in a electrostatic field

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so the closed part is from a to B

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N B to A

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so this means

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if we apply the famous strokes theorem to find out

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or to relate disclose a line intrigue

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well with the open surface integral

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by taking the help of this curve

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of the of this rector field quantity electric field

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then we can find it out that this close interval

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it is equal to the surface integral of that cross

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e dot DS and if we equate this equation

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this equation with this equation right

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so this means this this is equal to 0

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so how this can be equal to 0

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this can only be equal to zero

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if the core of electric field

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the inner quantity is equal to zero

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because your surface element

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it cannot be equal to zero right

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so the only thing that is equal to 0

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that is your attel crossey or otherwise

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otherwise the angle between this call and the normal

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to this surface element is equal to 90 degree

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than in that case

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also this quantity can be equal to zero

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but for the time being

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we are considering a general case

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so for the general case

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this dark cross is equal to zero

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because this quantity is going to be equal to zero

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so what we what this implies

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this implies that electrostatic field is conservating

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or irritational

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so that is uh your institution from here

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and then you can uh we also denote this uh

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differential equation as a second

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maximum equation for static electric field

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so picking on uh

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the research that we concluded on the previous light

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uh the uh curl of a vector

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it is equal to zero depend

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only if this electric quantity is a gradient

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of some scaler function

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so if we take the help of this vector identity

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that curve of gradient of a scale is equal to 0

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so we can do not like this way

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that this gradient of v is equal to electric cream

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in this case so

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here

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we assume that this E is equal to minus gradient of v

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so why we we choose this minus

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so this was the question

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that was only asked during the class

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so that at that point

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I told the student to reserve this question

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for future lecture so here's the answer

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so what is the answer first of all

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electric fantastic is the gradient of E right

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so it is coming out from this vector identity

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and then a negative sign shows that

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direction of E is opposite to

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the direction in which we increases

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right to the potential

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increases in the direction opposite to the electrical

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so for this case if you see here

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uh the electric field

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it is uh

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going away from this positively chart sphere right

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and if you you are moving a chart

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if you are moving a chart from a positive chart

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from here towards here

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then basically you are going to increase the potential

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right because you are working against the electricity

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so because of this that against thing

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because of that against thing

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this electric field so

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this negative sign means the sign is coming into play

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so negative sign

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shows that the direction of E is opposite to

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the direction in which V increases

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so he is in if you're uh

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removing this Q positive point

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charge outside towards this positively charged sphere

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then you're gaining the potential

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the potential differences being gained

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so that's why

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and is you are moving in the direction of

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direction opposite to the electric grid

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so that's why this negative sign is coming into play

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he is a directed from higher to lower levels

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and you can see here

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that he is moving in the opposite direction uh

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to the uh uh to the direction in which uh

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as your uh

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it is potential difference going to increase

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now let's solve this example of your book 4

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14 to further clarify the concepts of this work

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being done in an electric field

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or in or in the presence of that potential difference

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okay so what we need to find out

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we are given the potential potential

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electric potential like this

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so it is since it is our Tita and phi

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so it is your spherical coordinate system

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and we need to find out the electric proxanity

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at this particular point

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and then we need to work calculately

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well done in moving a positive charge

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and microcoulums from point at this to point this

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okay so first of all

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the electric proxanity

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we know that it is equal to absolute

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not electric field

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and electric field is equal to minus ingredient of v

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v we are given with

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since it is in spherical coordinate systems

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also

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we have to involve this gradient equation of spherical

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coordinate system and then uh

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we can change the uh partial derivative of this uh

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electric potential scaler

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scaler quantity with respect to R

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with respect to data and with respect to file

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and again take care of these cautions

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along this partial derivatives

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and then you can simplify these things

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so first derivation is with respect to R

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so we know that the derivative

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so the R

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r minus r r scale will be going in the numerator

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so it will be R r minus 2

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then you involve this P r rule

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so it will be coming out with minus 1 or r cube

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and minus minus will be cancelled with this minus

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so it will be a positive function

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for the second case

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the division is with this fact to this theatre

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and scientific we know that it is a cause of theatre

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and for the third case it is five

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so cause of five we know that it's a delevation

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it is equal to minus sign of uh sign of uh 5

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so that's why it is coming out to be positive

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minus minus positive

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okay then we have to evaluate this

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we know that it is r which is t

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r and this is 5 like this thing

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and then you can evaluate this expression

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to find out the electric flux tendency

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at this vertical point

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we're gonna let solve this uh part 2

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which is uh

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very important in this case because it will

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uh clarify your concept

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how to find out the work done

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while you are moving a particular charge

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in the presence of some electric grid

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from one point to another point

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so again the points given are in sperical coordinates

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system electric field we have found out

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using the information of electric potential

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and so the first method first method

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will involve our standard integral equation

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and uh from there

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we know that well done is equal to minus q times uh

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edot DL right

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this

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we have already contributed in our previous structures

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so uh since the electric field is conservated

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the part of integration is immaturial

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so this is very

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very important Assumption in simplifying this question

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the integration in this question

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because you can see here

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while moving from point a to B

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all three coordinates are changing

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right

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so in order to simplify this into a line integral

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so what we do we break down this part into three

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so

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since we know that electrostatic field is conservated

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for the part of integration in material

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so either will go from a to B like this way

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or we break up this part into uh

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three parts and that parts are depending upon the uh

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the variation of one variable at one time

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so in the first grade

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we are wearing this r from 1 to 4

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and we're keeping this Peter and five costume

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for the second case we are wearing this cheetah

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and we're keeping this by constant

261

00:11:43,166  -->  00:11:45,733
and we are keeping this R constant

262

00:11:45,733  -->  00:11:47,800
which was from the previous time

263

00:11:48,366  -->  00:11:51,200
and in the test time we are keeping this R question

264

00:11:51,200  -->  00:11:52,900
we are keeping this uh

265

00:11:52,900  -->  00:11:57,433
teacher question and we are only changing this 5

266

00:11:57,600  -->  00:11:58,933
so these are the three uh

267

00:11:58,933  -->  00:12:01,000
parts that we are going to integrate

268

00:12:01,333  -->  00:12:02,866
so if we see here

269

00:12:02,866  -->  00:12:05,800
and the first part only this R is changing

270

00:12:05,800  -->  00:12:10,900
so there a line uh element it is d R a R right

271

00:12:10,900  -->  00:12:13,200
this we have Covenant chapter number two

272

00:12:13,733  -->  00:12:16,400
and in the second case your teacher is wearing

273

00:12:16,400  -->  00:12:19,466
you know that if teacher is wearing then uh

274

00:12:19,466  -->  00:12:19,866
then

275

00:12:19,866  -->  00:12:23,600
then the line in line segment will be to equal to R

276

00:12:23,600  -->  00:12:25,233
d Peter B Peter

277

00:12:25,700  -->  00:12:29,766
so here you can involve that uh art rule as well

278

00:12:30,766  -->  00:12:34,366
okay the third case only 5 vary is 5 vary

279

00:12:34,366  -->  00:12:36,400
and we know that line in line segment

280

00:12:36,400  -->  00:12:40,900
for this one is our scientifica D85A5

281

00:12:43,300  -->  00:12:46,766
let's uh integrate these uh three segments one by one

282

00:12:47,100  -->  00:12:50,000
and uh we know that we are given with this electrical

283

00:12:50,066  -->  00:12:52,300
we know these uh line segments

284

00:12:52,400  -->  00:12:54,133
uh for which we have to integrate

285

00:12:54,133  -->  00:12:57,000
so the first uh first part first uh first line segment

286

00:12:57,000  -->  00:13:01,900
this one artist Jenny from 1 to uh 4 right

287

00:13:01,933  -->  00:13:06,066
and uh we integrate this with respect to Dr so again

288

00:13:06,366  -->  00:13:13,366
uh remember if we take this R R R R cube as a variable

289

00:13:13,366  -->  00:13:17,400
we take this quantities uh out our side T10 crossfire

290

00:13:17,400  -->  00:13:20,266
we evaluate at the constant at t 10 5

291

00:13:20,333  -->  00:13:22,366
we integrate this one or r cube

292

00:13:22,366  -->  00:13:24,366
so the resultant will be

293

00:13:26,100  -->  00:13:30,100
minus 1 hour minus 1 hour are four I guess

294

00:13:30,266  -->  00:13:31,233
and then you can

295

00:13:32,133  -->  00:13:34,200
you can simplify this integration

296

00:13:34,200  -->  00:13:36,100
and then you will see that oh okay

297

00:13:36,100  -->  00:13:38,033
this minus sign is coming over here right

298

00:13:38,133  -->  00:13:39,866
and then you can plug in these values

299

00:13:40,000  -->  00:13:40,766
for the second case

300

00:13:40,766  -->  00:13:43,233
only the integration is where it's back to this

301

00:13:43,266  -->  00:13:46,433
the Peter so we know that cost Peter's integration is

302

00:13:46,666  -->  00:13:49,000
is sign of Peter right

303

00:13:49,066  -->  00:13:56,366
so that's why you are going to see what you are

304

00:13:56,700  -->  00:13:58,166
yes so you can uh

305

00:13:58,166  -->  00:14:02,100
further put in these values of uh qualify and uh

306

00:14:02,100  -->  00:14:04,500
R constant you can find out this value

307

00:14:04,500  -->  00:14:05,566
and then in the last step

308

00:14:05,566  -->  00:14:10,233
you can uh integrate with respect to 5 signify

309

00:14:10,333  -->  00:14:12,666
we know that it is equal to minus cause of 5

310

00:14:12,666  -->  00:14:13,666
and then you can spot this

311

00:14:13,666  -->  00:14:16,033
putting these constant values for that part

312

00:14:16,066  -->  00:14:17,533
to simplify this expression

313

00:14:17,533  -->  00:14:20,233
so this is your uh your method number one

314

00:14:20,900  -->  00:14:23,300
once you are given the electric related formation

315

00:14:23,566  -->  00:14:26,000
and in case you are directly given

316

00:14:26,000  -->  00:14:28,533
the information of electric potential

317

00:14:28,533  -->  00:14:29,533
as in this case

318

00:14:29,533  -->  00:14:32,733
we were initially given the information of electric

319

00:14:32,733  -->  00:14:35,000
potential so what we need to do

320

00:14:35,200  -->  00:14:37,966
we need to simply involve this formula

321

00:14:37,966  -->  00:14:40,000
for the work that with er

322

00:14:40,000  -->  00:14:44,600
we derived in our previous lectures

323

00:14:44,600  -->  00:14:48,933
so that was work is equal to q times we a b right

324

00:14:48,933  -->  00:14:49,733
so we a B

325

00:14:49,733  -->  00:14:52,900
we know that it is the difference of b b minus we a

326

00:14:53,300  -->  00:14:57,033
so the potential at point B and potential at point a

327

00:14:57,133  -->  00:14:59,200
so scaler value we are a scaler function

328

00:14:59,200  -->  00:15:01,766
we are given electric for the collective potential

329

00:15:01,766  -->  00:15:05,366
we just need to simply plug in the values of this point

330

00:15:05,366  -->  00:15:06,166
a

331

00:15:07,300  -->  00:15:10,866
and this point B into this uh okay

332

00:15:10,866  -->  00:15:14,300
this point uh rather you can take it like this

333

00:15:14,300  -->  00:15:17,566
uh uh yes you can take it like this

334

00:15:17,566  -->  00:15:19,933
so point B will be coming out over here

335

00:15:19,933  -->  00:15:23,766
yes four like this point B will be coming out over here

336

00:15:24,866  -->  00:15:27,000
point I said a less a less erased

337

00:15:27,000  -->  00:15:29,666
less erased so that it is more clear

338

00:15:29,666  -->  00:15:32,800
so point B is coming out over here

339

00:15:32,866  -->  00:15:35,766
and point a is coming out over here right

340

00:15:35,766  -->  00:15:38,300
so B b minus v a just solve this

341

00:15:38,800  -->  00:15:41,900
this right this equation and we will point out B

342

00:15:41,900  -->  00:15:44,033
same answer that we found out early

343

00:15:44,166  -->  00:15:45,266
so these are the two ways

344

00:15:45,266  -->  00:15:47,600
so remember once the electric grid is given

345

00:15:47,600  -->  00:15:49,366
you have to involve this uh

346

00:15:49,366  -->  00:15:51,000
integral and then uh

347

00:15:51,000  -->  00:15:53,166
break down the part of your integration

348

00:15:53,166  -->  00:15:55,600
to simplify your line integration

349

00:15:55,800  -->  00:15:58,700
and if your potential information is given and you

350

00:15:58,700  -->  00:15:59,800
you need to find out the world

351

00:15:59,800  -->  00:16:01,666
done between these two points

352

00:16:02,266  -->  00:16:05,033
then you can simply involve this equation

353

00:16:05,333  -->  00:16:07,866
and use this informational

354

00:16:07,866  -->  00:16:09,633
collective potential directly

355

00:16:12,366  -->  00:16:15,866
now let's understand what is uh an electric die pole

356

00:16:15,866  -->  00:16:18,300
so an electric die pole is form

357

00:16:18,300  -->  00:16:20,533
so you can see here so it is die

358

00:16:20,533  -->  00:16:23,766
die means two pole so two pole

359

00:16:23,766  -->  00:16:25,866
there are two poles so

360

00:16:25,866  -->  00:16:31,700
it is made up of two point charges of equal magnitude

361

00:16:31,800  -->  00:16:34,166
but opposite polarity

362

00:16:34,166  -->  00:16:38,000
opposite sign and they are separated by a very

363

00:16:38,000  -->  00:16:39,766
very small distance d

364

00:16:39,766  -->  00:16:42,066
we will see here what is this distance d

365

00:16:42,400  -->  00:16:43,933
so remember electrodetol

366

00:16:43,933  -->  00:16:47,000
it is formed by using two point charges

367

00:16:47,000  -->  00:16:51,166
of equal magnitude but opposite polarity

368

00:16:51,166  -->  00:16:53,300
and with a very small suppression

369

00:16:53,966  -->  00:16:55,900
so let's consider this figure

370

00:16:56,733  -->  00:16:59,633
so now win a potential at point P

371

00:17:00,266  -->  00:17:01,200
at this point

372

00:17:02,100  -->  00:17:03,733
due to an electric tripod

373

00:17:03,733  -->  00:17:07,900
is just the sum of potentials due to each point charge

374

00:17:07,900  -->  00:17:09,733
that is potential at point

375

00:17:09,733  -->  00:17:10,966
so this is your

376

00:17:11,166  -->  00:17:13,466
as you can as you can treat this point as a

377

00:17:13,666  -->  00:17:18,433
this one point at uh at coordinates r t 10 5

378

00:17:18,733  -->  00:17:22,066
so we need to find out the potential at this point

379

00:17:22,066  -->  00:17:26,433
to do this positive charge and this minus q charge

380

00:17:26,600  -->  00:17:29,466
so we know the definition of this

381

00:17:31,133  -->  00:17:32,800
the potential

382

00:17:32,800  -->  00:17:37,166
so basically the potential at this point P will be up

383

00:17:37,500  -->  00:17:39,000
superposition of potential

384

00:17:39,000  -->  00:17:41,566
due to these two opposite polarity charges

385

00:17:41,566  -->  00:17:44,300
so that's why we are summing up the potential of

386

00:17:44,300  -->  00:17:47,100
individual potential of these two charges

387

00:17:47,200  -->  00:17:49,666
which is caused at this point p

388

00:17:50,533  -->  00:17:55,166
okay so uh so far this uh uh

389

00:17:55,166  -->  00:17:58,600
since uh this negative sign is sign is coming to play

390

00:17:58,600  -->  00:18:01,066
right so that's why this uh

391

00:18:01,166  -->  00:18:04,600
minus sign is coming in because of this thing right

392

00:18:05,133  -->  00:18:05,400
so

393

00:18:05,400  -->  00:18:08,133
we are assuming this supposed to charge is at distance

394

00:18:08,133  -->  00:18:09,700
radio distance R1

395

00:18:09,700  -->  00:18:12,866
this minus Q charge is at radio distance R2

396

00:18:12,933  -->  00:18:15,100
and this point from the origin

397

00:18:15,100  -->  00:18:17,966
it is at radio distance R

398

00:18:18,100  -->  00:18:19,200
and uh

399

00:18:19,200  -->  00:18:21,966
this is the smallest station between these two charges

400

00:18:21,966  -->  00:18:25,600
that is d and uh this uh p

401

00:18:25,600  -->  00:18:29,000
it is making an angle of a t job with respect to this

402

00:18:29,000  -->  00:18:29,800
uh

403

00:18:30,133  -->  00:18:31,666
z uh z

404

00:18:31,666  -->  00:18:33,200
uh z uh

405

00:18:33,200  -->  00:18:36,400
z axis because it is in this vertical partner system

406

00:18:36,733  -->  00:18:39,900
and how does this d cause of teacher comes

407

00:18:40,500  -->  00:18:41,133
uh this

408

00:18:41,133  -->  00:18:43,300
the d cause of data come from here

409

00:18:43,300  -->  00:18:46,500
that uh uh this d you you if

410

00:18:46,500  -->  00:18:51,266
if we assume that it is also Tita 1 right

411

00:18:51,466  -->  00:18:52,900
and this is Tita 2

412

00:18:53,966  -->  00:18:55,933
so we will take help of this

413

00:18:55,933  -->  00:18:59,966
uh Assumption that if you are

414

00:19:00,066  -->  00:19:02,466
your distance from this origin is very

415

00:19:02,466  -->  00:19:05,600
very greater than the expression between these two uh

416

00:19:05,600  -->  00:19:08,266
opposite charger point charges

417

00:19:08,766  -->  00:19:11,300
then uh what are these options

418

00:19:11,300  -->  00:19:14,700
so we can assume that this R1

419

00:19:15,300  -->  00:19:23,366
R1 so we can write it R1 is approximately equal to R2

420

00:19:23,666  -->  00:19:25,100
right so we can

421

00:19:25,666  -->  00:19:28,066
once you take the Altium of this first quantity

422

00:19:28,066  -->  00:19:30,933
like this one then you can plug in this thing

423

00:19:30,933  -->  00:19:34,566
that R1 is approximately equal to r

424

00:19:34,566  -->  00:19:37,066
2 is approximately equal to r

425

00:19:37,066  -->  00:19:39,766
so this will be equal to r r

426

00:19:39,766  -->  00:19:41,733
so that is r skip right

427

00:19:41,733  -->  00:19:45,366
so it is first uh first simplification okay

428

00:19:45,366  -->  00:19:49,700
the second simplification is the projection of this r

429

00:19:49,766  -->  00:19:51,200
on this uh

430

00:19:51,466  -->  00:19:54,800
other axis so how to find out this projection

431

00:19:55,733  -->  00:20:00,500
so we can assume that the difference between this R2

432

00:20:00,500  -->  00:20:04,933
and R1 if we assume that this is the difference

433

00:20:04,933  -->  00:20:09,000
for example we assume that this is the the projection

434

00:20:10,133  -->  00:20:11,466
this is the difference

435

00:20:11,466  -->  00:20:15,566
video difference between this R2 and R1

436

00:20:15,666  -->  00:20:17,800
so how to find out this thing

437

00:20:17,800  -->  00:20:22,200
it is d is very small we can consider that this cheetah

438

00:20:23,166  -->  00:20:27,300
it is approximately equal to Peter 1

439

00:20:27,666  -->  00:20:30,766
is approximately equal to Peter 2

440

00:20:31,566  -->  00:20:33,300
so we just take this a theater

441

00:20:33,300  -->  00:20:37,333
we we make the projection of this d on this axis

442

00:20:37,333  -->  00:20:39,066
we treat it as an axis

443

00:20:39,066  -->  00:20:44,033
we can find out that this R go minus or 1

444

00:20:44,066  -->  00:20:46,066
it is the projection of the

445

00:20:46,366  -->  00:20:48,700
on this separation or

446

00:20:48,700  -->  00:20:52,800
of this difference between these two radio distances

447

00:20:52,966  -->  00:20:55,300
so since we are projecting on this difference

448

00:20:55,300  -->  00:20:55,366
so

449

00:20:55,366  -->  00:20:58,966
that's why we are taking the cost component of this d

450

00:20:59,133  -->  00:21:01,700
so there is a d cause of data right

451

00:21:01,700  -->  00:21:03,833
so plug in here so this is your

452

00:21:05,800  -->  00:21:08,900
electric potential for this electric typo

453

00:21:11,333  -->  00:21:12,366
let's uh find out

454

00:21:12,366  -->  00:21:14,200
the electric feed for this typo

455

00:21:14,200  -->  00:21:16,666
is using the information that we drive on this

456

00:21:16,933  -->  00:21:18,433
on the previous slide

457

00:21:18,566  -->  00:21:22,300
so since it is in um uh cerical cord system

458

00:21:22,366  -->  00:21:24,000
we just uh need to uh

459

00:21:24,000  -->  00:21:26,966
take the portion derivative of this uh

460

00:21:27,300  -->  00:21:31,100
human quantity with respect to R t 10 5 okay

461

00:21:31,100  -->  00:21:34,600
and it is having only variable r and t r

462

00:21:34,600  -->  00:21:36,900
so this is going to be zero

463

00:21:36,900  -->  00:21:39,900
right because this will be acting as a constant

464

00:21:39,900  -->  00:21:42,200
with respect to five so for this thing

465

00:21:42,200  -->  00:21:43,666
uh it is uh

466

00:21:43,966  -->  00:21:44,800
uh Oscar

467

00:21:44,800  -->  00:21:45,800
so Oscar is uh

468

00:21:45,800  -->  00:21:48,733
one or Oscar derivative is minus 1 or R2

469

00:21:48,733  -->  00:21:49,966
so minus minus well

470

00:21:49,966  -->  00:21:51,200
then we cancel out over here

471

00:21:51,200  -->  00:21:52,800
and it could be a plus over here

472

00:21:52,800  -->  00:21:53,600
right

473

00:21:53,966  -->  00:21:55,533
and the question for the second thing

474

00:21:55,533  -->  00:21:58,366
it is causal tita we know that it's derivative

475

00:21:58,366  -->  00:22:00,900
it is minus sign of tita so again

476

00:22:00,900  -->  00:22:02,833
b it is coming out to be positive

477

00:22:02,866  -->  00:22:04,966
so that is the uh

478

00:22:04,966  -->  00:22:07,766
it's electric feed of the electric typo

479

00:22:07,966  -->  00:22:10,600
and then we can simplify by taking the common uh

480

00:22:10,600  -->  00:22:13,666
this QD divider for perhaps or not our cube

481

00:22:13,666  -->  00:22:17,233
and that is the simplified form of this electric feed

482

00:22:17,266  -->  00:22:19,366
due to an electric typo

483

00:22:20,900  -->  00:22:24,866
now let's further define some of the other definitions

484

00:22:24,866  -->  00:22:27,033
related to this electric typo

485

00:22:27,600  -->  00:22:30,800
first of all let's define what is this a d cost data

486

00:22:31,266  -->  00:22:34,000
but if you assume that this d

487

00:22:34,133  -->  00:22:37,133
d is a line vector line segment

488

00:22:37,133  -->  00:22:38,166
right uh

489

00:22:38,166  -->  00:22:39,666
since it is oriented along

490

00:22:39,666  -->  00:22:42,400
this is the XS so we can uh

491

00:22:42,400  -->  00:22:44,200
we can in the vector uh

492

00:22:44,566  -->  00:22:45,000
notation

493

00:22:45,000  -->  00:22:49,500
we can denote it like a d a Z first do not replace a Z

494

00:22:49,733  -->  00:22:54,300
an a Z will be directed from minus q two q

495

00:22:54,300  -->  00:22:57,600
so remember this thing for the convention of this book

496

00:22:57,600  -->  00:22:59,500
so it is directed from for the tripod

497

00:22:59,500  -->  00:23:03,800
it is directed from minus Q 2 q k

498

00:23:03,900  -->  00:23:06,866
then uh we define another terminology

499

00:23:06,866  -->  00:23:09,200
that is the dipal movement

500

00:23:09,200  -->  00:23:10,666
so what is dipal movement

501

00:23:10,666  -->  00:23:11,533
so dipal movement

502

00:23:11,533  -->  00:23:16,866
is the product of the magnitude of charge time b

503

00:23:18,266  -->  00:23:20,800
separation between these two charges

504

00:23:20,866  -->  00:23:24,766
so in this case it would consider it a

505

00:23:24,766  -->  00:23:27,166
d as a erector quantity right

506

00:23:27,566  -->  00:23:29,566
like this way okay

507

00:23:29,566  -->  00:23:32,866
so this is you can treat it like d

508

00:23:32,866  -->  00:23:34,266
a Z right

509

00:23:34,533  -->  00:23:38,333
and then we can we can involve what

510

00:23:38,333  -->  00:23:41,966
we can involve this this thing right

511

00:23:42,266  -->  00:23:44,966
this thing so we can involve this thing

512

00:23:44,966  -->  00:23:47,166
and this definition of the Z

513

00:23:47,333  -->  00:23:50,300
if you see here if you see here the magnitude of a r

514

00:23:50,300  -->  00:23:53,333
it is equal to one so basically

515

00:23:53,333  -->  00:23:56,366
we are taking the dark part out of this d

516

00:23:56,366  -->  00:23:58,800
along this radial direction

517

00:23:58,800  -->  00:24:00,100
along this radial direction

518

00:24:00,100  -->  00:24:02,966
because we are taking this projection

519

00:24:03,166  -->  00:24:05,366
we are taking this projection of d

520

00:24:05,366  -->  00:24:08,733
on this difference between R2

521

00:24:08,733  -->  00:24:11,366
minus R1

522

00:24:12,066  -->  00:24:14,866
right and which is approximately equal to

523

00:24:16,133  -->  00:24:17,466
the cause of data

524

00:24:18,200  -->  00:24:20,266
okay so uh

525

00:24:20,266  -->  00:24:21,966
let's further play around with this

526

00:24:21,966  -->  00:24:25,966
this thing so we define the uh vector

527

00:24:25,966  -->  00:24:28,633
uh diaper moment that is equal to q

528

00:24:28,866  -->  00:24:30,766
uh this uh d

529

00:24:30,766  -->  00:24:32,366
uh vector d right

530

00:24:32,500  -->  00:24:35,500
and then we can uh write down this uh

531

00:24:36,366  -->  00:24:39,033
this thing right this thing that

532

00:24:40,333  -->  00:24:41,133
q

533

00:24:41,700  -->  00:24:43,933
q d q d right

534

00:24:43,933  -->  00:24:44,800
this thing

535

00:24:45,400  -->  00:24:47,933
so we multiply both sides with q right

536

00:24:47,933  -->  00:24:49,800
we multiply both sides with q

537

00:24:50,100  -->  00:24:53,466
so it is equal to q d cost data

538

00:24:53,666  -->  00:24:56,566
and must we simplify in the form of this

539

00:24:57,966  -->  00:24:59,366
drop product notation

540

00:24:59,366  -->  00:25:02,400
we can simplify like this thing right

541

00:25:02,400  -->  00:25:06,466
so you can play around that it is equal to

542

00:25:07,400  -->  00:25:10,000
so multiply q d right

543

00:25:10,466  -->  00:25:15,800
q d vector so multiplying both sides with this q

544

00:25:17,000  -->  00:25:24,366
Q and Q right so it would be Q d dot a r right

545

00:25:24,566  -->  00:25:27,466
and now we can uh involve this definition of your

546

00:25:29,300  -->  00:25:31,400
type of moment like this

547

00:25:31,866  -->  00:25:35,700
and if you if you open up this dot product

548

00:25:35,700  -->  00:25:36,966
so it is in other words

549

00:25:36,966  -->  00:25:43,266
it is equal to q d cause of Twp magnitude of this a

550

00:25:43,266  -->  00:25:47,633
a R unit factor is 1 and the magnitude of this p is

551

00:25:48,133  -->  00:25:51,433
p magnitude is equal to curing

552

00:25:53,933  -->  00:25:56,533
like this okay so

553

00:25:56,533  -->  00:25:59,966
this is another form of this electric vector potential

554

00:25:59,966  -->  00:26:02,300
because of this electric iPhone

555

00:26:02,300  -->  00:26:05,200
since we know that unit vector is the vector

556

00:26:05,200  -->  00:26:07,000
divided by gratitude of the vector

557

00:26:07,000  -->  00:26:09,633
so we can further donate

558

00:26:11,366  -->  00:26:12,400
denote this equation

559

00:26:12,400  -->  00:26:15,466
in this dot product of the vector form

560

00:26:18,166  -->  00:26:23,400
so if your diple centre is at our bar

561

00:26:23,400  -->  00:26:26,266
this means if your diple is not at this origin

562

00:26:26,266  -->  00:26:30,100
rather it is at some another position back to our bar

563

00:26:30,100  -->  00:26:32,100
so you just be extend your

564

00:26:32,100  -->  00:26:36,933
simply extend your case to the standard case

565

00:26:36,933  -->  00:26:40,066
that you can take the distance

566

00:26:40,266  -->  00:26:42,133
you can treat this position

567

00:26:42,133  -->  00:26:46,533
vector as a distance vector between this R

568

00:26:46,533  -->  00:26:50,100
and R but it is just like the same case

569

00:26:50,100  -->  00:26:52,700
that we considered in our coolants law

570

00:26:53,133  -->  00:26:53,966
for once

571

00:26:53,966  -->  00:26:57,833
your point charge is not located at the origin

572

00:26:59,700  -->  00:27:03,566
k and then uh what else so again

573

00:27:03,566  -->  00:27:06,166
we can uh write this uh

574

00:27:07,000  -->  00:27:09,433
electrical in the vector form

575

00:27:09,733  -->  00:27:12,500
like this thing that are in the diaper moment form

576

00:27:12,500  -->  00:27:15,366
like this thing so this QD will define

577

00:27:15,366  -->  00:27:19,466
it is equal to this magnitude of diaper movement

578

00:27:19,466  -->  00:27:23,600
like this thing and rest things are staying the same

579

00:27:23,766  -->  00:27:24,633
okay

580

00:27:25,400  -->  00:27:28,966
okay so only way define this thing additionally

581

00:27:28,966  -->  00:27:31,500
so here so you can

582

00:27:31,566  -->  00:27:34,133
you can summarize this line like this way

583

00:27:34,133  -->  00:27:36,200
that it is the

584

00:27:37,933  -->  00:27:40,233
electric potential form in the dipole

585

00:27:40,966  -->  00:27:42,600
right in the dipole moment

586

00:27:42,600  -->  00:27:44,566
using the dipole movement information

587

00:27:44,700  -->  00:27:48,733
and it is the uh this second equation

588

00:27:48,733  -->  00:27:54,433
is the electric feeder of the dipole

589

00:27:54,600  -->  00:27:56,900
using the dipole moment in formation

590

00:28:00,066  -->  00:28:04,500
okay so there are some notes about the electric

591

00:28:04,700  -->  00:28:07,933
electric Dibol so our point charge is a monopole

592

00:28:07,933  -->  00:28:08,966
so it is a single

593

00:28:08,966  -->  00:28:13,166
single charge and its electric grid varies inversely

594

00:28:14,733  -->  00:28:16,366
with respect to our skill

595

00:28:16,366  -->  00:28:18,700
so it is in mostly proportionate to our skill

596

00:28:18,700  -->  00:28:20,566
that is electric wheel

597

00:28:21,333  -->  00:28:24,300
electric fee for the single point charge

598

00:28:24,300  -->  00:28:26,600
it is inversely proportional to R scale

599

00:28:26,600  -->  00:28:28,800
and we found out that it's potential

600

00:28:28,800  -->  00:28:30,900
it is inversely proportional to R

601

00:28:31,200  -->  00:28:35,266
so it is one part less than the R scale right

602

00:28:35,666  -->  00:28:38,600
so it is one part less than the electric fee right

603

00:28:39,666  -->  00:28:40,333
so likewise

604

00:28:40,333  -->  00:28:43,400
if we see over here from the last page equations

605

00:28:43,400  -->  00:28:44,266
that the electrode

606

00:28:44,266  -->  00:28:47,000
page due to a dye ball varies inversely

607

00:28:47,866  -->  00:28:50,700
uh as RQ right

608

00:28:50,766  -->  00:28:51,866
RQ this we saw

609

00:28:51,900  -->  00:28:55,200
and then we saw that the potential it varies

610

00:28:55,200  -->  00:28:57,600
uh inversely with respect to RQ

611

00:28:57,600  -->  00:29:00,333
so again it is one less than the uh

612

00:29:00,333  -->  00:29:02,500
this inverse relationship is

613

00:29:02,500  -->  00:29:05,366
one less than the electrician relationship

614

00:29:05,900  -->  00:29:08,033
and then we can extend this case to the

615

00:29:08,166  -->  00:29:10,900
to the higher order multi multiples

616

00:29:10,966  -->  00:29:14,766
so electrically due to successive high order multiples

617

00:29:14,766  -->  00:29:17,200
such as a quadruple quadruple

618

00:29:17,200  -->  00:29:18,633
which says quadruple as well

619

00:29:18,966  -->  00:29:22,233
consisting of two die poles or an octable

620

00:29:22,300  -->  00:29:23,866
uh consisting of two uh

621

00:29:23,866  -->  00:29:27,500
quadruples and a dear anniversary proportion to ask uh

622

00:29:27,500  -->  00:29:28,766
R R4

623

00:29:28,800  -->  00:29:31,400
R5 and R R6 onwards

624

00:29:31,400  -->  00:29:35,000
and likewise the potentials they will be 1

625

00:29:35,000  -->  00:29:39,533
they will be inversely proportional to all our part

626

00:29:39,533  -->  00:29:41,433
which is one less than the

627

00:29:42,266  -->  00:29:44,900
in worst relationship for the electric field

628

00:29:45,700  -->  00:29:47,966
so that is the extension of these two

629

00:29:47,966  -->  00:29:49,233
uh first two cases

630

00:29:50,900  -->  00:29:53,166
let's solve this example for the

631

00:29:53,400  -->  00:29:56,700
for the clarification of this electric dipole

632

00:29:56,700  -->  00:29:59,933
so two dipoles with dipole movement

633

00:29:59,933  -->  00:30:02,100
so here two dipoles are given

634

00:30:02,166  -->  00:30:03,933
and dipole movement is given right

635

00:30:03,933  -->  00:30:05,266
p is given in this case

636

00:30:05,733  -->  00:30:08,700
and uh for the first case it is having minus 5

637

00:30:08,700  -->  00:30:10,900
easy for the second dipole it is 9

638

00:30:10,900  -->  00:30:13,866
80 and the location of these two dipole is 0

639

00:30:13,866  -->  00:30:17,466
0MINUS2 and 0 0 3 so we see here

640

00:30:17,466  -->  00:30:20,900
so the first iPhone it is at if it is X

641

00:30:20,900  -->  00:30:25,466
y Z corner so it is X and y both are 0 only Z having

642

00:30:25,466  -->  00:30:27,200
so this is your first iPhone

643

00:30:27,900  -->  00:30:28,700
right

644

00:30:29,666  -->  00:30:32,433
this is your second iPod okay

645

00:30:32,666  -->  00:30:37,466
and we simply involve this summation of

646

00:30:37,766  -->  00:30:40,666
we need to find a differential act origin

647

00:30:40,700  -->  00:30:43,600
so this is our point of interest right

648

00:30:44,066  -->  00:30:48,466
so we simply add up the potential because of these two

649

00:30:48,533  -->  00:30:50,366
uh electric die poles

650

00:30:50,500  -->  00:30:54,466
uh uh at this point at this origin

651

00:30:54,466  -->  00:30:57,700
so we find out the potential due to this electric die

652

00:30:57,700  -->  00:30:59,366
the first diepole at this origin

653

00:30:59,366  -->  00:31:03,533
we find out the potential due to this second diepole

654

00:31:03,533  -->  00:31:06,300
at this region and then we sum some them up

655

00:31:06,300  -->  00:31:10,966
these two diepoles diepoles potential okay

656

00:31:10,966  -->  00:31:16,300
so for P1P1 is given as minus 5 and Nano Coolance a Z

657

00:31:16,400  -->  00:31:22,166
and what is R1 so R1 is your discuss factor

658

00:31:22,566  -->  00:31:26,066
uh for this first iPhone with respect to this origin

659

00:31:26,066  -->  00:31:28,133
so this is your destination right

660

00:31:28,133  -->  00:31:29,666
this is your destination

661

00:31:30,400  -->  00:31:35,033
this is your source 1 this is your source 2

662

00:31:36,566  -->  00:31:38,666
and you just need to subtract the destination

663

00:31:38,666  -->  00:31:40,733
coordinate from the source coordinates

664

00:31:40,733  -->  00:31:42,300
and likewise for the second one

665

00:31:42,300  -->  00:31:45,266
and you can find out that it is 2 a Z and minus a Z

666

00:31:46,166  -->  00:31:49,266
okay then simply plug in these uh values

667

00:31:49,400  -->  00:31:51,466
and since it is a dot product right

668

00:31:51,466  -->  00:31:54,666
so both are uh easy so easy dotted with easy

669

00:31:54,666  -->  00:31:58,666
easy go to one and then minus 9 is coming into place

670

00:31:58,666  -->  00:32:03,066
so it is minus 10 nanoquodoms and then what else

671

00:32:03,066  -->  00:32:06,866
okay R1 I one okay I one I 1

672

00:32:06,866  -->  00:32:09,533
2 so I one magnitude is 2 so it can

673

00:32:09,533  -->  00:32:12,933
it will come out to be 8 and R2 magnitude is 3

674

00:32:12,933  -->  00:32:15,100
so it will come out to be 27

675

00:32:15,100  -->  00:32:17,866
and then you can simplify to find out the potential

676

00:32:17,866  -->  00:32:22,233
that is minus 20.25 voltz

677

00:32:24,133  -->  00:32:25,466
so I'm here today's lecture

678

00:32:25,466  -->  00:32:27,166
so first of all

679

00:32:27,166  -->  00:32:30,966
we found out our relationship between electric and

680

00:32:30,966  -->  00:32:32,700
electric field and electric potential

681

00:32:32,700  -->  00:32:33,700
that how

682

00:32:33,700  -->  00:32:37,800
if we are given the information of electric potential

683

00:32:37,800  -->  00:32:41,133
then how we can find out be this electrical

684

00:32:41,133  -->  00:32:45,133
and if we are given the information of this electrical

685

00:32:45,133  -->  00:32:47,166
then how we can find out the

686

00:32:47,166  -->  00:32:50,000
this potential electric potential

687

00:32:50,000  -->  00:32:53,166
so these are the two very useful relationships

688

00:32:53,200  -->  00:32:56,100
then based on this property of

689

00:32:57,666  -->  00:33:01,533
of this consegrated property of electrostatic fee

690

00:33:01,533  -->  00:33:04,100
we drive this Maxwell's second equation

691

00:33:04,100  -->  00:33:06,233
that current of electrical is equal to zero

692

00:33:06,566  -->  00:33:11,000
then uh we discussed this uh electric uh typo

693

00:33:11,066  -->  00:33:12,466
and then we found out it's uh

694

00:33:12,466  -->  00:33:15,833
potential and the electric free equations

695

00:33:15,900  -->  00:33:19,766
uh and uh and this standard equation of these uh typos

696

00:33:20,533  -->  00:33:21,300
in the next class

697

00:33:21,300  -->  00:33:26,166
we are going to study this electric flux lines and uh

698

00:33:26,166  -->  00:33:28,200
what are the equiperential surfaces

699

00:33:28,200  -->  00:33:30,200
and what are the equiperential lines

700

00:33:30,200  -->  00:33:34,066
and then we will also determine the energy density

701

00:33:34,066  -->  00:33:35,833
in an electrostatic fee

702

00:33:37,966  -->  00:33:40,300
here I thank you all if you have any questions

703

00:33:40,300  -->  00:33:44,100
that will be entertained to your emails or online

704

00:33:44,100  -->  00:33:44,600
synchron

705

00:33:44,600  -->  00:33:47,833
a session that we have arranged at the department