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this is election No. 22 of a V232

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electromagnetic field theory today

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uh we just build on our uh previous lecture

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uh that was how to determine this uh

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resistance and capacity of uh

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registers and capacitors using the boundary problem

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and in this in this lecture

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we are going to solve the examples of that

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previous case and this will again cover your section

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as remaining part of your Section 6.5

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so we will uh will solve the

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some of the case examples

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remember this uh slide that we discussed in details

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in our previous lecture that if uh

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if you assume uh for the capacitor

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if you assume that discharge is constant

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and you find out the V in terms of Q

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okay then you can find O d capestinence

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and how to find O d v uh

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this is a potential

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uh across that uh parallel capestial that is

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uh for that we know that the relationship is uh

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the line integral of your electric field

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and how to determine this electric field

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you can take help of either causes

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law or decoulents law

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okay and then uh what was the second uh

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second uh method that you

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you assume that the voltage is constant across the

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by the plate capacitors uh

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and then you find out Q in terms of

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q in terms of your E

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in terms of your V okay

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in terms of your V not okay

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and how to find out the eh eh

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this eh queue in terms of v not

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you again need to take out of this eh

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eh flux equation

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and for this you need to find out electric field

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electric field is related to the

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there by this gradient

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relationship with the elective potential

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and how to powerfinder this potential

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then based then according to these scenarios

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either as you can apply this lapse equation

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or the portions equation

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so you need to solve this boundary problem

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uh for that uh for these particular cases for

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for which you have this volume circumstancety

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or you don't have this volume circumstancety

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and how to find the resistance of the capacitor

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for this case the first matter

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presumed that your vote is across that resistor

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it is constant

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you need to determine this current using the

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this grand density and we're using this homeslaw

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you can find out the tendency for this

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again you need to find out this electric field

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and it is related to the elective potential

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by using this relationship

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and again you need to solve this

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the class and pause

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and equation to find out this potential

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uh this again uh

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recap of your previous structure that uh

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these are the capacitances uh

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uh of different types of uh capacitors that we have uh

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drived for the generalized cases

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okay and then accordingly

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you can find out

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the legal distance between the players

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that is uh

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your uh due to your medium

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uh uh that uh that can be

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that can be found out using this relationship

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that was your relaxation time relationship

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that is the ratio of your primitivity

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and the connectivity of your medium

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and using this relationship

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and using this capacity over here

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you can find out the resistance

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like like this

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like this in this case but remember

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that this resistance is the leakage resistant

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it is not the distance of the capacity it played

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but it is the leakage resistance between the plates

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first of all this example 6.8 of your textbook

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so a metal bar of conductivity

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Sigma is bent of form of flat

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90 degree sector

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so it is 1/4 of the 1/4 of your complete circle and

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its inner radius is a this is your inner radius

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its outer outer radius is B

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that is this one this one outer radius

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inner radius

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and its thickness is thickness is t okay

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and we need to find out the resistance of the bar

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between the vertical curve surfaces

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at r is equal to a n r is equal to B

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so that is the

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you need to find out the resistance between these two

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okay between these two surfaces okay

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this sensor inner surface and the outer surface

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and what is the second thing

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this is chance between two horizontal surfaces

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that is your this one and this one

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top one okay

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between your top and the bottom surface

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so that is your second thing

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our bar

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our bar and our SO2 resistances you need to find out

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um between a vertical curved

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and located at this first radius

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roads equal to a and roads equal to B

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the bar has a non uniform cross section area

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because you can as you can see here that uh uh

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that uh this area is going to change right

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this area is going to change between these two uh

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between these two vertical

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these two vertical these these are the verticals okay

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inner inner surfaces okay inner surfaces interceptors

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so between vertical curved ends

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located at rose equal to a and rose equal to B

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the bar has non uniform cross section area okay

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because it is a see here it is a

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it is sector of row is equal to B

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and sector of row is equal to a

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so let's assume the potential difference

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be not constant

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potential difference between these two curve

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that row is equal to a and

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row is equal to B

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that we assume that at rose equal to a

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and rose equal to B so

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that we just take the Assumption that

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if this potential difference

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we not is constant between them

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then add inner radius you potentially 0 okay

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at this your potentially 0

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and at the outer radius

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you are taking the potential as we not

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okay

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and now we need to find a

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since we don't we don't have the uh wallet identity

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uh so uh

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we need to solve this labrass equation

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in this Slendrical Partners okay

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why Slender Partners you can see this material here

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this Dmitry is your slendricals material here okay

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the thing is so

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so the things are changing with this back to your row

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the variable is row okay

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so you uh you take its first integral

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so you you just need to recall your uh

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this uh laplash integration in terms of your uh

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slander regarding assistant from your earlier chapters

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okay

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and uh then integrate it once take this rubber here

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integrate it twice

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you know that the integral of this uh

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one or row it is and enough

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and then off row like this one OK

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with respect to this G row

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okay so as you're having two constants of integration

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you need to find out

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these two constructs of integration

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so you need two boundary conditions

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always remember the number of unknowns

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for the unknown under and for the

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for same

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for the number of boundary conditions required

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to determine the number of unknowns

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must be same of

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same numbers two unknowns

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so two boundary conditions

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so the first boundary condition is at inner radius

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your what is 0

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and the second boundary condition is your

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at your outer radius your what

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radius constant that is v not so plug in these two

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um boundary condition so for

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for the first one you will find out this B constant

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which you use this uh you plug in this into your hair

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and then you plug in this second

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boundary condition to find out this a

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okay so in the

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at this stage

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we have found out these two construct of integration

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uh no

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uh no

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you you uh you need to uh just rearrange the things

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okay just uh rearrange the things

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plug in these constant values and uh

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you you also know this uh

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rule of your love with me

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rule of subtraction that

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subtraction of love

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of two terms is equal to the love of their ratios okay

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like this one okay

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okay so now you know the potential

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you just take its gradient to find out electric feed

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so this is your electric feed

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and did you have found out this

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so I did there there are two things

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either you just take the gradient of this thing

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or you can just use this uh

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uh first

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the things that we have found out

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after taking the first interior of blood

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cash and equation on the previous life

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you can directly plug in that as well

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but take care of these symbols and things like this

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so it is always better it is always better to

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take the gradient of this this thing okay

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to avoid any confusion of the corner system

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no involve this homes bar that your surface condense t

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it is equal to the stigma e

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and here see that your d this

219

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surface surface area differential surface area

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it is having minus sign so y minus sign

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y minus sign because this is your positive

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and the inner is your 0 0 negative

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so electric field it is moving from outward to inward

224

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so that's why is your

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this radial direction of the electric field

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is in minus a row direction

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so this is very very important point over here

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that's why it is minus sign over here

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because your electric field

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it is against this radial direction

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okay so plug in this in this relationship and what is

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what is your left with

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everything is constant over here okay

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and uh

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you just need to integrate it with respect to defy

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so you will have power 2 okay

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why power 2 because it is a 90 degree sector okay

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so here 5 it is 5 by 2

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the second thing that it is raining is your DZ okay

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the DZ in this direction okay

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in this direction

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because your changes are in the radio direction

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so your surface idiot is differential surface idiot

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it is equal to this just recall

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recall your chapter number two

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and the second variable is yours changes are to two

247

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2 variables

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the first one is your uh this uh English changing

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the second one is your length is changing

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like this one okay

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and uh what else

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the integration with respect to Z is Z only

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and then this will replace your TR okay

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the thing is already constant

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and this is your private dessert

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this in integration with respect to your fine

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so you have found out this I

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in terms of your we not that was

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which was your and objective

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so this we not and this we not will be cancelling out

261

00:12:03,466  -->  00:12:08,433
and this is your resistance that you need to follow

262

00:12:08,700  -->  00:12:11,600
for this between these two vertical plates

263

00:12:11,600  -->  00:12:15,633
between these two vertical plates in and out of plate

264

00:12:17,933  -->  00:12:20,700
let's not solve the second part of this example

265

00:12:20,700  -->  00:12:22,766
where you need to find out the resistance

266

00:12:22,766  -->  00:12:24,200
between these two

267

00:12:24,866  -->  00:12:27,300
horizontal horizontal surfaces okay

268

00:12:27,366  -->  00:12:30,066
and here we assume that at the bottom surface

269

00:12:30,066  -->  00:12:31,966
there is your Z equal to 0

270

00:12:31,966  -->  00:12:36,400
we are assuming that is potential is zero okay

271

00:12:36,600  -->  00:12:38,000
and at the top surface

272

00:12:38,000  -->  00:12:40,566
we are assuming that it is having constant potential

273

00:12:40,566  -->  00:12:41,866
that is v not okay

274

00:12:41,866  -->  00:12:44,100
and this and this surface it is v not

275

00:12:44,466  -->  00:12:46,733
and this surface it is the bottom surface

276

00:12:46,733  -->  00:12:48,233
it is zero okay

277

00:12:49,300  -->  00:12:51,600
and uh no the things are changing

278

00:12:51,600  -->  00:12:54,100
the things are changing with respect to your Z axis

279

00:12:54,100  -->  00:12:55,833
okay Z axis

280

00:12:55,933  -->  00:12:56,866
so that's why you have

281

00:12:56,866  -->  00:12:59,666
we are having variations with respect to your Z

282

00:13:00,533  -->  00:13:03,933
Z and the lip class equation

283

00:13:03,933  -->  00:13:08,200
again we will use this to find out the potential

284

00:13:08,200  -->  00:13:10,200
and according to the electric fee

285

00:13:11,266  -->  00:13:13,766
and uh again this very uh

286

00:13:13,766  -->  00:13:15,333
this is slendical uh

287

00:13:15,333  -->  00:13:18,000
uh slendical uh corner systems

288

00:13:18,000  -->  00:13:21,400
lap class equation and then integrating it twice

289

00:13:21,466  -->  00:13:22,666
then this is uh uh

290

00:13:22,933  -->  00:13:25,966
v in terms of potential in terms of your uh

291

00:13:26,200  -->  00:13:28,966
uh constant uh in integration

292

00:13:29,500  -->  00:13:31,200
conscience of integrations

293

00:13:31,200  -->  00:13:32,333
and now you need to apply this

294

00:13:32,333  -->  00:13:35,066
simply the boundary conditions to find out the

295

00:13:35,066  -->  00:13:38,266
these constants that is your b equal to 0 and a 0

296

00:13:38,266  -->  00:13:40,766
b not divided thickness and uh

297

00:13:40,766  -->  00:13:41,866
this is your potential

298

00:13:41,866  -->  00:13:43,766
there's your potential in terms of here

299

00:13:45,066  -->  00:13:47,600
you're evading parameter that year Z

300

00:13:47,900  -->  00:13:50,933
and as you take the gradient of this and uh

301

00:13:50,933  -->  00:13:52,466
that is your electric field

302

00:13:55,266  -->  00:13:56,666
now if you know the electric field

303

00:13:56,666  -->  00:13:58,733
then you can find out the current density

304

00:13:58,733  -->  00:14:02,566
using this Homs law and not again

305

00:14:02,900  -->  00:14:08,100
to find out its surface integral with respect to

306

00:14:09,000  -->  00:14:11,400
to find out this current you can

307

00:14:11,733  -->  00:14:12,600
you see your head

308

00:14:12,600  -->  00:14:16,233
and you have taken the bottom surface at zero

309

00:14:16,366  -->  00:14:18,200
the top surface at me not

310

00:14:18,200  -->  00:14:20,266
so you electric feel

311

00:14:20,933  -->  00:14:22,266
electric feel it is

312

00:14:22,266  -->  00:14:25,400
it is from top to bottom

313

00:14:25,400  -->  00:14:27,766
that is your minus easy direction

314

00:14:27,766  -->  00:14:29,866
okay so the normal here

315

00:14:29,866  -->  00:14:32,500
it is minus easy okay

316

00:14:33,533  -->  00:14:37,100
and so the things are wearing with this

317

00:14:37,100  -->  00:14:38,266
so what is changing over here

318

00:14:38,266  -->  00:14:39,600
what is changing over here

319

00:14:39,600  -->  00:14:42,766
your radius is changing your fire is changing

320

00:14:42,766  -->  00:14:46,300
and your indeed this normally is your

321

00:14:47,666  -->  00:14:49,000
easy okay

322

00:14:50,300  -->  00:14:52,500
normal to these two separate surfaces

323

00:14:53,266  -->  00:14:54,900
and the wide way of taking minus easy

324

00:14:54,900  -->  00:14:56,266
because your electric feel

325

00:14:56,266  -->  00:14:59,266
it is from top to bottom okay

326

00:15:00,266  -->  00:15:02,066
no the things are very simple here

327

00:15:02,066  -->  00:15:03,466
again you need to uh

328

00:15:03,466  -->  00:15:05,400
integrate this with respect to fire

329

00:15:05,400  -->  00:15:09,433
that is your fiber tube the sector is your fiber tube

330

00:15:09,600  -->  00:15:10,600
and uh

331

00:15:10,600  -->  00:15:14,200
then integration with respect to row it is your row

332

00:15:14,200  -->  00:15:16,000
scale on 2

333

00:15:16,800  -->  00:15:19,866
the image they are from a to B and uh

334

00:15:19,866  -->  00:15:22,200
just plug in these two things and uh

335

00:15:22,566  -->  00:15:24,133
not use this current over here

336

00:15:24,133  -->  00:15:25,800
and then you can simplify the things

337

00:15:25,800  -->  00:15:27,200
so this this is the

338

00:15:27,500  -->  00:15:30,300
and this is the uh expression that you

339

00:15:30,300  -->  00:15:31,566
you were required to dry

340

00:15:33,166  -->  00:15:35,000
now alternate alternatively

341

00:15:35,000  -->  00:15:38,366
for this thing is using just basic formulation

342

00:15:38,766  -->  00:15:40,700
because for the top surface

343

00:15:40,766  -->  00:15:42,466
the cross section area

344

00:15:42,466  -->  00:15:45,533
it is uniform both top and bottom surfaces

345

00:15:45,533  -->  00:15:47,800
they are having the same same

346

00:15:48,700  -->  00:15:51,700
same surface area you see overhead this

347

00:15:51,700  -->  00:15:54,566
this is the same surface area

348

00:15:54,566  -->  00:15:57,700
the same as as I'm sure bottom one

349

00:15:58,166  -->  00:16:00,966
so you can apply the basic formula as well okay

350

00:16:01,000  -->  00:16:04,533
the only difference is that here the surface area

351

00:16:04,533  -->  00:16:06,333
so as here over here the thickness

352

00:16:06,333  -->  00:16:08,800
what is the suppression between them

353

00:16:08,933  -->  00:16:10,300
the distance between them

354

00:16:10,500  -->  00:16:13,233
it is your t okay

355

00:16:14,366  -->  00:16:18,200
and what is your uh what is your uh this uh

356

00:16:19,566  -->  00:16:21,100
what is your uh this uh

357

00:16:21,300  -->  00:16:22,333
uh that is uh

358

00:16:22,333  -->  00:16:25,500
surface area since it is out one fourth of your

359

00:16:27,100  -->  00:16:30,266
circle so that is your 1/4 of your

360

00:16:30,600  -->  00:16:35,066
one fourth of your by a square okay

361

00:16:36,266  -->  00:16:39,566
and what is over here that is your B minus E

362

00:16:40,800  -->  00:16:42,600
the effective the effective

363

00:16:42,866  -->  00:16:45,700
this your effective radius

364

00:16:47,533  -->  00:16:49,433
so that is the another method

365

00:16:51,200  -->  00:16:52,633
let's solve another example

366

00:16:52,700  -->  00:16:54,366
6.9 a coexial cable

367

00:16:54,366  -->  00:16:59,000
contains an insulating material of conductor desigma

368

00:16:59,000  -->  00:17:01,266
if the radius of the central y is a

369

00:17:01,366  -->  00:17:03,666
and that of the sheet that is your most

370

00:17:03,666  -->  00:17:06,166
it is a ground and that is a b

371

00:17:06,166  -->  00:17:07,366
so we need to uh

372

00:17:07,366  -->  00:17:10,400
show that the conductors of cable per unit land

373

00:17:10,400  -->  00:17:12,000
is this one okay

374

00:17:12,800  -->  00:17:14,866
conductors per unit land for the dialectic

375

00:17:14,866  -->  00:17:20,066
it is denoted by G and it is the recipe proclub here

376

00:17:20,133  -->  00:17:22,500
this leakage resistant so actually

377

00:17:22,500  -->  00:17:24,900
we need to find out this are here in this case

378

00:17:26,900  -->  00:17:29,000
okay so again we uh

379

00:17:29,000  -->  00:17:31,566
we use this uh uh

380

00:17:31,566  -->  00:17:35,766
boundary problem case that for this particular uh

381

00:17:35,766  -->  00:17:39,166
cylindrical quaxia case that uh

382

00:17:39,166  -->  00:17:40,800
the inner radius the inner

383

00:17:40,800  -->  00:17:45,066
the inner inner one is at your potential uh 0

384

00:17:45,066  -->  00:17:46,300
as you have seen in your

385

00:17:46,300  -->  00:17:48,366
in your previous slide that the inner

386

00:17:48,533  -->  00:17:51,666
inner thing at it is at inner radius conductor

387

00:17:51,666  -->  00:17:53,266
it is at differential 0

388

00:17:53,266  -->  00:17:55,133
and the outer one is at potential v

389

00:17:55,133  -->  00:17:59,766
not nothing's are changing with respect to your role

390

00:17:59,866  -->  00:18:02,366
for this slendical partner system

391

00:18:02,400  -->  00:18:05,666
no as you involve again delapasin equation

392

00:18:05,733  -->  00:18:09,566
take its double integral okay

393

00:18:09,566  -->  00:18:11,566
and so it is just like your example

394

00:18:11,566  -->  00:18:14,166
6.8 where we found all the voltages

395

00:18:14,166  -->  00:18:16,566
and the current and electric wheel

396

00:18:16,933  -->  00:18:18,633
as per these expressions

397

00:18:19,666  -->  00:18:22,266
not involve this electric field over here

398

00:18:22,933  -->  00:18:27,000
not here again again this is your zero okay

399

00:18:27,000  -->  00:18:28,300
this is your positive

400

00:18:28,600  -->  00:18:33,866
we not this is in your minus 0 direction okay

401

00:18:34,500  -->  00:18:38,300
that's why it is negative sign is coming over here okay

402

00:18:38,300  -->  00:18:41,366
so what you need to do that

403

00:18:41,366  -->  00:18:43,000
you just take the drop product of this

404

00:18:43,000  -->  00:18:45,233
with this differential surface area

405

00:18:45,266  -->  00:18:46,800
take its integration

406

00:18:46,933  -->  00:18:52,066
so the fire it is changing from 0 to 2 pie the Z

407

00:18:52,066  -->  00:18:55,400
your this Z direction you can take it as a C direction

408

00:18:55,400  -->  00:18:56,200
okay

409

00:18:57,300  -->  00:19:03,466
it is changing from zero to L and 0 to L okay

410

00:19:04,000  -->  00:19:07,733
and you just intigrate this thing okay

411

00:19:07,733  -->  00:19:09,300
so just intigrate this

412

00:19:09,300  -->  00:19:11,066
so you're going to have this to pie

413

00:19:11,066  -->  00:19:13,400
you're going to have this L over here

414

00:19:13,400  -->  00:19:15,900
and the other restings at your constant okay

415

00:19:16,800  -->  00:19:18,033
either constant okay

416

00:19:20,300  -->  00:19:24,300
okay so this Roy is going to cancel out with this one

417

00:19:24,300  -->  00:19:27,933
so resistance per unit land resistance per unit land

418

00:19:27,933  -->  00:19:30,066
you can just take it like this that resistance

419

00:19:30,066  -->  00:19:34,033
what was the resistance it was we not where I okay

420

00:19:34,166  -->  00:19:37,466
just plug in over here and for portion of land

421

00:19:37,466  -->  00:19:40,066
you just need to divide the distance with the R

422

00:19:40,066  -->  00:19:43,200
divided by this complete length okay

423

00:19:43,200  -->  00:19:46,000
so this will provide you the distance portion of plan

424

00:19:46,000  -->  00:19:47,766
and just take your stress approval to find out

425

00:19:47,766  -->  00:19:49,366
deep conductor conductor plan

426

00:19:51,566  -->  00:19:53,200
yeah another is useful example

427

00:19:53,200  -->  00:19:54,933
the connecting sphere steady

428

00:19:54,933  -->  00:19:56,966
connecting sperical shell with your dye

429

00:19:57,333  -->  00:19:58,900
is equal to 10 centimeter

430

00:19:58,900  -->  00:20:00,733
and b is equal to 30 centimeter

431

00:20:00,733  -->  00:20:01,400
are maintained

432

00:20:01,400  -->  00:20:03,466
at a potential difference of hundred volt

433

00:20:03,666  -->  00:20:06,100
such that your alter radius

434

00:20:06,100  -->  00:20:08,066
in this case your alter radius

435

00:20:08,066  -->  00:20:12,800
see alter radius see that 0 degree and 0% 0 volt

436

00:20:12,800  -->  00:20:16,200
and the innerry is at your this hundred volt

437

00:20:16,200  -->  00:20:20,200
so electric creates in Europe this a r direction

438

00:20:20,200  -->  00:20:21,966
pause to be our direction okay

439

00:20:23,200  -->  00:20:25,100
you need to determine this voltage

440

00:20:25,100  -->  00:20:26,133
and the electric wheel

441

00:20:26,133  -->  00:20:28,166
in the region between the shells

442

00:20:28,266  -->  00:20:31,866
and where this homogeneous dialect medium is

443

00:20:31,866  -->  00:20:33,700
having this relative permitivity

444

00:20:34,300  -->  00:20:36,266
you need to determine the total charge

445

00:20:36,266  -->  00:20:37,466
induction of the shells

446

00:20:37,466  -->  00:20:39,366
and the capacity of the capacitors

447

00:20:40,766  -->  00:20:41,333
let's uh

448

00:20:41,333  -->  00:20:42,600
proceed one by one

449

00:20:42,866  -->  00:20:45,200
since the things are changing in a videos

450

00:20:45,200  -->  00:20:48,700
back to your radio directions for this uh spherical

451

00:20:48,700  -->  00:20:51,433
um capacitor

452

00:20:52,000  -->  00:20:52,466
uh

453

00:20:52,466  -->  00:20:56,166
so you need to involve this lip Russian equation again

454

00:20:56,200  -->  00:21:00,366
um because there's Ruby's equal to 0 0 here okay

455

00:21:00,366  -->  00:21:02,166
Ruby equal to 0

456

00:21:02,733  -->  00:21:06,933
just take your stimulant trigger and so this is

457

00:21:06,933  -->  00:21:08,900
this is your potential

458

00:21:08,900  -->  00:21:11,400
in terms of the integration constant

459

00:21:11,400  -->  00:21:11,666
now

460

00:21:11,666  -->  00:21:15,000
apply this boundary conditions that at inner radius

461

00:21:15,000  -->  00:21:15,900
your voltage is

462

00:21:15,900  -->  00:21:19,400
we not at the outer radius your voltage is 0

463

00:21:19,666  -->  00:21:21,900
and once you will find out

464

00:21:21,900  -->  00:21:23,800
the construct of integration

465

00:21:23,800  -->  00:21:27,466
just plug in into your this main equation

466

00:21:27,466  -->  00:21:29,466
so this is your potential for this

467

00:21:30,666  -->  00:21:36,266
this concentrate uh spherical uh spherical capacture

468

00:21:36,600  -->  00:21:37,866
triplet capacture

469

00:21:39,533  -->  00:21:41,700
but you have found out the potential

470

00:21:41,700  -->  00:21:44,900
take its gradient to find out the electric field

471

00:21:44,900  -->  00:21:47,366
once you have found out this electric field

472

00:21:47,666  -->  00:21:50,466
put this electric field into this equation

473

00:21:51,000  -->  00:21:53,400
now you to determine this capacity

474

00:21:54,800  -->  00:21:59,266
we need to find our BV on either of the shell okay

475

00:21:59,500  -->  00:22:02,866
uh q discharge on either of the shell okay

476

00:22:03,466  -->  00:22:05,266
so here you have a zoom d

477

00:22:05,266  -->  00:22:06,466
we not as a constant

478

00:22:06,466  -->  00:22:09,500
remember we not as a constant you have a zoomed it

479

00:22:10,133  -->  00:22:11,733
and so you need to find out the

480

00:22:11,733  -->  00:22:13,066
this queue in terms of

481

00:22:13,066  -->  00:22:17,100
we not remember this relationship that the charge

482

00:22:17,100  -->  00:22:21,333
it is the surface integral of your electric feed no

483

00:22:21,333  -->  00:22:22,966
what things are changing over here

484

00:22:22,966  -->  00:22:24,800
your tita tita for this

485

00:22:24,800  -->  00:22:26,733
you know that in this vertical coordinate system

486

00:22:26,733  -->  00:22:28,300
it changes from 0 to pi

487

00:22:28,866  -->  00:22:31,766
your this Fi it is changing from 0 to pi

488

00:22:32,266  -->  00:22:36,033
and they change that with respect to your defined data

489

00:22:36,166  -->  00:22:38,566
and this is in your surface

490

00:22:39,933  -->  00:22:42,366
the normal was your er okay

491

00:22:42,366  -->  00:22:45,066
the normal unitractor was your er

492

00:22:46,400  -->  00:22:55,500
so what you need to interview

493

00:23:03,933  -->  00:23:05,900
okay so as you

494

00:23:05,900  -->  00:23:09,000
so what we have assumed that being with constant

495

00:23:09,000  -->  00:23:11,700
and we need to determine q in terms of v

496

00:23:11,900  -->  00:23:16,033
involving this solving this Lapras is equation

497

00:23:16,666  -->  00:23:19,000
no just simply put this into your hair

498

00:23:19,000  -->  00:23:20,533
so you're not really canceled out

499

00:23:20,533  -->  00:23:25,300
so this is your capacitors for this radical capacitor

500

00:23:28,866  -->  00:23:32,200
no you need to evaluate this

501

00:23:33,200  -->  00:23:35,466
find out this capacity for this particular case

502

00:23:35,466  -->  00:23:37,933
that radius are given and even notice

503

00:23:37,933  -->  00:23:41,700
given just all these voltages and electric read

504

00:23:41,700  -->  00:23:44,733
and the charge on these uh

505

00:23:44,733  -->  00:23:46,300
inner and outer shells

506

00:23:46,566  -->  00:23:50,666
and accordingly deep capacity between these two shells

507

00:23:53,133  -->  00:23:54,466
okay another example

508

00:23:54,700  -->  00:23:55,800
uh 6.11

509

00:23:55,800  -->  00:23:58,533
uh in section 6.5 we uh

510

00:23:58,533  -->  00:24:03,100
we solve the same example but there we assume that uh

511

00:24:03,100  -->  00:24:04,800
Q was uh we

512

00:24:04,800  -->  00:24:06,566
so what was that previous guess

513

00:24:06,566  -->  00:24:08,933
uh previous guess was that uh

514

00:24:08,933  -->  00:24:10,566
there are true matters uh

515

00:24:10,566  -->  00:24:11,800
contestants can be founded

516

00:24:11,800  -->  00:24:16,700
by either as human Q and finding out v or by either uh

517

00:24:16,700  -->  00:24:20,366
assuming V as a constant and finding out Q

518

00:24:21,000  -->  00:24:25,333
the formal approach was used in Section 6.5

519

00:24:25,333  -->  00:24:27,800
where we assume the q as a constant

520

00:24:27,800  -->  00:24:28,733
but here in this case

521

00:24:28,733  -->  00:24:31,566
we're going to assume that your V not is constant

522

00:24:31,566  -->  00:24:35,500
and we will find out the q in terms of your V not

523

00:24:35,600  -->  00:24:37,200
so this is the little matter

524

00:24:38,766  -->  00:24:40,900
and here again the boundary conditions are given

525

00:24:40,900  -->  00:24:43,866
that at your X is equal to 0

526

00:24:43,866  -->  00:24:45,333
that is your changing variable

527

00:24:45,333  -->  00:24:46,800
your X is equal to 0

528

00:24:47,066  -->  00:24:49,833
and your potential is taken to be 0

529

00:24:50,066  -->  00:24:52,066
and X is X is equal to d

530

00:24:52,066  -->  00:24:53,966
the suppression between these two players

531

00:24:54,000  -->  00:24:56,366
the potential is we not okay

532

00:24:56,866  -->  00:24:59,533
so that's why you're this electric wheel

533

00:24:59,533  -->  00:25:02,566
it is in your minus a X direction

534

00:25:04,333  -->  00:25:07,833
okay so again solve this lapation equation

535

00:25:08,000  -->  00:25:09,466
take the double in trigger

536

00:25:10,766  -->  00:25:12,300
put in this boundary conditions

537

00:25:12,300  -->  00:25:16,233
to find out the constants of integrations

538

00:25:16,400  -->  00:25:17,600
and once you have found out

539

00:25:17,600  -->  00:25:19,100
so this is your

540

00:25:19,366  -->  00:25:22,966
and voltage in terms of your changing variable acts

541

00:25:24,466  -->  00:25:25,800
is to find the questions

542

00:25:25,800  -->  00:25:28,733
first to find the charge on either of the plate

543

00:25:28,733  -->  00:25:31,266
so here you remember that your charge

544

00:25:31,266  -->  00:25:32,466
it is equal to your

545

00:25:34,133  -->  00:25:37,266
that uh the charge it is equal junior

546

00:25:37,500  -->  00:25:39,833
and this uh circus integral of here

547

00:25:39,866  -->  00:25:44,666
uh this uh density this is circuit chart density

548

00:25:44,866  -->  00:25:47,100
and you remember that from your boundary condition

549

00:25:47,100  -->  00:25:50,233
that your surface chart density

550

00:25:50,600  -->  00:25:53,466
your surface identity it is the drop product of here

551

00:25:54,566  -->  00:25:56,066
the electric proxanity

552

00:25:56,200  -->  00:25:58,266
with the normal genitomimal factor

553

00:25:58,333  -->  00:26:01,800
so that's why you can you can also treat it like that

554

00:26:01,800  -->  00:26:04,266
it is equal to d dot DS okay

555

00:26:05,200  -->  00:26:08,100
and here's your D uh

556

00:26:08,100  -->  00:26:10,500
here's your d is uh

557

00:26:10,500  -->  00:26:13,400
your judges is target in terms of your uh

558

00:26:14,666  -->  00:26:17,700
in terms of your basic definitions that your charge

559

00:26:17,700  -->  00:26:21,400
it is a surface integral of the charge dynasty

560

00:26:21,400  -->  00:26:24,500
and it planned it destroys the protigue detour

561

00:26:24,500  -->  00:26:26,166
you can equently you can

562

00:26:26,166  -->  00:26:30,000
you can equently treat it treated like this that uh

563

00:26:30,000  -->  00:26:33,266
the using the boundary conditions for the conductors

564

00:26:33,266  -->  00:26:35,866
your surface density is equal to the uh

565

00:26:35,866  -->  00:26:38,500
top product of your unitomal and e lactic uh

566

00:26:38,500  -->  00:26:39,433
flux tensity

567

00:26:40,066  -->  00:26:42,866
okay now simplify this thing okay

568

00:26:43,166  -->  00:26:45,100
and what you need to do

569

00:26:47,300  -->  00:26:50,933
you need to just take the gradient of this thing

570

00:26:50,933  -->  00:26:52,633
so this is your electric field

571

00:26:53,066  -->  00:26:55,533
okay and uh for low place

572

00:26:55,533  -->  00:26:58,200
uh you know that your uh uh a normal

573

00:26:58,200  -->  00:27:00,500
it is in the post 2 direction for the upper plate

574

00:27:00,500  -->  00:27:03,500
your normal is in B minus the X direction

575

00:27:04,700  -->  00:27:10,500
so uh your raw s your raw s is equal to uh minus aplon

576

00:27:10,500  -->  00:27:12,666
we not do advantage okay

577

00:27:12,800  -->  00:27:16,266
and uh so if this is the case

578

00:27:16,266  -->  00:27:18,666
if this is the case you see here

579

00:27:18,666  -->  00:27:20,133
so how you're playing around that

580

00:27:20,133  -->  00:27:23,666
your rule as it is equal to appslon right

581

00:27:23,666  -->  00:27:26,033
appslon e dot a n okay

582

00:27:26,500  -->  00:27:30,500
and where's your E it is equal to this one okay

583

00:27:30,500  -->  00:27:31,766
so that's why your

584

00:27:31,866  -->  00:27:34,700
this race is coming out to be minus ABS 1

585

00:27:34,700  -->  00:27:37,266
b not d for this bottom plate

586

00:27:37,766  -->  00:27:40,833
and if we are solving this equation

587

00:27:40,866  -->  00:27:42,400
this equation okay

588

00:27:42,900  -->  00:27:44,700
this equation so what

589

00:27:44,700  -->  00:27:47,066
what is using a simplification

590

00:27:47,066  -->  00:27:53,100
that this integral s is equal to your simply s okay

591

00:27:53,100  -->  00:27:55,433
so that is another simplification over here

592

00:27:56,466  -->  00:28:01,000
and so you are just just playing with this equation

593

00:28:01,000  -->  00:28:03,166
so this equation is no reduced to

594

00:28:03,166  -->  00:28:08,666
your Q is equal to raw s and s

595

00:28:08,733  -->  00:28:10,433
that is the thing okay

596

00:28:10,966  -->  00:28:14,800
where your raw ass it is coming out from here

597

00:28:14,966  -->  00:28:17,700
and this raw ass is coming out from here

598

00:28:17,966  -->  00:28:20,366
and this E is coming out from here

599

00:28:20,366  -->  00:28:22,833
so everything is interrelated

600

00:28:23,766  -->  00:28:26,333
so this is the charge on 1

601

00:28:26,333  -->  00:28:28,700
1 of the plate and if you find one

602

00:28:28,700  -->  00:28:30,800
need to find out the capacitys

603

00:28:30,800  -->  00:28:32,233
between these two players

604

00:28:32,333  -->  00:28:35,866
just take the magnitude of this charge

605

00:28:35,900  -->  00:28:37,666
divided with the constant

606

00:28:37,666  -->  00:28:39,166
potential that we have resumed

607

00:28:39,166  -->  00:28:42,466
so this is the same thing that we have derived

608

00:28:42,466  -->  00:28:44,200
using our first method

609

00:28:44,200  -->  00:28:47,133
where we assume the charge constant

610

00:28:47,133  -->  00:28:49,233
and found out the potential

611

00:28:51,700  -->  00:28:55,133
let's solve another example that is 6.12

612

00:28:55,133  -->  00:28:55,866
here

613

00:28:55,866  -->  00:29:00,100
just need to determine the capacity of the capacitors

614

00:29:00,200  -->  00:29:03,200
and here there are two mediums okay

615

00:29:03,400  -->  00:29:05,266
X1 R 1 and X1

616

00:29:05,266  -->  00:29:08,466
2 and the separation between them

617

00:29:08,466  -->  00:29:10,333
these two plates the capacitors

618

00:29:10,333  -->  00:29:11,133
the plates

619

00:29:12,000  -->  00:29:14,300
the only thing is there are two conductors okay

620

00:29:14,400  -->  00:29:15,500
and there are two

621

00:29:15,533  -->  00:29:17,966
two mediums to instead of single medium

622

00:29:17,966  -->  00:29:19,466
there are two mediums between them

623

00:29:19,466  -->  00:29:21,000
okay that is the difference

624

00:29:21,900  -->  00:29:23,866
and these two mediums are

625

00:29:24,200  -->  00:29:26,066
they can be changing in this

626

00:29:26,066  -->  00:29:28,800
in this manner or they can be changing in this manner

627

00:29:28,800  -->  00:29:29,600
okay

628

00:29:30,066  -->  00:29:31,100
next next

629

00:29:32,066  -->  00:29:34,300
solve this two cases one by one

630

00:29:34,300  -->  00:29:36,700
that what is the impact of this configation

631

00:29:36,700  -->  00:29:39,833
on the capacitors of this capacitor

632

00:29:40,333  -->  00:29:42,566
so in the first case your

633

00:29:42,700  -->  00:29:44,100
this electric field

634

00:29:44,100  -->  00:29:47,600
they are normal to the dialectic interfaces okay

635

00:29:47,600  -->  00:29:49,200
normal you can see here

636

00:29:49,300  -->  00:29:51,166
it is originating from here

637

00:29:51,166  -->  00:29:52,866
and it is terminating from here

638

00:29:52,866  -->  00:29:54,233
it is normal okay

639

00:29:54,666  -->  00:29:56,866
it is normal to the dialectic interface

640

00:29:56,866  -->  00:29:58,633
what is dialecting interface

641

00:29:59,100  -->  00:30:01,100
that is your dialectic interface

642

00:30:01,100  -->  00:30:03,033
okay that is this one

643

00:30:04,466  -->  00:30:07,400
or on the other hand on the other hand

644

00:30:07,400  -->  00:30:11,133
you can assume that the charge is constant

645

00:30:11,133  -->  00:30:14,500
across this capacitor okay

646

00:30:15,333  -->  00:30:18,600
and the capacitor can be treated

647

00:30:18,600  -->  00:30:19,400
so in this case

648

00:30:19,400  -->  00:30:22,866
the capacitor can be treated like two capacitors

649

00:30:23,133  -->  00:30:25,000
in theory okay

650

00:30:25,100  -->  00:30:26,700
because this charges

651

00:30:27,466  -->  00:30:30,266
the charge is constant

652

00:30:30,333  -->  00:30:33,500
the charge is constant across these two capacitors

653

00:30:34,600  -->  00:30:36,400
whereas in this second case

654

00:30:36,400  -->  00:30:39,300
your voltage is constant across these two

655

00:30:40,933  -->  00:30:44,533
across these two mediums or the inner

656

00:30:44,533  -->  00:30:45,633
where the capacitors

657

00:30:47,000  -->  00:30:49,300
okay so the capacity

658

00:30:49,300  -->  00:30:52,166
and Figure 1 can be treated as two capacitors

659

00:30:52,166  -->  00:30:54,633
C1 and C2 this is your C1

660

00:30:55,166  -->  00:30:58,700
this is your C22 capacitances

661

00:30:58,733  -->  00:31:00,666
and they are treated to be in series

662

00:31:00,666  -->  00:31:02,733
because the charge is constant

663

00:31:02,733  -->  00:31:04,633
remember q is constant

664

00:31:07,866  -->  00:31:10,300
in this configuration and also

665

00:31:10,300  -->  00:31:13,200
the electric feed is normal to the dialectic interface

666

00:31:13,733  -->  00:31:18,000
so if this is normal than for C1C1

667

00:31:18,000  -->  00:31:20,833
you see here the surface area

668

00:31:20,933  -->  00:31:24,300
the surface area seem s okay

669

00:31:26,166  -->  00:31:29,033
however the depth the distance between these two

670

00:31:31,300  -->  00:31:32,466
the plate and the interface

671

00:31:32,466  -->  00:31:36,300
it is d by 2 and likewise 4C2

672

00:31:38,100  -->  00:31:40,666
now you just need to involve the formulation

673

00:31:40,666  -->  00:31:41,266
uh

674

00:31:41,266  -->  00:31:45,133
for the Cds capacitors to find out the total capacitor

675

00:31:45,133  -->  00:31:46,200
and solve this thing

676

00:31:46,200  -->  00:31:48,666
and we'll wait our deep particular given values

677

00:31:49,500  -->  00:31:51,300
and if the electrical

678

00:31:51,300  -->  00:31:53,900
they are parallel to the dialecting interest of

679

00:31:53,900  -->  00:31:54,933
what did that is your

680

00:31:54,933  -->  00:31:58,033
this case was to play it not to play it

681

00:31:58,333  -->  00:31:59,333
and in this case

682

00:31:59,333  -->  00:32:02,600
the voltage is same across these capacitors

683

00:32:02,766  -->  00:32:06,300
so if the voted is same across these two capacitors

684

00:32:06,600  -->  00:32:09,933
then they will be treated as battle capacitors okay

685

00:32:09,933  -->  00:32:12,666
C1C2

686

00:32:14,533  -->  00:32:17,133
and in this case the only difference is

687

00:32:17,133  -->  00:32:18,400
another difference is

688

00:32:18,600  -->  00:32:21,200
the distance is d between these two plates

689

00:32:21,266  -->  00:32:23,866
but the surface area of these two plates

690

00:32:23,866  -->  00:32:26,900
it is passed by to

691

00:32:46,666  -->  00:32:47,866
really last example

692

00:32:48,533  -->  00:32:51,666
uh a slink like a pastor has a red eye of uh

693

00:32:51,666  -->  00:32:55,333
inner red eye of uh 1 cm or 2.5 cm

694

00:32:55,333  -->  00:32:58,200
if the space between plate is filled with a homogeneous

695

00:32:58,400  -->  00:33:00,366
in homogeneous remember this is no

696

00:33:00,366  -->  00:33:03,100
this is different case in homogeneous dialect media

697

00:33:03,333  -->  00:33:06,566
so it it is going to exhibit the uh

698

00:33:06,566  -->  00:33:09,000
variations in its relative permittivity

699

00:33:09,000  -->  00:33:11,333
with respect to this variable

700

00:33:11,333  -->  00:33:14,233
that is your role the radial radial direction

701

00:33:15,600  -->  00:33:18,366
now we need to find out the capacitys per metre land

702

00:33:18,366  -->  00:33:21,466
in this for this capacity again

703

00:33:21,466  -->  00:33:25,666
Jews this guardians involved the squads law

704

00:33:25,700  -->  00:33:26,366
so we

705

00:33:26,366  -->  00:33:30,200
we already studied that the electric field for these

706

00:33:30,266  -->  00:33:32,400
these types of the electric field

707

00:33:33,166  -->  00:33:35,366
electrically in between these two

708

00:33:36,566  -->  00:33:40,366
concentric shells for this aesthetic slendical pasture

709

00:33:40,366  -->  00:33:43,600
it is varying in this manner okay

710

00:33:43,900  -->  00:33:44,900
and if we know this

711

00:33:44,900  -->  00:33:47,300
we can find out the voltage using this

712

00:33:47,533  -->  00:33:50,666
e dot g relationship okay

713

00:33:51,866  -->  00:33:56,066
and here you just need to take care of this thing

714

00:33:56,066  -->  00:33:58,800
that absline is not absline

715

00:33:58,800  -->  00:34:00,066
not in this case

716

00:34:00,766  -->  00:34:04,733
rather it is actually not time digital permittuity

717

00:34:04,733  -->  00:34:06,400
and where the relative permittuity

718

00:34:06,400  -->  00:34:07,700
in this particular case

719

00:34:07,700  -->  00:34:10,066
it is our function of very well

720

00:34:10,066  -->  00:34:12,000
that is your role okay

721

00:34:12,266  -->  00:34:15,333
so here is going to cancel out this thing

722

00:34:15,333  -->  00:34:17,400
this thing and your left out with this thing

723

00:34:17,800  -->  00:34:20,166
and the integration of this thing

724

00:34:20,400  -->  00:34:21,733
integration of this thing

725

00:34:21,733  -->  00:34:23,800
it is equal to Eleanor

726

00:34:24,466  -->  00:34:30,633
this 10 + 2 because you remember that one of a row

727

00:34:30,766  -->  00:34:33,966
it is equal to Eleanor Row

728

00:34:36,766  -->  00:34:38,866
because it's derivative is

729

00:34:38,866  -->  00:34:40,800
it is available in the numerator

730

00:34:40,800  -->  00:34:43,000
and again it's derivative which is one

731

00:34:43,000  -->  00:34:44,833
it is available in the numerator

732

00:34:48,133  -->  00:34:51,200
now just put put put put my put in this uh

733

00:34:51,400  -->  00:34:55,366
uh radius to find out this uh voltage

734

00:34:55,366  -->  00:34:57,300
and then we'll wait it at the given

735

00:34:57,900  -->  00:35:00,566
at the given values of the radius

736

00:35:00,566  -->  00:35:03,300
and since we need to find out the capacity per metre

737

00:35:03,300  -->  00:35:06,200
so just put this allergy for two 1 bit

738

00:35:07,766  -->  00:35:11,500
so this is this is the assignment for those from the

739

00:35:11,500  -->  00:35:12,933
this chapter No. 6

740

00:35:12,933  -->  00:35:16,466
so all these three and these practice

741

00:35:16,466  -->  00:35:18,966
practice exercises of your chapter No. 6

742

00:35:19,500  -->  00:35:21,800
for this second edition book

743

00:35:23,333  -->  00:35:25,166
so all campus students can be conserved

744

00:35:25,166  -->  00:35:27,033
with your own campus students

745

00:35:27,066  -->  00:35:30,300
for the exact problems that you need to solve

746

00:35:31,800  -->  00:35:33,100
so I'm here today's lectures

747

00:35:33,100  -->  00:35:37,100
so today we just build on our previous uh lecture

748

00:35:37,200  -->  00:35:40,333
uh where we determine the resistance and capacitys

749

00:35:40,333  -->  00:35:45,133
for standard uh standard types of the capacitys

750

00:35:45,133  -->  00:35:48,166
and we solved here uh some examples

751

00:35:48,166  -->  00:35:50,800
some case the case examples to uh

752

00:35:50,800  -->  00:35:53,666
determine the capacitys and the resistances

753

00:35:53,666  -->  00:35:56,300
using either the causes law or the

754

00:35:56,566  -->  00:35:59,000
by solving these labrasians equations

755

00:35:59,000  -->  00:35:59,600
and then

756

00:35:59,600  -->  00:36:02,966
based upon these different matters that there are

757

00:36:02,966  -->  00:36:05,500
if we if we keep the voltage cost

758

00:36:05,500  -->  00:36:07,966
and then we need to find out the charge

759

00:36:07,966  -->  00:36:10,066
in terms of the water is and if

760

00:36:10,100  -->  00:36:12,266
and vice versa now we if we need

761

00:36:12,266  -->  00:36:16,600
if we keep the uh this uh current cost

762

00:36:16,600  -->  00:36:18,600
and then we need to find out the voltage

763

00:36:18,600  -->  00:36:19,700
in terms of year

764

00:36:20,966  -->  00:36:23,366
current to find out this uh

765

00:36:24,133  -->  00:36:25,833
resistance for the conductors

766

00:36:27,533  -->  00:36:32,166
in the next class we're going to study another method

767

00:36:32,400  -->  00:36:34,133
to find out the electric potential

768

00:36:34,133  -->  00:36:35,933
the electric field for the charges

769

00:36:35,933  -->  00:36:38,566
and that is known as the method of images

770

00:36:42,866  -->  00:36:44,100
here I thank you all

771

00:36:44,100  -->  00:36:45,700
if you have any questions

772

00:36:45,700  -->  00:36:47,166
uh they will be entertained

773

00:36:47,166  -->  00:36:49,766
uh through your emails or online

774

00:36:49,766  -->  00:36:52,266
synchronous sessions that we have arranged

775

00:36:52,266  -->  00:36:53,466
at the department