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this is election No. 14 of a V232 electromagnetic fifty

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today we'll study electric flux lines

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equi potential services

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and energy density in electro static feed

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this will cover your section 4

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point 10 and 4 point 11 of your textbook

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the learning objectives of today

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uh

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today's lecture will be to understand what is the uh

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this electric fox lines and the equiferential services

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and to determine the energy density in the presence of

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and electrostatic field

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first of all electric flux lines

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the idea of electric flux lines

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was first introduced by Michael Ferrari

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in his experimental investigation

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as a way of visualizing the electric field

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so as you can see

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it is some hypothetical or imaginary part

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so an electro electric uh

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flux line is an imaginary part

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or hypothetical part of line or surface

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and is a direction at any point

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indicates the direction of electric feed

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at that particular point so basically it is uh

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it is an imagery part that is uh utilized uh

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to uh represent uh the direction of electric field

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at some particular point

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and what is equal potential services

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so any surface on which deep potentially seem true

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so the potential is same

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short on this equal potential surface

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or this equal potential line

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uh in case if it is a equal potential uh line

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then the potential will be same on that line

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and we know that

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the intersection of equal potential surface

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with a plane reserves

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in a pot

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so that is a line

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so that is a one dimensional pot that is a 9

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so if your equal potential surface

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it is being cut by that one plane

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then we know that the intersection of two planes

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or two surfaces is a line

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and in this case

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that line will be known as equal potential line

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also we know that uh

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no work is done

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in moving a charge from one point to another

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along equal potential line or surface

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why because two of the surface is potentially same

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this v a is equal to v b

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okay and that is equal to V

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so there no work will be done in this case if we move

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uh from one point to another point

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on this equiprentious surface

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or a line so

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this implies that electric plus nines

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or the direction of E

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are always normal to electric surface

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or e equiponential surface

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all nines so what

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how can we say that it is normal to this uh

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these flat lines are normal

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to the equiperential surface

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9 surfaces

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because this start product is going to be the 0

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and this start product is only equal to 0

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if equal is equal to

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90

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right

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I repeat that electric Fox lines or in others detail

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the direction of electric fuel

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are always normal to equal potential services or

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or lines because this difference

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which is equal to the line integral of e dot TL

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is equal to 0 so that's why

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and this dot product is only equal to 0

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if and only if

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the angular displacement between this equal

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potential line and the electric feed is equal to Z

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is equal to uh zero

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and that is only once

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their angular displacement is equal to ninety

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okay in the following uh diagram

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uh we can see that these green lines

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the dash lines

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they are going to represent the ECU potential lines

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and the story lines

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they represent the electric free line

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so for a positively charged

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for a positively charged point charge

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we know that the electric free lines

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they are moving away from this source

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and if we see that these green lines

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they will represent these circular lines

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in this case it is a circular line

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so this circle will represent r

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will represent n equal potential line

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or equal potential circle in this case

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and as you can assume like that

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the closer to the point source

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the for example

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the potential is 30 volt than away from that

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so the potential is decreasing

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so it is 20 volt and again away it is 10 volt

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so it is again

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as you can see here that is equal to minus

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that'll be so again the potential is increasing

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in the opposite direction of electric field

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so if we are having the two parallel plate conductors

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that is your uh uh capacitor

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so two parallel plate uh conductors of uh

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oppositely charged uh uh polarity

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we know that again the electric field uh

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start from positive uh terminate on the inactive plate

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and closer to this policy

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the potential is later and then it is going to reduce

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and uh okay this and this is this is going to be it

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this is going to be

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it's equal potential line in this case

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now let's consider the case of two point charges

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so the first case is once they are oppositely charged

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for opposite polarity charges

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so they are going to attract each other

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so this is your electrify lines

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and you can assume that these greens

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they are going to represent the eco potential line

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uh lines in this case

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so the potential is going to be steam

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the resultant potential

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due to these two point charges is going to stay same

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on these uh

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hypothetical uh dotted lines green lines

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and at the point charges they are of same quality

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then they are going to repel each other

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and in this case you can see that

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this is going to be the equal potential

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the result and equal potential lines

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due to this oppositely charge charge quality charges

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and uh let's see

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what are the electric plus lines

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and differential services

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for the dipole

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so this is the same case that we discussed for the uh

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fine charge in the previous time

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so for the dipole

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this is uh for example it is it's positive chart right

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and this is negative chart

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then we know that it will start from positive charge

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and terminate on the negative charge

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like this way right

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and so these are the flux lines or electrical lines

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and what are the equal potential

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the resulted equal potential lines

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due to this die call will be represented like this

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dashed circle like this one

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know that the lines of force very very important

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take away from these three force lines

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that lines of force

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electrical lines or electric front lines

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or the direction of E field

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is always normal to equal potential surface

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you can see here that it is normal

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it is normal

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it is normal to the secret potential surface hotline

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well again it is normal at the start

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it also it is going to normal

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it is also going to normal so we're gonna do that

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we can take away from this slide

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that electric free lines or electric press lines

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they're always normal to the equi potential services

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uh some applications of this uh

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ecubential services are we know that uh

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if we place uh if we place these conducting bodies

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or conducting spheres in the electric field

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then uh the surface of this conducting body

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it is going to represent this equal potential volume

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because for this conducting bodies

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due to its smallest skin bag

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the charges are going to stay on the surface of these

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uh radical type of conducting bodies

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due to the metallic property

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or the higher conductivity property

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or the least is or the minimum skin effect of the uh

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conductors

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these charges will stay at the surface of conductors

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and charge will flow on this surface of conductor

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so we can treat this conductor as a

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as an ECU potential volume

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in the presence of electric clip

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the second example is very famous and practical example

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that is your electro cardio diagram

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electrocardio diagram

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that is your ECG and you're a medical

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and it is bi

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medical application of this ecubrancial services

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so a typical application of field mapping

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or electric flex lines or ecubrancias surgery

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found in the diagnosis of human heart

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our human heart beads in response

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in response to an electric speed potential

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that we apply using the electrodes

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uh Pisces electrodes across the uh heart

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uh on the uh uh once we plant

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plant those electrodes on the chest across that

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across the boundaries of that heart and the then heart

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then heart therefore the heart

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so basically the heart will beat

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in response to the electric fuel potential

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which is being applied across this heart

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and then what we are going to monitor

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uh the heart can be characterized at the typo

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as a typo with the field map

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similar to that on last line

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uh such a field map is useful

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in detecting abnormal heart position

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for example electro cardiogram

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so the uh the uh we can

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if we see some of the abnormalities in this uh

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uh Bible pattern uh

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which is generated by this uh

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heart beating against the standard uh

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uh

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type of pattern that we started on our earliest slides

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then

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there will be some abnormalities in the heartbeat of

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and the heart so first so what we do

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we apply the electric feed potential

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using the electrodes across this uh

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heart and then heartbeats

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heartbeats and generate electric feed lines right

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generate electric feed lines that we study

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that we basically read using the sensors out

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using the electrodes and then we plot that on the Ecgs

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now or in this case

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it is going to give you a pattern of like a typo

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national study how to determine this energy density

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in the presence of electrostatic feeds

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so first of all

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we consider our region free of elect electric field

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so there's no point chart presence in this region

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for example this region

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there's no point chart present in this region

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and now we are going to move three point charges Q1

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Q2 Q3 from infinity in this charge free region

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to determine the energy present

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in the assembly of charges

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we have to determine the amount of work necessary

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to assemble them so we basically

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we are going to calculate the amount of work

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to determine this energy

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required to assemble these three charges

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first of all

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no work is required to transfer Q1 from infinity to P1

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why

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because we are not working against any electric field

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which is present in this sorcery or charge free region

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and uh

261

00:12:34,400  -->  00:12:37,400
so because the space is initially charged sorcery

262

00:12:37,400  -->  00:12:38,566
so there's no electricity

263

00:12:38,566  -->  00:12:42,300
we are not we are not going to perform any of work

264

00:12:44,166  -->  00:12:47,900
and in the second case of work is being done

265

00:12:47,900  -->  00:12:53,033
once we transfer Q2 from infinity to point P2

266

00:12:53,266  -->  00:12:54,700
and in this case

267

00:12:54,700  -->  00:12:57,633
we are going to work against the electric field

268

00:12:57,766  -->  00:13:03,666
which has been created due to the presence of this Q1

269

00:13:03,866  -->  00:13:04,100
so

270

00:13:04,100  -->  00:13:07,800
we are working against electric field due to this Q1

271

00:13:08,000  -->  00:13:11,900
and we know that uh uh we know that that

272

00:13:11,900  -->  00:13:14,266
that work is equal to q b

273

00:13:14,500  -->  00:13:16,766
that we have already drive earlier

274

00:13:16,966  -->  00:13:19,066
so that work is equal to Q2

275

00:13:19,066  -->  00:13:21,366
since we are moving this uh Q2

276

00:13:21,366  -->  00:13:23,066
and then potential uh

277

00:13:23,500  -->  00:13:26,866
uh difference between this point P 2 and uh uh

278

00:13:26,866  -->  00:13:30,833
this P1 is going to be v 2 uh v 2 1

279

00:13:31,566  -->  00:13:32,566
in the third case

280

00:13:32,566  -->  00:13:35,966
uh now we are uh moving the uh point P3

281

00:13:37,333  -->  00:13:42,766
point er point chart Q2 Q3 er at point P3

282

00:13:43,766  -->  00:13:48,000
and we are placing this point charge to 3

283

00:13:48,000  -->  00:13:53,433
from infinity from infinity to point P3

284

00:13:53,766  -->  00:13:55,733
in the presence of electric feed

285

00:13:55,733  -->  00:13:59,066
which is now being created by the presence of Q

286

00:13:59,566  -->  00:14:05,100
2 and Q1 so we are working against two

287

00:14:05,100  -->  00:14:06,533
against an electric feed

288

00:14:06,533  -->  00:14:08,366
which is the resultant of electric feed

289

00:14:08,366  -->  00:14:10,366
due to these two point charges

290

00:14:10,366  -->  00:14:12,900
now and now the potential difference will be

291

00:14:13,100  -->  00:14:19,100
the potential difference between this point P3 and P2

292

00:14:19,566  -->  00:14:20,533
and the potential difference

293

00:14:20,533  -->  00:14:23,600
between this point B3 and B1

294

00:14:24,866  -->  00:14:27,633
now the total work done

295

00:14:27,766  -->  00:14:30,900
while positioning the three charges can be summed up

296

00:14:30,900  -->  00:14:32,666
like this one so far

297

00:14:32,666  -->  00:14:35,666
uh first Q1 charge it is equal to zero

298

00:14:35,666  -->  00:14:39,066
for the Q2 charge it is q 2 times v 2 1

299

00:14:39,066  -->  00:14:43,500
for Q3 charge q q q uh w 3

300

00:14:43,500  -->  00:14:44,300
that is q

301

00:14:44,800  -->  00:14:51,600
q 3 charge it is uh q 3 times uh v 3 1 plus v 3 2

302

00:14:51,933  -->  00:14:53,700
and these are the definitions of the

303

00:14:53,733  -->  00:14:57,766
which 1 is the potential at point 2 due to Q1

304

00:14:57,766  -->  00:15:00,500
potential at point 3 due to point a Q

305

00:15:00,566  -->  00:15:01,900
point charge Q1

306

00:15:01,933  -->  00:15:06,500
and potential at point 3 due to point charge Q2

307

00:15:07,933  -->  00:15:10,966
now if we place these charges in a reverse order

308

00:15:10,966  -->  00:15:13,766
that we if we place this charge Q3 fault

309

00:15:14,366  -->  00:15:15,266
then again

310

00:15:15,333  -->  00:15:18,200
since no charge is present in that free charge

311

00:15:18,200  -->  00:15:20,566
free zone so the well done will be zero

312

00:15:20,733  -->  00:15:22,200
and now we move this point

313

00:15:22,200  -->  00:15:22,700
charge Q

314

00:15:22,700  -->  00:15:25,766
in the presence of point charge 3 in that region

315

00:15:25,766  -->  00:15:28,000
so this will be equal to Q2

316

00:15:28,000  -->  00:15:30,033
Q2 times 3 to 3

317

00:15:30,100  -->  00:15:34,666
and then likewise 40 Q1 in the presence of charge q 3

318

00:15:34,666  -->  00:15:40,000
n Q2 so it could be q 1 times q 1 v 1 2+ v 1 3

319

00:15:41,266  -->  00:15:43,666
uh now we add up these two equations

320

00:15:44,166  -->  00:15:47,633
and we take these Q ones common

321

00:15:48,666  -->  00:15:52,200
then we can simplify this equation like this

322

00:15:53,500  -->  00:15:57,766
that's v 1 what is v 1 so v 1 is the potential

323

00:15:58,466  -->  00:16:01,433
potential difference or the potential at

324

00:16:01,666  -->  00:16:03,866
at this uh point uh

325

00:16:03,866  -->  00:16:07,766
B1 due to this point charge Q2 1

326

00:16:07,766  -->  00:16:14,733
Q3 and so Jordan potential at point B at point B1

327

00:16:14,733  -->  00:16:17,966
P2 and P 3 r V1 V2 and V3

328

00:16:17,966  -->  00:16:21,333
so V1 is the total potential at point V1

329

00:16:21,333  -->  00:16:24,633
due to point charges Q2 and Q3

330

00:16:24,733  -->  00:16:25,366
and likewise

331

00:16:25,366  -->  00:16:28,900
this V2 is the total potential at point P2

332

00:16:28,900  -->  00:16:32,000
where we are we are bringing this Q2

333

00:16:32,100  -->  00:16:34,700
uh and that is differentially to point uh

334

00:16:35,366  -->  00:16:36,900
point charge Q1 and Q3

335

00:16:36,900  -->  00:16:40,200
and likewise that point Q3 may be a point P3

336

00:16:40,200  -->  00:16:40,700
the protector

337

00:16:40,700  -->  00:16:44,666
prevention is due to the point charge at Q1 and Q2

338

00:16:44,700  -->  00:16:48,000
and then we can set bring this to on this side

339

00:16:48,000  -->  00:16:52,300
so this is the total well done total well done

340

00:16:52,300  -->  00:16:54,766
once we are moving three charges

341

00:16:55,866  -->  00:16:59,633
three charges to a free to a source predision

342

00:16:59,800  -->  00:17:02,866
and then we can simplify this equation

343

00:17:02,866  -->  00:17:04,600
in a smashing promise

344

00:17:06,300  -->  00:17:10,033
in case we are having the continuous chart distribution

345

00:17:10,200  -->  00:17:10,566
then

346

00:17:10,566  -->  00:17:15,466
we simply replace this summation with this integral

347

00:17:15,900  -->  00:17:18,766
and this cue by this

348

00:17:20,166  -->  00:17:23,466
uh this densities so in this first case

349

00:17:23,466  -->  00:17:26,600
so it is uh line chart density

350

00:17:26,600  -->  00:17:29,166
in the second case it is surface chart density

351

00:17:29,333  -->  00:17:32,100
and in the third case it is the walling chart density

352

00:17:33,200  -->  00:17:36,466
and this mission is being replaced by this line

353

00:17:36,466  -->  00:17:38,633
in trigger surface in trigger line

354

00:17:39,100  -->  00:17:42,633
this volume in trigger to find out the total charge

355

00:17:42,700  -->  00:17:46,500
to find out the total charge present in that volume

356

00:17:46,500  -->  00:17:48,100
surface or line

357

00:17:51,533  -->  00:17:52,533
no uh

358

00:17:52,533  -->  00:17:57,800
we uh further uh take the last uh this volume charger

359

00:17:58,666  -->  00:18:02,266
uh continuous uh uh country chart dance to example

360

00:18:02,333  -->  00:18:05,600
uh that total uh uh energy dance report this case

361

00:18:05,600  -->  00:18:07,200
the total work done in this case

362

00:18:07,500  -->  00:18:10,466
and we uh we take the help of this uh

363

00:18:11,500  -->  00:18:14,700
first equation of Maxwells that is Robizi

364

00:18:14,700  -->  00:18:18,266
put that dot e we just plug in this thing and we flip

365

00:18:18,266  -->  00:18:20,200
apply some of the vector identities

366

00:18:20,200  -->  00:18:22,933
the divergence theorem and some simplifications

367

00:18:22,933  -->  00:18:24,766
that you can consult in the book

368

00:18:24,766  -->  00:18:26,900
then it can simplify

369

00:18:27,500  -->  00:18:32,833
it can be simplified to this uh uh dropper.com

370

00:18:33,000  -->  00:18:36,800
so the energy density are total work done for this

371

00:18:36,800  -->  00:18:38,300
walling chart density

372

00:18:38,866  -->  00:18:43,700
so is the volume check it out d dot e

373

00:18:43,966  -->  00:18:44,700
but in other words

374

00:18:44,700  -->  00:18:47,233
we know that these equal to astronaut E

375

00:18:47,333  -->  00:18:49,533
and we can also simplify this dot term

376

00:18:49,533  -->  00:18:51,666
as the astronaut E square

377

00:18:52,000  -->  00:18:52,266
so

378

00:18:52,266  -->  00:18:55,033
this is the energy store that we work on in this case

379

00:18:56,333  -->  00:18:58,000
now we define another termology

380

00:18:58,000  -->  00:19:00,566
that is the electrostatic energy density

381

00:19:00,566  -->  00:19:04,366
so this is the well done or energy

382

00:19:04,500  -->  00:19:06,600
or energy per unit volume

383

00:19:06,733  -->  00:19:08,900
you can see here jewels uh

384

00:19:08,900  -->  00:19:13,100
your uh this uh jewels per uh cubic meter

385

00:19:13,366  -->  00:19:16,766
and also you can see it uh as a rate of change of uh

386

00:19:16,766  -->  00:19:17,766
this uh

387

00:19:18,933  -->  00:19:21,700
energy density with respect to volume

388

00:19:21,700  -->  00:19:22,766
differential volume

389

00:19:23,566  -->  00:19:24,533
so what we need to do

390

00:19:24,533  -->  00:19:28,166
we only we cut out this in children in this case

391

00:19:28,266  -->  00:19:30,633
simply cut out this in children in this case

392

00:19:30,666  -->  00:19:33,366
because differential is going to cut this

393

00:19:33,366  -->  00:19:35,633
or eliminate this in children

394

00:19:35,900  -->  00:19:39,866
and this is going to stay like half of d dot e

395

00:19:39,866  -->  00:19:43,866
or half of astronaut E squared or is equal to d squared

396

00:19:45,600  -->  00:19:48,533
divided to ABS cannot but in other words

397

00:19:48,533  -->  00:19:51,500
we can see that the energy store

398

00:19:51,500  -->  00:19:54,766
or energy density is also the volume

399

00:19:54,766  -->  00:19:57,700
in trigger of electrostatic energy text

400

00:19:59,366  -->  00:20:02,966
let's solve this example of your book that is 4.17

401

00:20:03,300  -->  00:20:04,600
two point charges

402

00:20:05,866  -->  00:20:07,166
okay the 3 point charge

403

00:20:07,166  -->  00:20:10,900
3 point charges are located at origin

404

00:20:12,466  -->  00:20:14,466
at this point and this point

405

00:20:15,133  -->  00:20:17,466
and we have to find out the energy in the system

406

00:20:18,866  -->  00:20:23,800
now we first ways do is to utilize this mission formula

407

00:20:24,566  -->  00:20:28,166
and what we can do here so we want

408

00:20:28,166  -->  00:20:32,166
we want when we know that it is equal to the potential

409

00:20:32,166  -->  00:20:34,200
due to charge Q2

410

00:20:34,200  -->  00:20:35,833
and Q3

411

00:20:36,533  -->  00:20:38,866
at this point P1

412

00:20:39,400  -->  00:20:44,366
so we take this s P1 we take this s P2

413

00:20:44,700  -->  00:20:47,566
we take this s P3 right

414

00:20:49,100  -->  00:20:52,500
and Q2 it is at the distance of 1

415

00:20:52,500  -->  00:20:56,500
we can see here this point is at the distance of

416

00:20:58,066  -->  00:21:00,433
0 0 1 right 0

417

00:21:01,800  -->  00:21:02,600
0

418

00:21:03,533  -->  00:21:04,333
1

419

00:21:05,533  -->  00:21:08,833
so p is P1 is actually at the origin

420

00:21:08,866  -->  00:21:13,500
and this P3 is at the distance of 1 0

421

00:21:13,566  -->  00:21:15,800
0 like this way

422

00:21:16,400  -->  00:21:17,500
so the distance between

423

00:21:17,500  -->  00:21:19,300
that is using the distance formula

424

00:21:19,300  -->  00:21:22,766
it is only simply the z axis that is the 1

425

00:21:23,166  -->  00:21:26,666
and for this Q3 it is also the quatu one right

426

00:21:26,666  -->  00:21:29,833
because only axis coming into play in this case

427

00:21:30,800  -->  00:21:32,066
therefore this V2

428

00:21:32,066  -->  00:21:36,933
V2 is the potential at this point V1 due to discharges

429

00:21:36,933  -->  00:21:39,066
Q1 a Q3

430

00:21:40,266  -->  00:21:41,366
no uh

431

00:21:41,366  -->  00:21:42,700
this is your yeah

432

00:21:42,700  -->  00:21:45,300
this is your elevation point right

433

00:21:45,400  -->  00:21:47,200
and from here to here again

434

00:21:47,200  -->  00:21:49,066
the distance is 1 however

435

00:21:49,066  -->  00:21:52,466
from here to here the distance can be found out

436

00:21:52,466  -->  00:21:55,266
using your Pytaborus theorem right

437

00:21:55,666  -->  00:21:57,100
uh so that is equal to

438

00:21:57,800  -->  00:21:59,700
that is very diagonal in this case

439

00:21:59,700  -->  00:22:01,066
right and we know that is

440

00:22:01,066  -->  00:22:02,200
it is it is the

441

00:22:02,333  -->  00:22:06,366
some of the scales of the two sides scheduled

442

00:22:06,966  -->  00:22:10,233
scheduled of the some of the scales of the two sides

443

00:22:10,966  -->  00:22:13,700
and that will be at route 2 square root 2

444

00:22:14,000  -->  00:22:15,666
and for the third case uh

445

00:22:15,666  -->  00:22:17,033
for this case again

446

00:22:17,366  -->  00:22:18,733
uh for this Q1

447

00:22:18,733  -->  00:22:21,900
it is distances 1 and for Q2

448

00:22:21,933  -->  00:22:24,300
the distance is again the diagonal that is Q2

449

00:22:24,300  -->  00:22:27,133
so it is at the potential at point P3

450

00:22:27,133  -->  00:22:30,366
due to this point chart Q1 and Q2

451

00:22:30,600  -->  00:22:32,866
then you can simply plug in these value 2 1

452

00:22:32,866  -->  00:22:33,366
2 2 1

453

00:22:33,366  -->  00:22:34,200
2 3

454

00:22:34,700  -->  00:22:37,300
take care of these polarities

455

00:22:37,933  -->  00:22:40,966
and here you can see the polarities as well

456

00:22:42,166  -->  00:22:47,566
and you can simplify that it is 13.37 nanojoles

457

00:22:49,333  -->  00:22:50,466
no alternatively

458

00:22:50,466  -->  00:22:51,666
what is the second way

459

00:22:51,866  -->  00:22:56,166
so the second way is instead of the submission formula

460

00:22:56,166  -->  00:22:58,833
you involve your basic uh

461

00:22:59,900  -->  00:23:03,100
definition of finding all this electrostatic energy

462

00:23:03,266  -->  00:23:06,166
so what was that you bringing

463

00:23:06,600  -->  00:23:12,100
bringing this charge Q1 from infinity till this point

464

00:23:12,533  -->  00:23:15,500
right so for this case no work will be done

465

00:23:16,100  -->  00:23:18,000
secondly you bring this charge

466

00:23:18,000  -->  00:23:20,466
Q2 in the presence of the electric field of

467

00:23:20,466  -->  00:23:24,500
due to this Q1 and then you bring this charge Q2

468

00:23:24,500  -->  00:23:27,600
Q3 in the presence of electric field of this Q1

469

00:23:27,600  -->  00:23:31,700
N Q2 so uh for the second case

470

00:23:31,700  -->  00:23:33,133
uh for first grade zero

471

00:23:33,133  -->  00:23:36,233
for the second case it is uh Q1

472

00:23:36,300  -->  00:23:38,933
uh Q2 okay so potential

473

00:23:38,933  -->  00:23:40,566
we know that it is Q1

474

00:23:40,700  -->  00:23:42,466
and what is this distance formula

475

00:23:42,466  -->  00:23:44,233
so the distance formula is d

476

00:23:44,500  -->  00:23:49,266
uh distance formula is D so here uh the source

477

00:23:49,266  -->  00:23:50,200
the source source

478

00:23:50,200  -->  00:23:52,100
okay source and destination

479

00:23:52,100  -->  00:23:56,533
so destination is your uh destination is your

480

00:23:56,533  -->  00:23:58,433
so your decision P1

481

00:23:59,300  -->  00:24:02,033
P2 and P3 right

482

00:24:03,533  -->  00:24:04,800
so this is your destination

483

00:24:04,800  -->  00:24:07,766
P2 is your destination and this is P1 is your source

484

00:24:07,766  -->  00:24:09,900
so this is your distance from order in this case

485

00:24:11,166  -->  00:24:14,366
and uh for the uh for this third case

486

00:24:15,000  -->  00:24:18,166
your uh distance formula for this one

487

00:24:18,266  -->  00:24:20,900
uh for the first case it is uh okay

488

00:24:20,900  -->  00:24:23,566
this is your destination in this case please

489

00:24:23,566  -->  00:24:23,933
please

490

00:24:23,933  -->  00:24:27,500
your destination and you have to come from P1 and P

491

00:24:27,500  -->  00:24:33,800
P1 to P3 it is simply uh 1 1

492

00:24:33,800  -->  00:24:35,200
0 - 1 0

493

00:24:35,200  -->  00:24:36,000
0 - 0 0

494

00:24:36,000  -->  00:24:39,966
0 and for this one it is destination

495

00:24:39,966  -->  00:24:41,200
we know that it is 1 0

496

00:24:41,200  -->  00:24:44,766
0 and the source we know that it is 0

497

00:24:44,766  -->  00:24:48,533
0 1 and then you can simplify these two uh

498

00:24:48,533  -->  00:24:51,766
things and put back these values and uh

499

00:24:51,766  -->  00:24:52,700
it will come out to you

500

00:24:52,700  -->  00:24:54,733
so take care of this more as well right

501

00:24:54,733  -->  00:24:58,200
this transformer uh this modulation modesty

502

00:24:58,700  -->  00:25:01,300
uh once you take the difference of these two position

503

00:25:01,300  -->  00:25:03,966
rectors then you can take it a scale root

504

00:25:03,966  -->  00:25:08,300
uh uh modern and the answer is same

505

00:25:09,933  -->  00:25:11,500
another example of your book

506

00:25:11,733  -->  00:25:12,700
so in this case

507

00:25:12,700  -->  00:25:16,233
a chart distribution which vertical symmetry is given

508

00:25:16,400  -->  00:25:19,566
so this vertical chart distribution given in this case

509

00:25:19,733  -->  00:25:22,966
and you're serving the volume chart

510

00:25:22,966  -->  00:25:25,800
dance to your pro v is equal to constant

511

00:25:25,800  -->  00:25:29,333
per or not once you are within this sphere

512

00:25:29,333  -->  00:25:32,633
that is your radial distance from 0 to r

513

00:25:32,766  -->  00:25:37,200
and out of this uh sperical entity

514

00:25:37,200  -->  00:25:39,466
your chart entity is equal to 0

515

00:25:40,066  -->  00:25:42,000
and we need to determine the energy store

516

00:25:42,000  -->  00:25:43,700
stored in this region

517

00:25:44,700  -->  00:25:46,000
we know that elected

518

00:25:46,000  -->  00:25:48,466
fever's radical chart distribution from uh

519

00:25:48,466  -->  00:25:49,966
Gaza's law uh

520

00:25:49,966  -->  00:25:53,166
in uh this 4 point d section

521

00:25:53,166  -->  00:25:57,666
that it is equal to runaut r divided t F snaraut a r

522

00:25:58,066  -->  00:25:59,666
I did magnitude of this one

523

00:25:59,666  -->  00:26:01,433
we know that it is a simply

524

00:26:01,766  -->  00:26:03,366
I take take away this uh

525

00:26:03,366  -->  00:26:05,500
junitractor and magnitude scared

526

00:26:05,500  -->  00:26:07,766
simply takers scared of this fan

527

00:26:08,533  -->  00:26:12,366
so we need to find out this uh energy density right

528

00:26:13,700  -->  00:26:17,333
that is equal to half up B volume trigger lock

529

00:26:17,333  -->  00:26:19,700
astronaut is clear right

530

00:26:20,466  -->  00:26:22,333
and what we need to do

531

00:26:22,333  -->  00:26:24,233
we need to plug in this value right

532

00:26:24,700  -->  00:26:28,400
now the only variable in this case is our scale

533

00:26:28,400  -->  00:26:31,100
because there's no cheater time

534

00:26:31,100  -->  00:26:32,166
there's no fight time

535

00:26:32,166  -->  00:26:34,433
with respect to select the partner system

536

00:26:34,666  -->  00:26:38,000
so these things are being that as a constant

537

00:26:39,266  -->  00:26:40,900
uh for this vertical corner system

538

00:26:40,900  -->  00:26:44,200
we know that DV is equal to r square sign Peter d

539

00:26:44,200  -->  00:26:52,000
r N d t 35 and uh we can uh simplify this integration

540

00:26:52,000  -->  00:26:55,866
these two integration that this is equal to 4

541

00:26:55,866  -->  00:26:58,400
5 you can solve this at your own time

542

00:26:59,333  -->  00:27:04,200
and since RSK is there so RS R it will become RSK

543

00:27:04,200  -->  00:27:07,500
RSK will come R4 its integration will be R5

544

00:27:07,500  -->  00:27:11,400
we have F5 and just plug in this range of 0 to R

545

00:27:11,400  -->  00:27:17,166
and this is your energy density in this complete

546

00:27:18,800  -->  00:27:22,000
sperical entity with charge distribution will be

547

00:27:23,500  -->  00:27:26,466
this is your assignment for week No. 7

548

00:27:26,466  -->  00:27:27,900
you have to follow these two weekly

549

00:27:27,900  -->  00:27:31,866
it is two question from chapter number four

550

00:27:32,300  -->  00:27:33,866
practice exercises

551

00:27:33,900  -->  00:27:36,866
and please follow this 2nd edition of the textbook

552

00:27:37,000  -->  00:27:38,266
and off campus student

553

00:27:38,266  -->  00:27:40,600
you can concern the on campus student

554

00:27:40,700  -->  00:27:42,433
for the exact questions

555

00:27:44,133  -->  00:27:46,533
so I'm here today's lectures to Posttrophone

556

00:27:46,533  -->  00:27:50,300
we understood what is uh this electric flex lines

557

00:27:50,300  -->  00:27:52,566
and what are the equipracial services

558

00:27:52,566  -->  00:27:54,400
and the equipracial lines

559

00:27:54,400  -->  00:27:57,100
and then we uh uh we uh

560

00:27:57,100  -->  00:28:00,633
we studied how to determine this electric density

561

00:28:01,333  -->  00:28:03,500
our electrostatic energy

562

00:28:03,700  -->  00:28:05,900
in the presence of electrostatic fees

563

00:28:05,900  -->  00:28:09,766
once we are bringing one or more than one charges into

564

00:28:09,766  -->  00:28:11,900
some charge fee is done

565

00:28:14,000  -->  00:28:15,066
in the next class

566

00:28:15,066  -->  00:28:18,033
we are going to start off with our chapter No. 5

567

00:28:18,100  -->  00:28:21,566
that is electric fees in material space

568

00:28:21,566  -->  00:28:24,400
instead of free space that we have been starting up

569

00:28:24,400  -->  00:28:25,800
you know absent not

570

00:28:25,800  -->  00:28:28,733
we will study some materials

571

00:28:28,733  -->  00:28:31,300
which are having properties other than the free space

572

00:28:31,300  -->  00:28:35,066
we will study in that properties of those materials

573

00:28:35,066  -->  00:28:36,533
in that particular lecture

574

00:28:36,533  -->  00:28:38,400
the next lecture and we will

575

00:28:38,400  -->  00:28:42,300
we will also understand what are these two types of

576

00:28:42,300  -->  00:28:44,733
the crimes that that are the convection

577

00:28:44,733  -->  00:28:46,433
current and the conduction

578

00:28:50,066  -->  00:28:52,466
here I thank you all if you have any questions

579

00:28:52,466  -->  00:28:56,200
they will be entertained to your emails or online

580

00:28:56,200  -->  00:28:56,700
synchron

581

00:28:56,700  -->  00:28:59,900
a session that will have arranged at the department