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this is election No. 10 of a V232

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electromagnetic field theory

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today we will solve examples from your previous trophy

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that is

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electric feed due to continuous charge distributions

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and electric first density

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this is again

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from your section 4.3 and 4.4 of your textbook

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the learning objectives of uh

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today's lecture are to solve

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the examples regarding degeneration of electric fields

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a due to continuous

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a chart distribution and find out the electric

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electricity due to these distributions

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earlier we studied point charge

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and electric vehicle intensity

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generated by this point charge

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at some elevation point using the coolant slot

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then we covered how to uh

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how to evaluate the electrified intensity

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at a given point due to a line charge distribution

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having symmetrical charge distribution on its surface

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and we uh we

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we studied that if there is a small

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differential element on this aligned charge density

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then the charge on the smaller differential

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length element would be equal to the charge density

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times e

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differential element length element

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and

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if you want to find out the total charge on this line

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charge density then

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we have to simply integrate that

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differential charge element

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on that

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complete complete line

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complete line chart dances and length like this one

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the second case the third case that we studied

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was regarding the surface chart dynasty

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that

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if the chart distribution is uniformly distributed

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on a surface

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then then

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then on on a differential surface element DS

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the charge would be raw as times d DS and again

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in order to find out the total charge on the surface

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of that surface job density

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we have to simply integrate that role as times DS

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or overly complete surface limits of that

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that surface job density

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the third one was the 3D volume charge density Roe v

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and for that the differential element

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charge element is equal to all times GV

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the differential volume element and again

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we if we want to find out the total charge

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present on the surface of this volume

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chart density then

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we have to simply integrate over the entire volume

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for that given volume charge density

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so that is your

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that will give you total charge on that

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residing on that volume charge density surface

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there on uh we uh

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we study that the electric intensity due to a point

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charge is given by couloms law

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that is queued about the call by ABS law

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not our sphere times

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and the unit rector and the unit distance rector

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and if you want to find out the elective intensity

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due to these different containers

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charge distribution

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we have to simply replace this Q by the total charge

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available on this line charge density

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or the total charge available on the circuit

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charge density

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or the total charge available on this wallet

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charge density

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and these are the three

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these are the three cases that we studied

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in our previous lectures

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and later on we studied that if you're aligned charge

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if your line charge is a finite finite land

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then this is this line charge equation

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it can be simplified to this equation

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and here we straight the ID alpha 1

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alpha 2 R

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d

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angular distributions

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are depended on the peniplear distance

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between the uh

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between the revelation point and the uh

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and this uh headline charge uh planning conductor

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and

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alpha 1 is from the starting point of that conductor

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and the alpha 2 is from the terminating point of that

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conductor with respect to that uh uh uh

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uh with respect to that uh perpendicular distance

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and in case our line chart distribution

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it is off in finite line

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then it will be extending in finitely in the positive

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z and negative z direction

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and again and in that case we

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we studied that the alpha 1 and alpha 2 would be plus

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minus pi by 2

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and electric create

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intensity would be simplified into this form

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then we discussed what happens if we

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we want to find out the electricity

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due to this surface identity

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then it is simplified to this rule

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as divided by two astronaut

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which is independent of the

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independent of the distance from the revolution point

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but that particular surface identity

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and for the volume charge dynasty

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we found out that it is equal to Q

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divide report by astronaut RSP AR

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where Q is d

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put a charge residing on that

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that volume charge dynasty

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then we studied what is electric plugs

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so it is a number of lines passing through a surface

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and if it is a junior surface

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then it is known as electric class density

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and with a studied relationship between the electric

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intensity and the electric prostency

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how we can find out electric prostency

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by simply multiplying that electric frequency

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with that permittivity characteristics of that medium

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for this free space is abstract

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which is the 8.85 inch tenders 4 minus twelve

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okay

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let's solve uh this uh first example of your uh book

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that is 4.5 so uh here uh uh finite sheet

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so it is uh truly problem

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and the final sheet is given which is having the

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so it is the final sheet

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which is having the dimensions of 0

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2 1 in x direction and 0 2

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1 in y direction and it is on Z plane

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so it is x y plane so that is Z plane so use the 0

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0 and it is having a surface chart density

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and here note that the surface chart density

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it is in terms of XY variables

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and what we need to find out

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we need to find out the total charge on the sheet

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that is Q we need to find out the electricity

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at this elevation point and then

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we have to find out the force being experienced by

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are given charge at this regulation point

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so

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these are the three things that we have to determine

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first of all the total charge residing on this warning

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charge density

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is formed by simply integrating that surface

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chart density over that complete surface

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so here it is very simple

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that you just need to put in that role

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as then since it is a cardigan partner system problem

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so your DS is equal to d x d y

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and further you can uh now

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you can simplify this integration

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by taking half of this substitution rule

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so you can uh replace the x square by zu

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you just take its differentiation

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your you know

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that extra x square differentiation is the 2 x

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and then along with it's a variable offer integration

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a differentiation and then from here

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you can read in to find out this X times d

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X is equal to the U2

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so if you use all these two these two uh replacements

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substitutions

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then you can simplify the equation into this like that

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you replace the X square by shoe

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you replace this X times d X with this U by 2

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like this one

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and now this this integration is very simple

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now you have to simply integrate with this

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back to this U and then you can

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since this thing is inside this

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so it's uh uh involved this uh pyro

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and it is the integration is equal to the

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whatever is the inside uh is part at 3 by 2 + 1

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that will come out to be 5 by 2 divided by N+ 1

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so again it is 5 by two and then the differentiation

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you know what it is again the inside integral

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the differentiation of inside integral is 1

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4 / 1 is the same thing

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so from here you put the uh

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the integration image that were 0 to 1

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for this again particular is you variable

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so for again you have to play around with the Inter

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with this integration limits as well

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so far 0 as for a value of a 0 for given x

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the g is again is equal to 0

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if the value of a parliament of x is 1

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then g is equal to 1

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because it is at x square function

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now you simply uh

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you simplify these uh uh these limits

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plug in these values and simplify this uh this thing

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then it will the first the upper limit will come out

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comes out to be 26 with all because it is 1

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and then plus 25 is 26

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and then the lower limit it is 0

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so it will comes out comes out to be as 0 + 25

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again it is while still plus 25

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now the things are left out in terms of variable why

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next uh against all this uh

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uh uh

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this uh integration with the spectrum y

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by taking help of this substitution again

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we replace this y square by variable v

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again we take it to differentiation

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again we find out this uh product y

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d y and what we need to do

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we need to replace this y divide with this d V about 2

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and this y square with this v

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no this is the integration that you have to perform

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and once you do the integration

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with respect to this thing

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then it will come

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come out to be this 5 by 2 will come out to be 7 by 2

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this will be 7 by

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2 divided by this is 7 by 2

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now you again put in the limits

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it will comes out to be 27 like this thing

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and the other thing it will it will come out to be 25

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again

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you can consult the book for the explanation as well

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and once you will simplify the things it will

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it will give you a total charge on this volume

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as a 32.15

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Newton coolants

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well let's uh

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find out the electric fuel intensity

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at the evolution point

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so here the evolution point is 0 0

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5 right and your source is a a fine at sheet

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so since the axis varying varying in this sheet

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in this sheet y is varying in this sheet

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and these constant that is equal to zero

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so involving the distance vector

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so this is your distance erector minus

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minus x minus y n 5

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and now you want to find out your magnitude

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so you can find out by this

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x square plus y square plus 25 square root no

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plug in this value in this given formulation

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you know that a r is equal to r vector

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divided by magnitude of r

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so it will be a r cube and due to the spare root

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it will be a 3 by two

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right so now you can simplify this thing

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that this thing will be cancelled

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this thing and what we are left with is

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this thing

259

00:11:34,733  -->  00:11:37,533
after the translation we are left with this thing

260

00:11:37,533  -->  00:11:39,800
now you can open up this product

261

00:11:39,800  -->  00:11:43,800
so the first product is x square right

262

00:11:43,800  -->  00:11:45,100
minus x square y

263

00:11:45,400  -->  00:11:49,133
then second product is minus y square x

264

00:11:49,133  -->  00:11:51,066
and the third product is x y

265

00:11:51,700  -->  00:11:53,133
now you simplify uh

266

00:11:53,133  -->  00:11:54,000
no use uh no

267

00:11:54,000  -->  00:11:56,766
solve this integration one by one using

268

00:11:56,766  -->  00:11:59,100
using this each term for the extra

269

00:11:59,100  -->  00:12:00,866
so since it can be seperable

270

00:12:00,866  -->  00:12:02,533
so you can solve this integration

271

00:12:02,533  -->  00:12:04,133
this integration separately

272

00:12:04,133  -->  00:12:06,266
and put in their respective limits

273

00:12:06,266  -->  00:12:08,600
so we know that the integration of extras

274

00:12:08,600  -->  00:12:10,500
with respect to a d x

275

00:12:10,500  -->  00:12:11,166
s x

276

00:12:11,166  -->  00:12:14,000
Q by 3 so put in this uh limit

277

00:12:14,000  -->  00:12:16,400
so it is 1 or 3 for this y

278

00:12:16,400  -->  00:12:19,333
it is y squared by 2 put in limits again

279

00:12:19,333  -->  00:12:22,966
uh for the second part it is uh x squared by 2 and y

280

00:12:22,966  -->  00:12:26,766
y q by 3 and put in limits simplified

281

00:12:26,766  -->  00:12:30,166
for this thing it is x squared by 2 and y squared by 2

282

00:12:30,166  -->  00:12:31,866
put in limits and simplify

283

00:12:31,866  -->  00:12:34,366
so this is the electric field at

284

00:12:35,700  -->  00:12:36,800
the given elevation

285

00:12:36,800  -->  00:12:41,166
point due to this finite surface identity

286

00:12:42,200  -->  00:12:44,966
now you want to find out the force being exacted

287

00:12:44,966  -->  00:12:46,700
at this evaluation point

288

00:12:46,700  -->  00:12:50,433
if some point charge of minus 1 million

289

00:12:50,700  -->  00:12:52,066
8 minus one

290

00:12:52,066  -->  00:12:55,533
mini column is placed at the same evolution point

291

00:12:55,533  -->  00:12:58,133
let simply involve this columns law

292

00:12:58,133  -->  00:13:00,666
and then it is equal to q and 2 e

293

00:13:00,666  -->  00:13:02,566
and then you can simplify this thing

294

00:13:05,933  -->  00:13:09,000
the second example of your book is uh

295

00:13:09,000  -->  00:13:10,866
that you have uh two planes

296

00:13:10,866  -->  00:13:13,866
that is X is equal to y and y is equal minus 3

297

00:13:14,066  -->  00:13:17,366
and intercaring uh the charge uh

298

00:13:17,366  -->  00:13:18,200
that's equal

299

00:13:18,200  -->  00:13:20,566
is equal to 10 nanoquents per meter square

300

00:13:20,566  -->  00:13:22,633
and 15 nanoquents per meter square

301

00:13:23,100  -->  00:13:25,800
and this 1/3 charge distribution

302

00:13:25,800  -->  00:13:27,700
that is a 9 charge distribution

303

00:13:27,866  -->  00:13:31,000
and here it is as you can assume that it is

304

00:13:31,300  -->  00:13:34,866
it is along y axis because it is it is an

305

00:13:34,866  -->  00:13:36,700
it is due to the intersection of two planes

306

00:13:36,700  -->  00:13:39,833
that is actually equal to 0 and is equal to two

307

00:13:40,100  -->  00:13:41,800
and it is carrying this uh

308

00:13:41,800  -->  00:13:44,700
10 pine and coolant square metre charge distribution

309

00:13:45,000  -->  00:13:45,966
and what we need to do

310

00:13:45,966  -->  00:13:48,100
we need to find out the lack of intensity

311

00:13:48,100  -->  00:13:49,400
at this given point

312

00:13:49,400  -->  00:13:52,300
little these three different charge distributions

313

00:13:54,266  -->  00:13:55,866
now you can assume like this

314

00:13:55,866  -->  00:13:59,400
that the electric field at this given point

315

00:13:59,400  -->  00:14:04,400
it is a solution of three electric fields due to this

316

00:14:05,766  -->  00:14:09,066
two two out of them are due to this plane

317

00:14:09,066  -->  00:14:10,133
these two planes

318

00:14:10,133  -->  00:14:13,066
and one is due to this crank and conductor

319

00:14:14,000  -->  00:14:17,400
so first of all the electric intensity

320

00:14:17,400  -->  00:14:21,366
and this evaluation point due to this sheet 1

321

00:14:21,366  -->  00:14:23,566
so it is actually equal to two

322

00:14:23,733  -->  00:14:26,866
so what you need to do is your access

323

00:14:26,933  -->  00:14:29,133
so it is your uh uh uh

324

00:14:29,133  -->  00:14:31,200
so you are currently focus on this uh

325

00:14:31,200  -->  00:14:34,200
pictorial diagram that your magical representation

326

00:14:34,200  -->  00:14:36,666
that it is your X y and Z and eh

327

00:14:36,666  -->  00:14:37,600
this first name

328

00:14:37,600  -->  00:14:41,666
it is a question distance of x is equal to 2

329

00:14:41,733  -->  00:14:43,800
so it is in finite sheet

330

00:14:43,866  -->  00:14:46,100
which is varying in the z direction

331

00:14:46,100  -->  00:14:50,966
and in the y direction and it's x coordinates constant

332

00:14:51,100  -->  00:14:53,000
so how to find out the normal

333

00:14:53,066  -->  00:14:54,733
towards the evaluation point

334

00:14:54,733  -->  00:14:56,266
so what is the valuation point

335

00:14:56,266  -->  00:15:00,800
it is 1 1MINUS1 so you can consider like this it is one

336

00:15:01,300  -->  00:15:08,933
see it is minus 1 right 1 - 1 and 1

337

00:15:08,933  -->  00:15:13,900
1MINUS1 so so it is you can take it like sorry

338

00:15:13,900  -->  00:15:14,666
if you can

339

00:15:14,666  -->  00:15:17,800
you can achieve it like this thing that it is

340

00:15:19,266  -->  00:15:22,033
uh like uh this is one right

341

00:15:22,066  -->  00:15:22,533
okay y

342

00:15:22,533  -->  00:15:24,900
X is one right like this one

343

00:15:24,900  -->  00:15:32,100
so this is your x y plane point and then it is minus 1

344

00:15:32,100  -->  00:15:34,500
so it is in the down direction

345

00:15:34,500  -->  00:15:38,066
so you can take it like it is the 1

346

00:15:38,300  -->  00:15:43,100
1 minus one regulation point and what you need to do

347

00:15:43,100  -->  00:15:45,266
you would need to find out the normal vector

348

00:15:45,266  -->  00:15:48,966
pointing from the this uh

349

00:15:48,966  -->  00:15:51,466
sufficient identity towards the evaluation point

350

00:15:51,466  -->  00:15:55,100
so since it's x x of the relation point is 1

351

00:15:55,100  -->  00:15:57,400
so it is it is towards the

352

00:15:57,866  -->  00:16:01,466
towards this negative E extraction

353

00:16:01,800  -->  00:16:04,100
the normals the normal from this

354

00:16:05,000  -->  00:16:07,366
explain towards the revolution point

355

00:16:07,366  -->  00:16:09,300
is towards the minus extraction

356

00:16:09,300  -->  00:16:11,166
so a n is equal to minus e x

357

00:16:11,500  -->  00:16:14,200
and since it is independent of the distance

358

00:16:14,200  -->  00:16:15,966
from the surface identity

359

00:16:15,966  -->  00:16:17,800
towards the revolution point

360

00:16:17,800  -->  00:16:20,600
you just need to put in this surface identity

361

00:16:20,600  -->  00:16:23,700
and divide by beam free space permittivity

362

00:16:23,700  -->  00:16:26,266
since northern medium is mentioned over here

363

00:16:26,333  -->  00:16:28,800
and then you can simplify like this

364

00:16:29,500  -->  00:16:33,000
the second uh second sheet is a y is equal to minus 3

365

00:16:33,000  -->  00:16:35,366
so this is falling over here

366

00:16:35,366  -->  00:16:38,466
right is your y is equal minus 3 however

367

00:16:38,466  -->  00:16:40,900
in this case your Z is in finite

368

00:16:40,933  -->  00:16:45,000
and your axis in finite in both directions

369

00:16:45,000  -->  00:16:46,666
like this like this

370

00:16:47,500  -->  00:16:50,766
and now again you need to find out this normal

371

00:16:50,766  -->  00:16:52,800
so here is your elevation point

372

00:16:52,800  -->  00:16:55,000
so this is your elevation point

373

00:16:55,133  -->  00:16:56,266
so in this case

374

00:16:56,300  -->  00:16:59,800
your normal is pointing the words this direction right

375

00:16:59,800  -->  00:17:03,100
so this is equal to a Y like this

376

00:17:03,700  -->  00:17:05,366
so this normally is why again

377

00:17:05,366  -->  00:17:06,500
put in this uh

378

00:17:06,733  -->  00:17:09,800
official identity and you can evaluate this E2

379

00:17:11,866  -->  00:17:15,933
let's find find out this third electric intensity

380

00:17:15,933  -->  00:17:18,166
and due to this line distribution

381

00:17:18,566  -->  00:17:20,333
and here first of all

382

00:17:20,333  -->  00:17:23,966
you remember that this hero and this rule

383

00:17:24,100  -->  00:17:27,866
they are due to the perpendicular distance vector

384

00:17:28,000  -->  00:17:31,300
the perpendicular distance vector from the

385

00:17:31,766  -->  00:17:34,466
from your source towards the destination

386

00:17:34,966  -->  00:17:37,266
and uh from the distance formula

387

00:17:37,266  -->  00:17:39,666
if you want to find out the perpendicular distance

388

00:17:39,666  -->  00:17:42,700
sector what it is that your line

389

00:17:42,700  -->  00:17:46,500
your line was comprising of x is equal to 0

390

00:17:46,500  -->  00:17:49,666
and z is equal to 2 10 intersection

391

00:17:49,766  -->  00:17:53,666
so here your uh 209+ your 9

392

00:17:53,700  -->  00:17:56,000
your X is equal to 0 right

393

00:17:56,466  -->  00:18:02,800
your y is varying and your Z is 2 like this one

394

00:18:02,933  -->  00:18:04,300
all the points are here

395

00:18:04,400  -->  00:18:07,866
only your y is varying from minus infinity to infinity

396

00:18:08,800  -->  00:18:11,400
so if you want to find out the botanical distance

397

00:18:11,400  -->  00:18:13,066
between this elevation point

398

00:18:13,066  -->  00:18:15,566
and the line what you need to do that

399

00:18:15,566  -->  00:18:18,700
you need to keep this variable y

400

00:18:19,333  -->  00:18:21,066
same as the

401

00:18:21,933  -->  00:18:26,600
same as the y coordinator of this evaluation point

402

00:18:26,600  -->  00:18:29,966
that is equal to one right

403

00:18:30,266  -->  00:18:31,733
so for this line

404

00:18:31,733  -->  00:18:34,933
you know that your x is equal to 0 constant

405

00:18:34,933  -->  00:18:38,500
your is equal to constant and since your y is variable

406

00:18:38,500  -->  00:18:38,866
no you

407

00:18:38,866  -->  00:18:42,466
if you keep the y is equal to by steam coordinate

408

00:18:42,466  -->  00:18:43,633
that is of uh

409

00:18:43,733  -->  00:18:46,266
your evaluation point in this case it is 1

410

00:18:46,266  -->  00:18:46,533
then

411

00:18:46,533  -->  00:18:49,200
you can find a different pendicular distance factor

412

00:18:49,200  -->  00:18:52,966
between these two uh uh this uh line uh line charger

413

00:18:53,100  -->  00:18:58,100
uh line charger uh density and this uh evolution point

414

00:18:58,100  -->  00:18:59,300
and if you see

415

00:18:59,733  -->  00:19:03,233
basically you are moving towards this point

416

00:19:03,300  -->  00:19:06,066
like this one and this is going to be your

417

00:19:06,300  -->  00:19:08,466
are in this case like this one

418

00:19:08,466  -->  00:19:11,266
and it is representing your perpendicular distance

419

00:19:11,266  -->  00:19:13,433
the magnitude is representing your

420

00:19:13,966  -->  00:19:16,466
a tangler distance between this line chart dance

421

00:19:16,466  -->  00:19:18,233
G and B evaluation point

422

00:19:18,866  -->  00:19:21,300
and if you want to drive in the 2

423

00:19:21,300  -->  00:19:22,966
deprecatory representation

424

00:19:22,966  -->  00:19:27,866
you can further explain it like this that on this

425

00:19:28,166  -->  00:19:28,966
on this

426

00:19:29,566  -->  00:19:31,766
on this X Z on the Z

427

00:19:31,766  -->  00:19:35,600
explain your rise equal to rise up to one constant

428

00:19:35,600  -->  00:19:37,333
and you can find out from here

429

00:19:37,333  -->  00:19:41,966
that revelation point is on the studio is 1 1

430

00:19:41,966  -->  00:19:45,066
1 common minus 1 that is your accent Z

431

00:19:46,100  -->  00:19:48,866
and here you just need to find out the distance

432

00:19:48,866  -->  00:19:50,466
rector between these two things

433

00:19:50,466  -->  00:19:53,400
and so this was your distance rector

434

00:19:53,400  -->  00:19:54,600
coming out from there in the

435

00:19:54,600  -->  00:19:56,400
in the form of your unit factors

436

00:19:56,400  -->  00:19:57,533
you take its magnitude

437

00:19:57,533  -->  00:19:59,266
it is the perpendicular distance

438

00:19:59,266  -->  00:20:01,800
between this linechardency

439

00:20:01,800  -->  00:20:03,300
and this revolution point

440

00:20:03,300  -->  00:20:05,400
you know that the unit factor is equal to factor

441

00:20:05,400  -->  00:20:07,366
divided by the magitude of the lecture

442

00:20:07,900  -->  00:20:10,300
and from here now you plug in these

443

00:20:10,666  -->  00:20:15,533
these values in this given formula and a like this

444

00:20:15,533  -->  00:20:19,900
a row is like this your row is the magnitude of per R

445

00:20:19,900  -->  00:20:22,600
and you can simplify this like this

446

00:20:24,100  -->  00:20:25,600
okay and

447

00:20:25,700  -->  00:20:28,466
since you have evaluated all the electric intensity

448

00:20:28,766  -->  00:20:30,566
now the resultant electric free

449

00:20:30,566  -->  00:20:34,133
in the intensity at this human evaluation point

450

00:20:34,133  -->  00:20:35,266
is the superposition

451

00:20:35,266  -->  00:20:37,266
or the summation of all these three

452

00:20:37,400  -->  00:20:38,966
electric free intensity

453

00:20:39,000  -->  00:20:41,000
and you can simplify like this one

454

00:20:43,300  -->  00:20:45,133
now let's solve this example

455

00:20:45,133  -->  00:20:49,600
that is regarding the determination of electric flux

456

00:20:49,600  -->  00:20:52,433
density at this given evaluation point

457

00:20:52,666  -->  00:20:57,900
and it is due to our No. 1 point charge

458

00:20:58,300  -->  00:21:01,166
point charge which is placed at this point four zero

459

00:21:01,166  -->  00:21:02,966
zero so if this is X

460

00:21:02,966  -->  00:21:06,366
y and d so this is your point charge location right

461

00:21:06,366  -->  00:21:12,300
4 0 0 0 your x coordinate is 4 and y and and y and 0

462

00:21:12,300  -->  00:21:17,066
so it is a longer x x is line the 9 chart distribution

463

00:21:17,066  -->  00:21:18,866
it is along your y axis right

464

00:21:18,866  -->  00:21:21,366
again your y is going to change

465

00:21:21,500  -->  00:21:23,866
your y is going to change right

466

00:21:23,866  -->  00:21:25,500
and since since it is on y axis

467

00:21:25,500  -->  00:21:28,133
so it's X is going to zero

468

00:21:28,133  -->  00:21:30,433
it's y is going to be variable

469

00:21:30,600  -->  00:21:33,666
and it's disease again going to be zero

470

00:21:34,666  -->  00:21:38,000
now your evolution point is 4 0 3

471

00:21:38,000  -->  00:21:42,066
so here again you see that it is your 4 right

472

00:21:42,066  -->  00:21:43,600
this was again before

473

00:21:43,700  -->  00:21:48,866
this is your uh rise again 0 and this is your 3

474

00:21:49,100  -->  00:21:50,266
so you are over here

475

00:21:50,266  -->  00:21:55,233
so this is your evolution point four zero and three

476

00:21:56,766  -->  00:21:58,800
okay now you first of all

477

00:21:58,800  -->  00:22:00,333
you need to find out the electrically

478

00:22:00,333  -->  00:22:03,933
due to this point charge which is at this 4

479

00:22:03,933  -->  00:22:05,500
0 0 point location

480

00:22:05,733  -->  00:22:09,666
just involve this coolum floor electrical intensity

481

00:22:09,666  -->  00:22:12,800
multiply by this absolute not

482

00:22:12,800  -->  00:22:17,600
you remember the formula it is equal to d absolute not

483

00:22:19,066  -->  00:22:20,800
E right

484

00:22:21,600  -->  00:22:24,600
okay this uh this decision rector

485

00:22:24,600  -->  00:22:26,800
you know that it is the uh

486

00:22:26,800  -->  00:22:29,300
position rector of this destination

487

00:22:29,300  -->  00:22:31,233
minus the position rector of this force

488

00:22:31,300  -->  00:22:34,233
and we know that this the position rector of this uh

489

00:22:34,766  -->  00:22:37,133
destination with respect to origin is 4 0

490

00:22:37,133  -->  00:22:42,500
3 and for this source point point charge it is 4 0

491

00:22:42,500  -->  00:22:43,300
0

492

00:22:43,733  -->  00:22:45,233
you can replace this

493

00:22:45,933  -->  00:22:48,600
this is unit factor by this distance vector

494

00:22:48,600  -->  00:22:50,666
divided by the magnitude of distance vector

495

00:22:50,666  -->  00:22:52,300
and it will simplify to this

496

00:22:52,300  -->  00:22:53,066
and distress vector

497

00:22:53,066  -->  00:22:56,300
divided by the magnitude tube of this a distance vector

498

00:22:57,766  -->  00:22:59,500
now you can uh take this magnitude

499

00:22:59,500  -->  00:23:03,000
so and uh you can put in this uh uh genetractor

500

00:23:03,000  -->  00:23:06,200
it will be the subject over here and you can simplify

501

00:23:06,200  -->  00:23:08,833
you can take take the cube of this magnitude

502

00:23:08,866  -->  00:23:13,766
and which will comes out to be 27 right

503

00:23:13,866  -->  00:23:14,866
and as you can see

504

00:23:14,866  -->  00:23:19,033
your book quality for the calculations as well and

505

00:23:20,866  -->  00:23:23,466
now this is your minus point one three

506

00:23:23,466  -->  00:23:25,433
eight in B a Z direction

507

00:23:26,000  -->  00:23:26,966
and you can see over here

508

00:23:26,966  -->  00:23:30,066
it is also in B pointing in B negative z direction

509

00:23:30,066  -->  00:23:31,066
like this thing

510

00:23:31,800  -->  00:23:32,600
okay

511

00:23:34,000  -->  00:23:37,366
no we need to find out the electrical identity

512

00:23:37,366  -->  00:23:39,566
due to this line charge distribution

513

00:23:39,566  -->  00:23:43,166
and again we need to find out this patentical distance

514

00:23:43,166  -->  00:23:45,300
between this line charge identity

515

00:23:45,300  -->  00:23:46,766
and this regulation point

516

00:23:46,800  -->  00:23:49,300
again your why is the rating in this thing

517

00:23:49,300  -->  00:23:51,466
and what is the why of this relation point

518

00:23:51,466  -->  00:23:54,100
that is zero may you keep

519

00:23:54,100  -->  00:23:57,166
may you need to keep this varying y of this 9

520

00:23:57,166  -->  00:24:00,766
chart density equal to this y of the regulation point

521

00:24:00,800  -->  00:24:02,800
so these two must be same

522

00:24:03,200  -->  00:24:06,900
and since it is a y aligned line chart density

523

00:24:06,900  -->  00:24:10,566
so it x and z r already equal to zero

524

00:24:10,566  -->  00:24:14,100
because it is being created in intersection of this x

525

00:24:14,100  -->  00:24:15,166
and z plan

526

00:24:16,300  -->  00:24:20,166
now phone this uh distance vector fourth minus this uh

527

00:24:20,166  -->  00:24:22,966
minus the uh destination minus the fourth

528

00:24:22,966  -->  00:24:24,466
divide with the magnitude of this uh

529

00:24:24,466  -->  00:24:27,366
distance vector so it will come out to be 5

530

00:24:27,400  -->  00:24:30,366
and then you uh need to plug in this uh

531

00:24:30,966  -->  00:24:35,966
this is unitractor the perpendicular distance

532

00:24:36,066  -->  00:24:37,200
distance factor

533

00:24:37,700  -->  00:24:42,066
unitractor divided by the magnitude of this unitractor

534

00:24:42,366  -->  00:24:46,233
so you can simplify this in terms of this again since

535

00:24:47,533  -->  00:24:49,666
again since it is a

536

00:24:50,166  -->  00:24:52,533
it is a superposition or a summation problem

537

00:24:52,533  -->  00:24:54,366
that you have to uh uh

538

00:24:54,366  -->  00:24:57,700
you have to add up the effect of this electric flex

539

00:24:57,700  -->  00:25:00,066
flexes due to this uh uh

540

00:25:00,066  -->  00:25:00,766
due to this uh

541

00:25:00,766  -->  00:25:02,866
point charge and the 9 chart distribution

542

00:25:02,866  -->  00:25:05,866
in order to find out the overall electric trucks uh

543

00:25:06,100  -->  00:25:07,900
uh at the electrical density

544

00:25:07,900  -->  00:25:09,600
had be given evaluation point

545

00:25:09,600  -->  00:25:11,766
due to these two different charge distributions

546

00:25:11,766  -->  00:25:13,533
and that is your resultant

547

00:25:13,533  -->  00:25:16,200
after dissermation of these two individual uh

548

00:25:16,200  -->  00:25:17,500
electric trucks density

549

00:25:20,400  -->  00:25:23,500
uh this is a very useful uh tutorial to understand uh

550

00:25:23,500  -->  00:25:25,166
columns law and uh

551

00:25:25,166  -->  00:25:27,200
it is uh developed by

552

00:25:27,533  -->  00:25:30,000
and the simulator is developed by a CP12

553

00:25:30,000  -->  00:25:31,333
exploration series

554

00:25:31,333  -->  00:25:33,966
and you can find all the simulator uh

555

00:25:33,966  -->  00:25:36,366
a link from this uh source

556

00:25:36,466  -->  00:25:37,133
uh and I

557

00:25:37,133  -->  00:25:40,300
I've also copied the simulation uh

558

00:25:40,300  -->  00:25:42,866
uh video on your uh shared folder

559

00:25:43,133  -->  00:25:48,566
and so what is the explain to do that in this tutorial

560

00:25:48,566  -->  00:25:49,733
and in the simulations

561

00:25:49,733  -->  00:25:51,466
you can create number of charges

562

00:25:51,466  -->  00:25:54,833
you can assign the platies you can find out there

563

00:25:55,566  -->  00:25:59,600
you can you can place them at any of these locations

564

00:25:59,600  -->  00:26:04,366
you can find you can adjust the good separation

565

00:26:04,366  -->  00:26:09,700
that scale on this front panel

566

00:26:09,800  -->  00:26:12,633
and then you can find out that

567

00:26:12,933  -->  00:26:16,000
how the electrical lines are generated from pulse

568

00:26:16,000  -->  00:26:17,466
to charge and electric charge

569

00:26:17,533  -->  00:26:19,533
and how the uh

570

00:26:19,533  -->  00:26:20,933
ecurrential lines are generated

571

00:26:20,933  -->  00:26:22,866
in the surrounding of an uh uh

572

00:26:22,866  -->  00:26:24,800
uh charge and uh

573

00:26:24,800  -->  00:26:27,566
how to find out the electricity in the uh

574

00:26:27,566  -->  00:26:29,233
at some particular uh

575

00:26:29,466  -->  00:26:32,566
point and give vicinity of these to uh

576

00:26:32,566  -->  00:26:34,633
be different types of the charges

577

00:26:34,666  -->  00:26:37,333
and you can also find out the elective potential

578

00:26:37,333  -->  00:26:40,300
which is a scale of quantity at the given charges

579

00:26:40,466  -->  00:26:41,600
so this complete tutorial

580

00:26:41,600  -->  00:26:44,633
will provide your brief overview

581

00:26:44,933  -->  00:26:47,000
and you can also play around with this

582

00:26:47,566  -->  00:26:48,700
uh to further uh

583

00:26:48,700  -->  00:26:52,066
get an understanding of the coolants now and how the uh

584

00:26:52,066  -->  00:26:57,000
things that uh are there in the area life once uh more

585

00:26:57,000  -->  00:26:57,800
more than uh

586

00:26:57,800  -->  00:27:00,200
more than one number of charges are present

587

00:27:00,300  -->  00:27:03,166
uh individuality of the another charge

588

00:27:05,533  -->  00:27:08,400
this is your assignment for week No. 5

589

00:27:08,400  -->  00:27:11,100
and you have to solve these two questions

590

00:27:11,100  -->  00:27:12,700
from your chapter No.

591

00:27:12,700  -->  00:27:16,666
4 of textbook and kindly follow this second edition

592

00:27:16,666  -->  00:27:18,566
if you don't have the second edition

593

00:27:19,500  -->  00:27:23,900
then you can mash the questions from your colleagues

594

00:27:23,900  -->  00:27:25,300
which are on campus

595

00:27:26,500  -->  00:27:28,333
and take the snapshot of those questions

596

00:27:28,333  -->  00:27:30,966
which are waiting in your 3rd edition

597

00:27:30,966  -->  00:27:33,100
with those factories 2nd edition

598

00:27:35,333  -->  00:27:36,666
so I'm here today's lecture

599

00:27:36,666  -->  00:27:38,900
so we uh after covering the

600

00:27:38,900  -->  00:27:41,800
a brief for introduction on this electric field

601

00:27:41,866  -->  00:27:42,600
due to this uh

602

00:27:42,600  -->  00:27:46,500
continuous charge distributions and the electric flux

603

00:27:46,500  -->  00:27:48,100
tensity due to this uh

604

00:27:48,400  -->  00:27:50,200
charge distributions uh

605

00:27:50,200  -->  00:27:54,800
we sold uh some of the examples of your book and uh

606

00:27:54,800  -->  00:27:55,166
the

607

00:27:55,166  -->  00:27:57,933
and we saw that how to find out the electric trucks

608

00:27:57,933  -->  00:28:01,366
intensity and electricals tensities due to uh

609

00:28:01,366  -->  00:28:02,833
due to the presence of uh

610

00:28:03,000  -->  00:28:05,366
uh more than one uh electric uh

611

00:28:05,533  -->  00:28:06,333
uh good uh

612

00:28:06,333  -->  00:28:08,500
presence of more than one uh

613

00:28:08,500  -->  00:28:10,133
electric charge distributions

614

00:28:10,133  -->  00:28:12,100
either it is in the form of point charge

615

00:28:12,100  -->  00:28:14,733
or it is in the form of line charge distribution

616

00:28:14,733  -->  00:28:17,666
or it is in the form of circus charge distributions

617

00:28:17,800  -->  00:28:19,900
so if they are more than one

618

00:28:19,900  -->  00:28:22,166
then we need to find out them individually

619

00:28:22,166  -->  00:28:23,766
for that evaluation point

620

00:28:23,766  -->  00:28:25,566
and then we'll have to sum them up

621

00:28:25,566  -->  00:28:27,833
to find out the oral impact of that

622

00:28:28,566  -->  00:28:29,833
child distribution

623

00:28:31,166  -->  00:28:32,500
in the next class uh

624

00:28:32,500  -->  00:28:36,066
we will cover that uh cause Islam or we can uh

625

00:28:36,066  -->  00:28:38,800
find out the electrical tensities uh

626

00:28:38,800  -->  00:28:41,166
due to this uh recharge distribution

627

00:28:41,166  -->  00:28:44,633
if they are exhibiting some sort of this smetry

628

00:28:45,000  -->  00:28:48,333
either they are aligned with some some access

629

00:28:48,333  -->  00:28:50,600
they are they are aligned with some surface

630

00:28:50,600  -->  00:28:53,966
or they are aligned with some given point

631

00:28:53,966  -->  00:28:56,400
including swear lightings

632

00:28:56,400  -->  00:28:58,266
then we we can involve to

633

00:28:58,266  -->  00:29:01,766
we will involve this Maxwell education to drive this

634

00:29:01,766  -->  00:29:05,500
uh gazes law and then we will utilize the uh

635

00:29:05,500  -->  00:29:06,933
gazes law uh

636

00:29:06,933  -->  00:29:07,800
where we can uh in

637

00:29:07,800  -->  00:29:09,333
in in different uh

638

00:29:09,333  -->  00:29:11,300
applications that how we can uh

639

00:29:11,300  -->  00:29:12,266
apply this uh

640

00:29:12,533  -->  00:29:15,700
gazes law to simplify the uh

641

00:29:15,700  -->  00:29:16,666
this uh

642

00:29:16,666  -->  00:29:17,900
finding out of uh

643

00:29:17,900  -->  00:29:21,000
electric trucks tensities and electric intensities

644

00:29:21,066  -->  00:29:23,233
uh with the help of this uh

645

00:29:23,300  -->  00:29:24,533
gazada in case

646

00:29:24,533  -->  00:29:26,400
if they are exhibiting this

647

00:29:26,400  -->  00:29:28,500
symmetrical distribution properties

648

00:29:34,266  -->  00:29:35,333
here I thank you all

649

00:29:35,333  -->  00:29:36,666
if you have any questions

650

00:29:36,666  -->  00:29:40,400
they will be entertained to your emails or online

651

00:29:40,400  -->  00:29:40,900
synchron

652

00:29:40,900  -->  00:29:44,100
a session that will have arranged at the department