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so here is my first example . i want to calculate
the electric field remember this is electric
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field or electric field intensity so i want
to calculate the electric field due to an
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infinite line charge i have an infinite line
charge that is simply meaning that there is
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a line charge of density rho l which is measured
in coulombs per meter located on a very very
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narrow or a very very thin wire like distribution
and this wire extends all the way from minus
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infinity to plus infinity
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so this line charge extends all the way from
minus infinity to plus infinity i can locate
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any coordinate system to perform this particular
calculation so let me say i am going to use
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the cartesian coordinate system and i am going
to line up my z axis along the line charge
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so the z axis is along the line charge and
i have two other axis out here for me sorry
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this is y axis and this is the x axis so the
line charge actually goes all the way towards
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from minus infinity along z to plus infinity
along z
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it has no other x or y dimensions it is such
narrow that it is fitting entirely on the
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line i want to find the electric field so
let me say i want to find the electric field
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at some point in the x y plane so i want to
find the electric field at some point in the
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x y plane so on this plane the value of z
will be equal to zero how do i find the electric
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filed well we are going to use this super
position rule that we have written down here
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except that this summation will now become
an integration integration over the line along
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which the charge is distributed
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now remember that in cartesian coordinate
systems we showed that line integral along
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the z direction is simply dz along that particular
direction so the line integral is actually
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along the z direction so this d l z is actually
along d z z prime of course in this case i
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dont really need the vector notations i am
not going to use the vector notation this
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is how the line charges changes next i need
to first locate particular point on the source
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so i am going to locate a point at z prime
height from the x y location
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so i have x y location out here and i am going
to locate a point at z prime and i am going
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to consider a small incremental distance dz
prime here what is the charge that is there
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in this particular small distribution a small
like differential length tz what is the total
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charge that this differential line segment
contains it is simply the line charge density
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times dz and we are really thankful that this
line charge density is uniform in the sense
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that it does not depend on x and y coordinates
in fact does not even depends on z coordinate
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it is constant everywhere so this is the amount
of charge the differential charge that this
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particular line segment dz prime actually
possesses now this is the source point so
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source point is given by the position vector
r prime which is z prime in the direction
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of z axis correct what about the field point
well to get the field point you have to find
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the posting vector at point x y in the z plane
so the field point is x x hat plus y y hat
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there is no z component out there so you can
clearly find out what is r minus r prime this
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vector will be x x hat plus y y hat minus
z z hat
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so this is r minus r prime vector which is
actually a vector that is directed from the
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source to the field point so from the source
to the field point you have vector which is
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this particular vector so what is the magnitude
of this vector or magnitude cube of this vector
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it would be x square plus y square plus sorry
this is z prime so z prime square to the power
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3/2 the magnitude itself x square plus y square
plus z prime square under root and then you
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have to simply raise it to the cubic power
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so it becomes x square + y square + z prime
square to the power 3/2 now i can put down
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all these quantities inside the integral and
i can find out what is the electric field
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at the position x y and zero that is at the
z plane at any point x and y so this will
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be equal to what is the source charge value
here the incremental charge will be rho l
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dz so rho l dz so dz prime is the incremental
charge correct divided by 4 pi epsilon zero
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this is the constant nothing to do with particular
thing and i did not state that but i am assuming
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that this line charge is located in free space
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so i have 4 pi epsilon zero and this quantity
x square plus y square plus y square plus
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z prime square to the power 3/2 and then multiplied
by the vector r minus r j prime remember so
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that vector will be x x hat plus y y hat minus
z prime z hat no wonder people do not like
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electro magnetics as much as we would like
to because there is these differential integrals
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that are sitting over here and these are not
very straight evaluate all the times
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but the beauty of electromagnetics is that
if you perform this calculation you will realize
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that or if you substitute these calculations
analytically with numerical methods you will
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realize that electromagnetics is actually
in itself very simple only thing is certain
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of these things will make it seem complicated
but they are not really complicated you can
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get hold of a very nice table of integrals
books and then you can perform all these integrations
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or you can get hold of a computer and with
a technics that i am going to describe to
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you in latter classes you can perform this
entire calculation using the numerical technique
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so lets move on so this electric field is
of course not the total electric field because
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to get the total electric field i have to
consider contributions of every segment of
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the line charge which is going all the way
from minus infinity to plus infinity in essence
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what i am going to do is to integrate this
whole thing from minus infinity to plus infinity
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now my objective for the next few minutes
is to evaluate this particular vary daunting
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looking integral but we will see that this
is not such a bad integral after all
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the important point to note here is that i
have some integration . with respect to certain
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variable that is happening and there are certain
factors in the denominator but there are factors
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in the numerator which are all vectors so
we have probably not seen these types of integral
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in our earlier study we had integrals in which
the inside –the ingredients was simply of
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function of whatever the integral variable
had it had some functions the functions could
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have some constants or something but thats
all right that inside ingredients was always
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some function it was not a vector
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now the question is what do i do with integrals
when there are vectors thankfully integral
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is a linear operation integration is a linear
operation which means that the three integrals
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which i showed you in the last slide can actually
be broken up into three different integrals
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so i actually can break up this into three
different integrals and i will call them as
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ix + iy + iz prime where ix will be the integral
rho l/ 4 pi epsilon zero is a constant
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so i am going to take this out of the integral
there is minus infinity plus infinity the
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integration variable is said and there is
x square plus y square plus z prime square
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to the power 3/2 and there is x into x hat
this is the vector similarly you can find
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out or you can write down expression for i
y bar and i z bar i am not going to do all
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those calculations i am just going to show
you how to do one integral and i am hoping
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that you can use the knowledge that you have
already studied in terms of integration and
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differentiation to obtain the values of the
other integrals
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so you have this particular integral you can
solve this integral first of all look at this
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as z is changing in the sense that physically
what is happening i am moving from z is equal
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to minus infinity all the way to z is equal
to plus infinity so at each point as i move
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what is happening to the direction of the
x vector the direction of the x vector is
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still the same at every point along z axis
the direction of the x vector is not changing
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the x vector is remaining in the same direction
as the x vector that would be there at the
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origin
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so the x vector is also not changing in magnitude
that is the length of this arrow is not changing
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as i move up and down along the z axis so
in another words in this particular case we
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are fortunate that this entire term turns
out to be constant with respect to z axis
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so i can take this entire constant out of
the integral now i will be left out with something
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that would look like this here itself have
minus infinity to plus infinity there is x
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dz prime divided by x square plus y square
plus z square to the power 3/2 now using the
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well- known techniques of integration i am
going to write down this square plus y square
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as sum r square
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this is constant because z is changing not
x and y changing if i have fixed x and y then
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the only way this is changing is because z
is changing so i can call this x square plus
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y square as sum r square does not really matters
what r is or you could call just as a square
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or s square whatever letter that you want
to use you can use here and then write down
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z prime is equal to r tan theta so that tz
prime will be equal to r sec square theta
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d theta and when i change the variable of
integration from z prime to theta of course
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i have to change the limits of integration
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what is the new limit of integration for theta
well theta will be equal to z prime/r or rather
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tan theta will be equal to z prime/r and when
z prime is equal to minus infinity tan theta
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will be equal to minus infinity and theta
will be equal to minus pi/ 2 and on the other
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hand when z prime is equal to plus infinity
then theta will be equal to plus pi/2 so the
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integral variables are from minus pi/2 to
plus pi/2 so the integral variables are from
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minus pi/2 to plus pi/2
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and all these constants i am not going to
write at this point so you can actually take
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x also out does not really matter so i have
rho l x x hat/ 4 pi epsilon zero and outside
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of the integral i am not really bothered about
that tz prime is r sec square theta d theta
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and in the denominator i have x square plus
y square which is r square r square plus z
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prime square z prime square is r square tan
square theta
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so r square is a common factor out there i
can take that out one plus tan square theta
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is sec square theta and there is a root 3/2
power that you have to raise both of them
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so that leaves you r to the power three sec
to the power 3 theta r cancels out with one
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of the r in the denominator sec square theta
cancels out with sec theta in the denominator
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leaving the integral from minus pi/ 2 to plus
pi/2 there is 1/r square here and 1/ sec theta
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is left out
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now i know that sec theta is nothing but 1/
cos theta therefore 1/ sec theta is cos theta
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d theta do you think this integral will be
equal to zero well see how the cos theta function
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is behaving so this is will be from minus
pi/ 2 to plus pi/2 so the area under this
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particular curve will not be equal to zero
and you can actually found out what the area
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of the curve is so if you carry out this integral
cos theta becomes minus cos theta becomes
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sin theta and then when you subtract this
as two
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so the result of all this for i x let me write
down the value of i x in the next slide . so
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ix is actually given by rho l/ 4 pi epsilon
zero there is an r square in the denominator
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and on the numerator i get 2 x x hat and what
is r square r square is nothing but square
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plus y square over here similarly you can
find out what is i y the integral over the
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y you are going to see that this would be
rho l 2y y hat divided by 4 pi epsilon zero
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square plus y square you can also show that
i z will be equal to zero so that the total
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electric field at the point x and y is given
by rho l/ 2 pi epsilon zero x square + y square
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x x hat plus y y hat volt/ meter dont forget
to use that volt/meter at the end because
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this is what the electric field is measured
in
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before going on to the next part let us see
what this equation is telling us what is the
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direction of the electric field in the x y
plane now if you look at this equation and
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you have your x and y directions over here
and the z axis is coming up this particular
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way so in the z is equal to plane where we
have calculated the electric field the direction
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of the electric field is always along that
line which is at x x hat plus y y hat so if
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for example x is equal to 1 y is equal to
1 the direction will be along y hat plus x
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hat which would be making an angle of 45 degrees
with respect to the x axis
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so you can actually see that the direction
of the electric field is always what we call
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is radially away from the line what is happening
to the magnitude of the electric field well
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the magnitude of the electric field is obtained
by looking at the magnitude of this particular
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vector so you have rho l/ 2 pi epsilon zero
x square plus y square there is no change
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here but the magnitude of this vector x x
hat plus y y hat is x square plus y square
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under root now there is square root fellow
you can cancel that out with one of this terms
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in the denominator so the magnitude is actually
going as rho l/ 2 epsilon zero r where r is
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the –so if this is my y and this is my x
axis r is a set of all points which are at
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a distance r from the origin
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so the magnitude of the electric field is
all constant over here in this particular
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at a distance r and more importantly the fields
are going away as not as 1/r square but they
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are going as 1/r this is the field of a line
charge and this is a field of a point charge
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point charge goes as 1/r square whereas line
charge goes as 1/r interesting why is the
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line charge going as 1/r well what is happening
as if you have the charge distribution over
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here so you are looking at the charge line
at a distance r from this one so you have
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outmost a 90 degree field of view
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so if this is r then your field of view is
approximately r as you go away r the field
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goes as 1/ r square but the charge will multiply
by r so the numerator will increase as r denominator
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will decrease as r square and this r increases
the numerator composite for the r decreasing
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the denominator or r square decreasing the
denominator leaving your with the field which
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is going as 1/r now if you have followed this
discussion so far you might have several questions
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one of the questions is why is this person
doing this in the hard way
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this let us admit is actually a hard way of
doing this particular problem why was i doing
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this problem in a hard way because i wanted
to illustrate that when you are faced with
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the problem in electromagnetics do not always
rush to mathematics mathematic will always
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give you correct results there is no doubt
about that one as long as your physical problem
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is correctly setup mathematic will always
give you correct results
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however if you start blindly applying mathematic
unless you are mathematical wizard some of
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these problems are very very hard to solve
if you neglect the physics side of the problem
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that is to say you forget how the charge is
distributed you forget how to choose the proper
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coordinate system you neglect what is major
aspect of these kind of problem is called
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symmetry you will be doing lot of work which
you could have done in a very short time
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if you had taken care of all these things
so you never forget to use symmetry never
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forget to setup the appropriate coordinate
system to reduce the amount of work that you
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are doing in order to get the same result
so this result is correct only thing i did
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it the hard way to show you that never do
these types of problems in this hard way look
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for symmetry look for an appropriate coordinate
system and your work will be done in a much
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faster way
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so the question is of course is this coordinate
system not suited for this type of charge
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distribution answer is no there are other
coordinate systems and we are going to study
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the next coordinated system now which is much
better suited for studying these types of
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problems and that coordinate system is what
is called as cylindrical coordinate system
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. so we are going to look at cylindrical coordinate
system and some surprises will be there if
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you are coming from cartesian coordinate system
and hopefully keep these surprises in your
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mind