1
00:00:13,670 --> 00:00:20,960
suppose that vector is r and i can write this
vector r as consisting of 3 vectors rx i can
2
00:00:20,960 --> 00:00:29,910
resolve this vector r as rx x hat plus ry
y hat plus rz plus z hat the same vector in
3
00:00:29,910 --> 00:00:35,910
the cylindrical co-ordinate system can also
be written as rr r hat where rr will be the
4
00:00:35,910 --> 00:00:43,940
component of the vector r along r hat vector
along the radial component plus r phi phi
5
00:00:43,940 --> 00:00:47,250
hat plus rz z hat
6
00:00:47,250 --> 00:00:51,390
you can guess from intuition that rz will
be equal to rz import cases and you will be
7
00:00:51,390 --> 00:00:55,750
right however we are now interested in rx
ry rr and r phi
8
00:00:55,750 --> 00:00:56,750
.
9
00:00:56,750 --> 00:01:07,440
so i know that r is equal to rx x hat plus
ry y hat plus rz z hat this is also equal
10
00:01:07,440 --> 00:01:19,020
to rr r hat plus r phi phi hat plus rz z hat
now if i try to find out the component of
11
00:01:19,020 --> 00:01:26,840
the r vector along the x hat vector that is
along the x axis i would be obtaining i can
12
00:01:26,840 --> 00:01:32,390
do that one by finding out the dot product
of car along x so if i do this one i am going
13
00:01:32,390 --> 00:01:37,880
to get the component of r along x correct
so if i do this operation i am going to get
14
00:01:37,880 --> 00:01:40,549
rx why
15
00:01:40,549 --> 00:01:46,189
what would happen when you take x dot with
respect to this one it stands out that x hat
16
00:01:46,189 --> 00:01:51,359
dot x hat will be equal to one because the
length of one vector upon itself will be giving
17
00:01:51,359 --> 00:01:58,270
the length of the vector itself however what
happened to x hat dot y hat this became equal
18
00:01:58,270 --> 00:02:04,600
to zero because x is perpendicular to y and
dot product will be equal to zero when theta
19
00:02:04,600 --> 00:02:06,430
ab is equal to 90 degrees right
20
00:02:06,430 --> 00:02:12,380
so this is another definition of perpendicularity
or normality or orthogonality two vectors
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00:02:12,380 --> 00:02:17,570
are said to be perpendicular to each other
or normal to each other or orthogonal to each
22
00:02:17,570 --> 00:02:25,959
other when the dot product between the two
vanishes similarly x hat dot z hat will also
23
00:02:25,959 --> 00:02:33,580
be equal to zero in fact the corresponding
vectors x y and z they form mutually perpendicular
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00:02:33,580 --> 00:02:38,310
set of vector something that we looked at
in the last class
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00:02:38,310 --> 00:02:46,550
r dot x hat will give you the rx component
now i have to dot this r vector in the cylindrical
26
00:02:46,550 --> 00:02:52,690
co-ordinate by the x hat vector right so if
i do that one this would also be equal to
27
00:02:52,690 --> 00:03:07,370
rr r hat dot x hat plus r phi phi hat dot
x hat plus rz z hat dot x hat this component
28
00:03:07,370 --> 00:03:13,750
is equal to zero or this value is equal to
zero so rx will be equal to what is r dot
29
00:03:13,750 --> 00:03:16,590
x this is nothing but cos phi
30
00:03:16,590 --> 00:03:21,890
so have cos phi rr and what about phi dot
x in the previous slide we have already seen
31
00:03:21,890 --> 00:03:28,590
that this is equal to minus sin phi r so therefore
this is minus sin phi multiply this one by
32
00:03:28,590 --> 00:03:35,530
r phi what will be ry you can show similarly
by considering the dot product of r with y
33
00:03:35,530 --> 00:03:48,159
this will be equal to sin phi rr plus cos
phi r phi plus zero times rz plus zero times
34
00:03:48,159 --> 00:03:50,170
rz just to complete the equations
35
00:03:50,170 --> 00:03:59,090
i can also write down rz as zero times rr
plus zero times r phi plus one time rz okay
36
00:03:59,090 --> 00:04:03,250
now this might be looking very suspiciously
like a set of linear equations and these are
37
00:04:03,250 --> 00:04:05,959
the set of linear equations and these are
the set of linear equations we can obtain
38
00:04:05,959 --> 00:04:19,209
rx ry and rz if i know the component value
of rr r phi and rz by a transformation matrix
39
00:04:19,209 --> 00:04:22,780
which takes me from cylindrical to rectangular
co-ordinate systems
40
00:04:22,780 --> 00:04:31,470
this is the cylindrical to rectangular co-ordinate
system and the matrix is cos phi minus sin
41
00:04:31,470 --> 00:04:41,340
phi zero sin phi cos phi zero one zero and
zero this is a 3 by 3 matrix which transforms
42
00:04:41,340 --> 00:04:46,070
any vector which is in the cylindrical co-ordinate
system into rectangular co-ordinate system
43
00:04:46,070 --> 00:04:51,750
if you are interested in finding what would
be the corresponding transformation from rectangular
44
00:04:51,750 --> 00:04:56,360
to cylindrical you can actually show that
this transformation is nothing but the inverse
45
00:04:56,360 --> 00:04:58,400
of this matrix t cr
46
00:04:58,400 --> 00:05:05,830
and you can show that this will be equal to
cos phi sin phi zero there is a zero here
47
00:05:05,830 --> 00:05:17,640
one here minus sin phi cos phi zero and zero
okay i will leave this as an exercise to you
48
00:05:17,640 --> 00:05:24,010
to show this one now these two matrices are
very useful to you in fact you can take a
49
00:05:24,010 --> 00:05:27,530
look at any electromagnetic text book they
are going to give you these kind of formulas
50
00:05:27,530 --> 00:05:33,350
which will allow you to convert of vectors
in one co-ordinate system to another co-ordinate
51
00:05:33,350 --> 00:05:34,350
system
52
00:05:34,350 --> 00:05:39,920
i will give you one simple example of conversion
how to do this particular conversion and i
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00:05:39,920 --> 00:05:44,680
will point out one very important aspect of
cylindrical co-ordinate systems to you what
54
00:05:44,680 --> 00:05:47,160
is that problem
.
55
00:05:47,160 --> 00:05:54,510
i have a vector a which is described in the
rectangular co-ordinate systems having components
56
00:05:54,510 --> 00:06:07,920
ax and ay okay ax and ay where ax is equal
to x by x square plus y square ay is equal
57
00:06:07,920 --> 00:06:16,680
to y by x square by y square now i want you
to find out the vector in the cylindrical
58
00:06:16,680 --> 00:06:20,660
co-ordinate systems how do i find the vector
in the cylindrical coordinate system i have
59
00:06:20,660 --> 00:06:24,410
to use the formula that i developed earlier
so what is that formula
60
00:06:24,410 --> 00:06:31,270
i have to use this transformation matrix i
know ay ay i know az a z in this case is zero
61
00:06:31,270 --> 00:06:38,240
and then i have to find out ar a phi and az
okay let me proceed to find that one out ar
62
00:06:38,240 --> 00:06:42,990
a phi will be equal to because there is no
z component there i am not going to write
63
00:06:42,990 --> 00:06:47,500
down the z component for you i mean because
there is no point in writing down that one
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00:06:47,500 --> 00:07:00,810
this will be equal to cos phi sin phi minus
sin phi cos phi times ax ay hopefully this
65
00:07:00,810 --> 00:07:03,120
is alright
66
00:07:03,120 --> 00:07:07,330
so let us go back to the matrix and check
cos phi sin phi minus sin phi and cos phi
67
00:07:07,330 --> 00:07:13,150
so we are on the right track now substitute
for ax and ay if you substitute for ax and
68
00:07:13,150 --> 00:07:20,560
ay you are going to get ar will be equal to
cos phi times ax and sin phi times ay but
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00:07:20,560 --> 00:07:25,220
i also know that x square plus y square is
actually equal to r square in cylindrical
70
00:07:25,220 --> 00:07:32,990
co-ordinate systems and substituting therefore
ax as x by r square and for y as y by r square
71
00:07:32,990 --> 00:07:33,990
what will be ar
72
00:07:33,990 --> 00:07:47,020
ar will be equal to cos phi x by r square
plus sin phi y by r square okay but what is
73
00:07:47,020 --> 00:07:59,160
x cos phi and y sin phi and what is a phi
a phi is minus sin phi x by r square plus
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00:07:59,160 --> 00:08:09,130
cos phi y by r square okay now let us right
down the vector in the cylindrical co-ordinate
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00:08:09,130 --> 00:08:13,690
system this vector in the co-ordinate system
will be equal to 1 by r square is a common
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00:08:13,690 --> 00:08:17,190
factor that is coming out everywhere so i
can take that common factor out
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00:08:17,190 --> 00:08:31,370
so i have x cost phi plus y sin phi along
the radial direction plus cos phi y cos phi
78
00:08:31,370 --> 00:08:43,050
minus x sin phi along the phi direction okay
but i also know that x is equal to r cos phi
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00:08:43,050 --> 00:08:50,089
y is equal to r sin phi so i can substitute
for x and y in this expression so if i substitute
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00:08:50,089 --> 00:08:56,470
for that one this expression x cos phi plus
y sin phi that is ar term will become x is
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00:08:56,470 --> 00:09:05,600
equal to r cos phi so that becomes r cos square
phi by r square plus y is r sin phi therefore
82
00:09:05,600 --> 00:09:15,670
this becomes r sin square phi by r square
which is equal to 1 by r and a phi is equal
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00:09:15,670 --> 00:09:19,550
to minus x is nothing but r cos phi
84
00:09:19,550 --> 00:09:33,129
so minus r cos phi sin phi by r square plus
y is r sin phi cos phi by r square which is
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00:09:33,129 --> 00:09:43,050
equal to a big zero so the vector a in cylindrical
co-ordinate system is simply given by r hat
86
00:09:43,050 --> 00:09:53,420
divided by r is this surprising to you if
you are not surprised then you should be surprised
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00:09:53,420 --> 00:09:59,810
why you should be surprised because in order
to represent a point which is originally represented
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00:09:59,810 --> 00:10:04,699
in cartesian coordinates as x and y or x by
x square plus y square and y by y square x
89
00:10:04,699 --> 00:10:10,740
square plus y square we had to give the values
of x and y both the values of x and y
90
00:10:10,740 --> 00:10:15,700
to represent the same point in cylindrical
co-ordinate system i still have to give the
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00:10:15,700 --> 00:10:20,879
values of r and phi i have to tell you what
distance that particular point is located
92
00:10:20,879 --> 00:10:26,379
and what is the angle with respect to the
x axis that i have to locate that point on
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00:10:26,379 --> 00:10:31,290
if i don’t give you r and phi you will not
be able to pin point that this point in the
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00:10:31,290 --> 00:10:35,850
cylindrical co-ordinates will correspond to
the same point in the rectangular co-ordinates
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00:10:35,850 --> 00:10:36,850
right
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00:10:36,850 --> 00:10:43,660
there are an infinite number of choices yet
my vector a which was originally ax x hat
97
00:10:43,660 --> 00:10:50,829
plus ay y hat is actually giving me component
only along r there is no mention of phi component
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00:10:50,829 --> 00:10:55,160
the phi component is zero in this case z of
course is also zero in this case does it really
99
00:10:55,160 --> 00:11:02,839
means that if phi hat is equal to zero or
if a phi is equal to zero that is any phi
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00:11:02,839 --> 00:11:06,680
component is equal to zero does this imply
that phi is equal to zero
101
00:11:06,680 --> 00:11:14,439
no phi is not zero just because there is no
phi component it does not mean that the phi
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00:11:14,439 --> 00:11:21,600
component is also equal to zero in fact the
phi component is given by inverse tan of y
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00:11:21,600 --> 00:11:26,990
by x if you know what is the value of ax and
ay or this is equivalent of writing this as
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00:11:26,990 --> 00:11:33,430
tan inverse of ay by ax if you know what is
ay and ax you should be able to find out what
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00:11:33,430 --> 00:11:40,779
is the value of phi similarly you know what
is the value of r or the ar component you
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00:11:40,779 --> 00:11:44,200
should be able to find phi and you will see
that this phi will not be zero
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00:11:44,200 --> 00:11:49,290
so this is a very important step for me to
highlight to you just because phi is equal
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00:11:49,290 --> 00:11:55,249
to zero it does not mean that the a phi component
is zero it does not mean that phi is also
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00:11:55,249 --> 00:12:01,750
equal to zero this is not correct do not think
that a phi component is zero means phi is
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00:12:01,750 --> 00:12:11,379
also equal to zero but this makes sense physically
because after all where is my vector pointing
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00:12:11,379 --> 00:12:16,839
on this particular radius let us say this
is the point where i had originally in cartesian
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00:12:16,839 --> 00:12:21,149
co-ordinate system represented as x and y
component and i had set up a particular vector
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00:12:21,149 --> 00:12:22,220
also
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00:12:22,220 --> 00:12:26,950
the corresponding vector in the cylindrical
co-ordinate would be directed from the origin
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00:12:26,950 --> 00:12:32,999
passing through this particular point right
this point no doubt is given by r and phi
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00:12:32,999 --> 00:12:37,850
values but the direction of the vector is
only along r direction because there is no
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00:12:37,850 --> 00:12:43,199
component of phi direction or there is no
component of the vector along phi direction
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00:12:43,199 --> 00:12:49,769
okay this you keep in mind it is very important
now we have seen this result let us put this
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00:12:49,769 --> 00:12:56,230
result into good use by resolving the same
problem of infinite line charge okay
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00:12:56,230 --> 00:12:57,230
.
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00:12:57,230 --> 00:13:01,720
we consider the infinite line charge problem
earlier we are going to revisit the same problem
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00:13:01,720 --> 00:13:07,639
and learn how to use the cylindrical co-ordinate
systems this is sometimes called as circular
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00:13:07,639 --> 00:13:14,309
cylindrical co-ordinates it doesn’t really
matter i have again an uniform line charge
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00:13:14,309 --> 00:13:20,319
of line charge density rho l kept on the z
axis going all the way from minus infinity
125
00:13:20,319 --> 00:13:26,260
to plus infinity i again consider a line segment
at a height of z prime from the z is equal
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00:13:26,260 --> 00:13:33,160
to zero plane starting from the horizontal
plane and i consider a small segment dz prime
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00:13:33,160 --> 00:13:43,360
the total charge there will be rho l multiplied
by dz prime okay now i have to find out the
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00:13:43,360 --> 00:13:49,899
field at any point in the plane which i am
now going to consider any general r phi and
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00:13:49,899 --> 00:13:55,420
z point okay i am not looking for z is equal
to zero although you can solve the problem
130
00:13:55,420 --> 00:14:01,040
with z is equal to zero let me consider any
z value and see what happens you will see
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00:14:01,040 --> 00:14:07,250
that the problem does not become complicated
at all so what is the unit vector for this
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00:14:07,250 --> 00:14:09,319
one
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00:14:09,319 --> 00:14:13,430
okay what is the unit vector for the field
point and what is the unit vector for the
134
00:14:13,430 --> 00:14:23,579
source point the vector for the source point
is r prime is z prime z hat this is at a height
135
00:14:23,579 --> 00:14:30,930
of z hat and what is the vector for the field
point it turns out that the vector for the
136
00:14:30,930 --> 00:14:36,319
field point has no phi component again remember
just because there is no phi component it
137
00:14:36,319 --> 00:14:45,019
does not mean that phi itself is zero so i
have this as r r hat plus z z hat
138
00:14:45,019 --> 00:14:55,399
again remember there is no phi component no
phi component does not mean phi is equal to
139
00:14:55,399 --> 00:15:00,480
zero in fact phi will be non zero we will
later see why phi does not enter into picture
140
00:15:00,480 --> 00:15:06,209
over here so what is this vector now from
the source the field point the vector is r
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00:15:06,209 --> 00:15:18,249
minus r prime which is r r hat plus z minus
z prime along z okay now we use the expression
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00:15:18,249 --> 00:15:22,610
for electric field we integrate in the appropriate
dimensions
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00:15:22,610 --> 00:15:27,930
i am looking at the electric field at a particular
point r phi z in the plane this will be equal
144
00:15:27,930 --> 00:15:33,410
to rho l by 4 pi epsilon zero i am going to
take this constant anyway outside so i am
145
00:15:33,410 --> 00:15:38,420
going to put that constant outside of the
integral first integral is going along z direction
146
00:15:38,420 --> 00:15:45,959
from minus infinity to plus infinity and inside
i have r minus r prime by magnitude of r prime
147
00:15:45,959 --> 00:15:48,019
to the power 3 right
148
00:15:48,019 --> 00:15:57,050
so i have r r hat plus z minus z prime multiplied
by z hat divided by r square by z minus z
149
00:15:57,050 --> 00:16:04,300
prime square to the power 3 by 2 and this
entire thing is actually integrated along
150
00:16:04,300 --> 00:16:10,839
z prime now i still have an integral which
have this component of 3 by 2 but i know how
151
00:16:10,839 --> 00:16:15,580
to solve that integral but what is the big
change i have done if you see what is the
152
00:16:15,580 --> 00:16:20,490
big change what has happened out of the 3
integrals one integral has dropped out
153
00:16:20,490 --> 00:16:26,579
now there are only two integrals for me to
work around however here i have to be careful
154
00:16:26,579 --> 00:16:31,610
a little bit again the integral will be the
sum of the integral for r and sum of the integral
155
00:16:31,610 --> 00:16:38,649
for z because there is an r hat inside the
integral z hat inside the integral however
156
00:16:38,649 --> 00:16:45,029
in this particular case it turns out that
at any point on the z axis i might be r hat
157
00:16:45,029 --> 00:16:50,970
does not change or does not depend on z z
also does not depend on z
158
00:16:50,970 --> 00:16:59,059
we remember that r hat and phi hat depend
only on phi and not on z they also don’t
159
00:16:59,059 --> 00:17:04,740
depend on r so in this case there is no integration
of r so r hat does not depend on z therefore
160
00:17:04,740 --> 00:17:10,170
that also becomes a constant and can easily
come out of the integral so when it comes
161
00:17:10,170 --> 00:17:16,690
out of the integral you can use the same ideas
that we develop in the last class to solve
162
00:17:16,690 --> 00:17:25,339
for this integral in the last class and you
can basically show that this integral this
163
00:17:25,339 --> 00:17:32,900
entire integral will turn out to be rho l
by 2 phi epsilon zero r in the r direction
164
00:17:32,900 --> 00:17:33,900
okay
165
00:17:33,900 --> 00:17:38,250
so the fields are in the radial direction
radially outward if it is a positive line
166
00:17:38,250 --> 00:17:44,930
charge radially inward if it a negative line
charge density okay so i have rho l by 2 pi
167
00:17:44,930 --> 00:17:50,820
epsilon zero r r hat and there is no phi component
also the integral for the z component drops
168
00:17:50,820 --> 00:17:59,270
out so this is the electric field at r phi
and z this is completely independent of z
169
00:17:59,270 --> 00:18:02,630
depends only on r and they are only going
away along r direction
170
00:18:02,630 --> 00:18:07,490
let me give you briefly why this particular
electric field has come out the way it has
171
00:18:07,490 --> 00:18:13,510
come out see we have chosen a proper co-ordinate
system but what we have not talked about is
172
00:18:13,510 --> 00:18:20,010
the symmetry of the problem symmetry is a
very very very important concept in electromagnetic
173
00:18:20,010 --> 00:18:24,720
if you look at the textbook that we are using
for this course electromagnetic engineering
174
00:18:24,720 --> 00:18:29,960
electromagnetic by hayt the way he solves
this problem the infinite line charge problem
175
00:18:29,960 --> 00:18:36,320
is that he gives you the first statement of
the solution saying that in solving such problems
176
00:18:36,320 --> 00:18:41,570
you have to look for symmetry because symmetry
simplifies calculation
177
00:18:41,570 --> 00:18:46,340
there are two things that we have to do which
are paramount important one is choosing the
178
00:18:46,340 --> 00:18:51,170
right co-ordinate system and then exploiting
the symmetry so if you choose the right co-ordinate
179
00:18:51,170 --> 00:18:56,250
system in this case instead of 3 integrals
i am now down to two integrals and if i had
180
00:18:56,250 --> 00:19:02,930
actually exploited the principle of symmetry
i will not even have worried about the z component
181
00:19:02,930 --> 00:19:07,480
let us see symmetry what does it do symmetry
asks two kinds of question
182
00:19:07,480 --> 00:19:14,370
it first says at any given particular point
which way the electric field will be pointing
183
00:19:14,370 --> 00:19:20,580
and then it will also ask on what variables
what co-ordinates does this electric field
184
00:19:20,580 --> 00:19:26,700
depends on imagine that there is no infinite
line charge out here and then you are at a
185
00:19:26,700 --> 00:19:30,430
particular distance may be you are in that
horizontal plane you are at a particular distance
186
00:19:30,430 --> 00:19:35,840
r and you look around this charge and you
move along the phi direction
187
00:19:35,840 --> 00:19:42,020
so imagine yourself moving along the phi direction
the line charge is uniformly infinitive and
188
00:19:42,020 --> 00:19:46,310
the line charge basically does not have any
dependence on the phi so as you keep moving
189
00:19:46,310 --> 00:19:51,410
along phi direction you won’t see any difference
in the line charge so the line charge looks
190
00:19:51,410 --> 00:19:57,610
the same no matter what value of phi you are
in therefore there is clearly no way the electric
191
00:19:57,610 --> 00:20:02,700
field is going to depend on phi it does not
matter if the line is infinity long or if
192
00:20:02,700 --> 00:20:04,130
the line is finitely long
193
00:20:04,130 --> 00:20:11,420
as long as the line is symmetric with respect
to the phi axis or the azimuthal direction
194
00:20:11,420 --> 00:20:17,770
there will be no dependence on the phi component
similarly for an infinite line charge we can
195
00:20:17,770 --> 00:20:23,320
move up and down there will again be no z
dependence because you move up you are going
196
00:20:23,320 --> 00:20:27,210
to see the same charge you move down you are
going to see the same charge you can move
197
00:20:27,210 --> 00:20:31,500
up and down and you are going to still see
the same uniform line charge
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therefore there cannot be any dependence on
the z component the only way you can have
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your electric field magnitude decrease is
if you move away from the charge so you have
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the charge and you keep moving away from the
charge and you can expect that as you move
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away from the charge there will be some drop
in the magnitude of the electric field so
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that is why the electric field is dependent
only on the r co-ordinate
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only along the radial distance from the charge
suppose you are at a particular point and
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now you say can there be a phi component in
the electric field the answer is no because
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to have a phi component remember electric
field is a force essentially in some sort
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of a force you can place the test charge at
a point and then see if there is a to be a
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component along phi there has to be some other
charge which has to be giving you the force
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in the phi direction
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there has to be a charge in the phi direction
but we have no such charge we have only one
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charge and that is along the z axis and that
would not give you the phi component so there
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is no phi component in the electric field
can there be a component along the z direction
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for an infinite line charge if i have to say
that okay if there is some non zero z hat
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direction it also means that there is a charge
somewhere which is giving you a field along
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z axis but that is not the case here
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the only electric charge that we have charge
distribution that i have is sitting on the
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z axis and it is sitting along all the way
from minus infinity to plus infinity if at
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this distance if this is the charge and therefore
this distance i need to have a z component
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then i need to have a force somewhere over
here which is pushing the charge test charge
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around in this upward direction or in the
downward direction right if i have a negative
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component
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so there is no z component there is no phi
component there can only be r component because
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for the r component there is a charge here
it will exert a force up there so if you had
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actually exploited all these symmetry you
would have concluded that we need not even
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have perform the integral for the z axis i
did not introduce this for a very specific
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reason there is another law which we are going
to study very shortly called as gauss’s
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law
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gauss’s law is specifically meant for exploiting
symmetries the problem that we took for two
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classes to solve two classes can actually
be solved very easily if you apply gauss’s
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law and we are going to do that one once we
get acquainted with that particular law so
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for now we are not putting any symmetry arguments
however we are only going to choose the appropriate
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co-ordinate systems that is what i have done
here
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there are couple of other problems that you
might be interested and these are very important
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as well at this point i will give them as
an exercise and therefore you can work them
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out and if time permits we will come back
to those distributions i will like to move
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on from solving these type of charge distribution
problems to introduce you to a different concept
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in our study of electromagnetics and this
concept is called potential