1
00:00:13,860 --> 00:00:20,170
now that was coordinate system which allowed
me to specify any point in space by with reference
2
00:00:20,170 --> 00:00:26,660
to a particular coordinate system by a set
of three numbers i was able to specify every
3
00:00:26,660 --> 00:00:31,660
point in space with respect to this coordinate
system of course i could have chosen any other
4
00:00:31,660 --> 00:00:36,250
coordinate system but you know that if i choose
the coordinate different system the numbers
5
00:00:36,250 --> 00:00:42,800
associated with the points will change but
the points themselves will be physically same
6
00:00:42,800 --> 00:00:46,660
the points will not change only the numbers
that you associated will change if you change
7
00:00:46,660 --> 00:00:54,570
the coordinate system okay consider associating
vectors to this coordinate system now that
8
00:00:54,570 --> 00:00:59,420
we have talked about how to associate point
in a three dimensional rectangular coordinate
9
00:00:59,420 --> 00:01:04,129
system i want to know associate vectors on
this plane or i have to describe vectors on
10
00:01:04,129 --> 00:01:12,330
this space so how do i describe vectors remember
all vectors are drawn with their origin sitting
11
00:01:12,330 --> 00:01:18,580
at the origin of the coordinate system such
that i can now specify any vector the end
12
00:01:18,580 --> 00:01:21,750
of point of any vector by 3 numbers
13
00:01:21,750 --> 00:01:27,200
so in this particular case i have a vector
r which is called as a position vector and
14
00:01:27,200 --> 00:01:33,080
this position vector has its origin at o and
it is terminating at some point which is intersection
15
00:01:33,080 --> 00:01:37,730
of 3 planes this plane that you are seeing
in the front which is slightly dark is the
16
00:01:37,730 --> 00:01:43,119
x is equal to constant plane this plane that
you are seeing here is the y is equal to constant
17
00:01:43,119 --> 00:01:48,040
plane and this plane that you are seeing here
on the top is the z is equal to constant plane
18
00:01:48,040 --> 00:01:53,020
so if you know bring all the 3 planes together
the intersection will be the point where the
19
00:01:53,020 --> 00:01:59,060
vectors r terminates so now i can draw a line
from the origin until i reach the point here
20
00:01:59,060 --> 00:02:06,530
and that will define a vector r to me in other
words given any vector whose origin is at
21
00:02:06,530 --> 00:02:10,670
the origin itself and which is terminating
at any other point in the space all i have
22
00:02:10,670 --> 00:02:16,620
to do is to set up the 3 planes so that i
can describe the end point of the vector okay
23
00:02:16,620 --> 00:02:23,510
and this vector can now be specified by just
3 numbers or i can resolve this vector into
24
00:02:23,510 --> 00:02:25,000
3 different vectors
25
00:02:25,000 --> 00:02:30,209
these 3 different vectors how do i resolve
that remember resolving a vector means you
26
00:02:30,209 --> 00:02:35,890
take a vector and then you lookay at the components
of the vector along x direction and along
27
00:02:35,890 --> 00:02:39,950
y direction this is what we did for the two
dimensional case now there is no reason why
28
00:02:39,950 --> 00:02:44,540
we cannot do this for the three dimensional
case so you take a vector okay and lookay
29
00:02:44,540 --> 00:02:49,160
at its component along the x direction lookay
at its component along the y direction and
30
00:02:49,160 --> 00:02:54,569
lookay at its component along the z direction
so you can actually resolve any vector into
31
00:02:54,569 --> 00:02:58,800
x y and z components
32
00:02:58,800 --> 00:03:02,440
so that is precisely what we are going to
do and when you resolve a vector you will
33
00:03:02,440 --> 00:03:08,900
actually be generating 3 mutually perpendicular
vectors just as in the two dimensional case
34
00:03:08,900 --> 00:03:17,380
you generated 2 mutually perpendicular vectors
here you are going to generate 3 mutually
35
00:03:17,380 --> 00:03:27,970
perpendicular vectors these vectors are x
x hat plus y y hat plus z z hat where x y
36
00:03:27,970 --> 00:03:34,020
and z without the hats are the magnitudes
or the points that are dependent on the particular
37
00:03:34,020 --> 00:03:38,569
vector r that you are considering and x hat
y hat and z hat are the unit vectors along
38
00:03:38,569 --> 00:03:43,779
x y and z directions
.
39
00:03:43,779 --> 00:03:50,120
so this is resolution of vector now we want
to add 2 vectors or subtract 2 vectors we
40
00:03:50,120 --> 00:03:54,489
have already seen how to do this graphically
but with the knowledge of coordinate systems
41
00:03:54,489 --> 00:03:59,660
is it possible for me to do all this without
drawing lines every time and without translating
42
00:03:59,660 --> 00:04:05,640
one vector onto the other vector yes it is
possible and to do that one you have to so
43
00:04:05,640 --> 00:04:08,080
this is an example in which we are going to
do that one
44
00:04:08,080 --> 00:04:15,159
you have start imagining that every point
or every point can be represented by a vector
45
00:04:15,159 --> 00:04:20,250
okay whose origin is at the origin and which
terminates at the desired point for example
46
00:04:20,250 --> 00:04:28,349
i have 2 vectors here r p and r q which are
defining 2 positions position p which is given
47
00:04:28,349 --> 00:04:34,790
by the coordinates 1 2 and 3 how do i get
to 1 2 and 3 i move 1 unit along x i move
48
00:04:34,790 --> 00:04:39,210
2 units along y and then i move 3 units along
z right
49
00:04:39,210 --> 00:04:46,159
so i get to the point p which is 1 unit along
x 2 units along y and 3 units along z so this
50
00:04:46,159 --> 00:04:53,490
point is described equivalently by a vector
r p so please note that this point in space
51
00:04:53,490 --> 00:05:00,800
which has its coordinate 1 2 3 can be equivalently
defined by this vector r p okay
52
00:05:00,800 --> 00:05:07,580
similarly if i take another position q which
is at 2 minus 2 and 1 okay i can define another
53
00:05:07,580 --> 00:05:13,879
vector r q whose origin again is situated
at the origin of the coordinate system and
54
00:05:13,879 --> 00:05:21,879
which is terminating or whose head is situated
at 2 minus 2 and 1 again 2 units along x minus
55
00:05:21,879 --> 00:05:28,330
2 units along y so you need to go to the left
so minus 2 units along y and 1 unit along
56
00:05:28,330 --> 00:05:29,330
z
57
00:05:29,330 --> 00:05:35,360
now you have a vector r p q okay which is
a vector which is directed from point p to
58
00:05:35,360 --> 00:05:42,469
point q this vector is not defined in terms
of the origin how i am going to describe this
59
00:05:42,469 --> 00:05:49,719
vector r p q here is where the vector addition
will come to my help i know that this vector
60
00:05:49,719 --> 00:05:59,189
r p plus this vector r p q is equal to the
vector r q okay this vector r p plus the vector
61
00:05:59,189 --> 00:06:03,029
r p q is equal to the vector r q
62
00:06:03,029 --> 00:06:11,039
what we said about all vectors being at the
origin means that every position is equivalently
63
00:06:11,039 --> 00:06:15,349
represented by a vector whose origin is at
the origin of the coordinate system and which
64
00:06:15,349 --> 00:06:20,889
terminates at the particular point however
the vector r p q is a vector which is not
65
00:06:20,889 --> 00:06:25,590
defined from the origin 2 it is actually defined
between the 2 points p and q and you have
66
00:06:25,590 --> 00:06:28,189
to specify that those points for this vector
67
00:06:28,189 --> 00:06:35,960
however you can specify the vector r p q in
terms of the position vectors r p and r q
68
00:06:35,960 --> 00:06:41,840
how do i do that one you start you know from
the vector addition that r p plus r p q is
69
00:06:41,840 --> 00:06:50,520
equal to r q that is r p plus r p q is equal
to r q which implies that r p q is equal to
70
00:06:50,520 --> 00:06:56,110
r q minus r p so it is a simple transferring
r p to the right hand side you get r p q is
71
00:06:56,110 --> 00:07:01,629
equal to r q minus r p and i already know
how to represent the vector r p
72
00:07:01,629 --> 00:07:07,009
the vector r p can be represented go back
here the vector r is represented by giving
73
00:07:07,009 --> 00:07:14,869
its 3 components x y and z vectors which was
x x hat y y hat and z z hat therefore r p
74
00:07:14,869 --> 00:07:22,300
can be written as x hat 1 into x hat is just
x hat plus 2 into y hat plus 3 into z hat
75
00:07:22,300 --> 00:07:30,999
similarly r q vector will be 2 x hat minus
2 y hat plus 1 z hat now if i subtract r q
76
00:07:30,999 --> 00:07:33,689
and r p i am going to get r p q
.
77
00:07:33,689 --> 00:07:38,949
how do i subtract that in a given coordinate
system a vector is represented by 3 points
78
00:07:38,949 --> 00:07:45,699
x 1 y 1 and z 1 or equivalently x 1 x hat
plus y 1 y hat plus z 1 z hat and similarly
79
00:07:45,699 --> 00:07:51,869
vector b the sum of the 2 vectors is given
by the vector whose components are simply
80
00:07:51,869 --> 00:07:57,789
added like this so you have the x component
added to get x 1 plus x 2 y component is y
81
00:07:57,789 --> 00:08:01,169
1 plus y 2 and z component is z 1 plus z 2
82
00:08:01,169 --> 00:08:05,619
similarly to get the difference between a
and b vector you have to subtract x 2 from
83
00:08:05,619 --> 00:08:11,969
x 1 y 2 from y 1 and z 2 from z 1 these 3
points define the end point of the difference
84
00:08:11,969 --> 00:08:17,349
vector a minus b so if you go back to the
previous slide r p q was a vector directed
85
00:08:17,349 --> 00:08:23,759
from p to q and that was given by r q minus
r p and you can subtract the individual components
86
00:08:23,759 --> 00:08:31,240
you are going to get a vector which is minus
x hat minus 4 y hat minus 2 z hat equivalently
87
00:08:31,240 --> 00:08:38,190
it can be represented as a vector with tail
at 0 0 0 and head at minus 1 minus 4 and minus
88
00:08:38,190 --> 00:08:39,370
2 okay
.
89
00:08:39,370 --> 00:08:45,889
this is how you are going to subtract the
2 vectors so we have seen how to define a
90
00:08:45,889 --> 00:08:51,300
vector in 3 dimensional cartesian coordinate
system so given a vector a we can resolve
91
00:08:51,300 --> 00:08:56,879
this vector in to 3 vectors which are mutually
perpendicular so when we say mutually perpendicular
92
00:08:56,879 --> 00:09:02,670
we mean by if you take pairs of vectors these
pairs of vector will be perpendicular with
93
00:09:02,670 --> 00:09:03,810
each other
94
00:09:03,810 --> 00:09:12,350
so this vector a can be resolved into 3 vectors
a x x hat plus a y y hat plus a z z hat and
95
00:09:12,350 --> 00:09:18,829
we call this a x a y and a z as the x component
of the vector a y component of the vector
96
00:09:18,829 --> 00:09:24,279
a and z component of the vector a and the
vector that we formed along the x direction
97
00:09:24,279 --> 00:09:27,379
will be the component times the vector
98
00:09:27,379 --> 00:09:32,519
now one of the things that we have seen in
coulomb's law is that we not only require
99
00:09:32,519 --> 00:09:37,990
the vector but we also require its magnitude
because the magnitude gives you the separation
100
00:09:37,990 --> 00:09:43,759
between the 2 charges so when you have 2 charges
separated by a distance r we define a unit
101
00:09:43,759 --> 00:09:46,760
vector along the line that joins the 2 charges
102
00:09:46,760 --> 00:09:51,060
but at the same time i also want to know what
is the magnitude of the separation to get
103
00:09:51,060 --> 00:09:56,149
the magnitude of the separation we need to
define the magnitude of vector we normally
104
00:09:56,149 --> 00:10:02,160
denote the magnitude of vector by putting
it in between 2 vertical lines so this denotes
105
00:10:02,160 --> 00:10:03,160
the magnitude of vector
106
00:10:03,160 --> 00:10:08,230
sometimes we simply write down the number
without vector notation that would also denote
107
00:10:08,230 --> 00:10:14,279
the magnitude of vector and this magnitude
of vector is given by square root of a x square
108
00:10:14,279 --> 00:10:20,459
plus a y square plus a z square which you
can think of as a three dimensional generalization
109
00:10:20,459 --> 00:10:23,899
of pythagoras rule
110
00:10:23,899 --> 00:10:28,639
so magnitude of the vector actually tells
you the distance of the vector from the origin
111
00:10:28,639 --> 00:10:35,279
that means at what point the head of the vector
is located with respect to the origin so the
112
00:10:35,279 --> 00:10:39,939
distance from the origin of the vector to
the tail point or the end point of the vector
113
00:10:39,939 --> 00:10:47,329
is the magnitude of the vector what is the
magnitude of the unit vector 1 so do not forget
114
00:10:47,329 --> 00:10:48,329
that
115
00:10:48,329 --> 00:10:54,720
how do i define a unit vector along a particular
direction so i have a vector a which is starting
116
00:10:54,720 --> 00:10:59,950
with the origin o and terminating at some
point in the three dimensional space how do
117
00:10:59,950 --> 00:11:05,480
i define a unit vector along that what is
a unit vector it is a vector which is directed
118
00:11:05,480 --> 00:11:10,260
at a particular direction any direction if
the vector is directed along x direction then
119
00:11:10,260 --> 00:11:15,540
it will be a unit vector x hat if it is directed
along y it would be a unit vector along y
120
00:11:15,540 --> 00:11:16,540
hat
121
00:11:16,540 --> 00:11:22,459
now if i have a unit vector directed in general
along a i can denote this as some a hat and
122
00:11:22,459 --> 00:11:29,480
what is that it is a vector so i need to give
the vector a itself but since the magnitude
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00:11:29,480 --> 00:11:34,730
of the vector has to be unit i will have to
divide this a by its magnitude it some sort
124
00:11:34,730 --> 00:11:42,470
of normalization okay so if i have 10 x hat
as the vector along x direction and if i divide
125
00:11:42,470 --> 00:11:48,000
the 10 x hat by 10 which is the magnitude
along x i will get x hat which is the unit
126
00:11:48,000 --> 00:11:49,199
vector along x
127
00:11:49,199 --> 00:11:55,310
so the same rule if i want to specify unit
vector a i specify this by writing a and putting
128
00:11:55,310 --> 00:12:00,930
a hat on top of it i can sometimes use bold
letter a with a subscript a or i can use u
129
00:12:00,930 --> 00:12:05,480
a all this different notation you will be
seeing in literature so unit vector along
130
00:12:05,480 --> 00:12:11,930
a particular direction is given by the vector
divided by its magnitude this is the unit
131
00:12:11,930 --> 00:12:18,240
vector along a vector a in its particular
chosen coordinate system okay
132
00:12:18,240 --> 00:12:24,660
here are some exercises for you i have given
you three points so one point is m at minus
133
00:12:24,660 --> 00:12:31,920
1 2 and 1 and the point n at 3 minus 3 and
0 and another point p which is minus 2 minus
134
00:12:31,920 --> 00:12:37,839
3 and minus 4 what i would like you to do
is to first find a piece of paper and draw
135
00:12:37,839 --> 00:12:44,240
the 3 axes and try to represent this points
m n and p on that piece of paper okay so i
136
00:12:44,240 --> 00:12:49,209
wanted to get some familiarity with representing
the point so take this as an exercise and
137
00:12:49,209 --> 00:12:51,779
locate the points m n and p
138
00:12:51,779 --> 00:12:58,290
now remember to every point m n or p i can
actually define a vector whose origin is at
139
00:12:58,290 --> 00:13:06,490
the origin and with whose head is at m n or
p i want you to find out this vector r m n
140
00:13:06,490 --> 00:13:13,930
what is this vector r m n it is a vector which
is directed from m to n okay and what is that
141
00:13:13,930 --> 00:13:20,819
vector given us r n minus r m this is in general
if i have a vector r i f okay it would be
142
00:13:20,819 --> 00:13:26,819
a vector which is directed from the initial
point to the final point f and then that vector
143
00:13:26,819 --> 00:13:32,329
will be given by the position vector r f minus
position vector r i
144
00:13:32,329 --> 00:13:38,050
so r m n i want you to find out similarly
i want you to find out r m p and then add
145
00:13:38,050 --> 00:13:44,810
to this r m p to r m n need to do these both
graphically as well as using the coordinate
146
00:13:44,810 --> 00:13:50,339
system values okay finally find the magnitude
of the vector m also find the unit vector
147
00:13:50,339 --> 00:13:55,240
along the direction m okay how do i find the
unit vector along the direction m you know
148
00:13:55,240 --> 00:14:00,309
what is the vector m and divide by its magnitude
and what is the magnitude you already found
149
00:14:00,309 --> 00:14:03,069
that out in the previous exercise okay
150
00:14:03,069 --> 00:14:11,139
so you do all this exercise to gain some familiarity
with using the vectors okay so we have taken
151
00:14:11,139 --> 00:14:17,759
a large . . from coulomb's law we were discussing
coulomb's law and then we started discussing
152
00:14:17,759 --> 00:14:23,029
vector analysis went to coordinate system
all this was required because electromagnetic
153
00:14:23,029 --> 00:14:27,519
is essentially a three dimensional subject
okay you need to know vector analysis you
154
00:14:27,519 --> 00:14:32,829
need to know how to setup coordinate systems
if you want to have good success in electromagnetics
155
00:14:32,829 --> 00:14:34,069
courses okay
.
156
00:14:34,069 --> 00:14:38,579
so we will now come back to coulomb's law
and try to work out a problem hopefully whatever
157
00:14:38,579 --> 00:14:44,209
that we have learnt in the vector analysis
and coordinate systems will come to our help
158
00:14:44,209 --> 00:14:51,329
and help us find the solution to this particular
problem what is the problem it is essentially
159
00:14:51,329 --> 00:14:57,730
a simple problem we have an origin okay and
we have 2 charges one charge at q 1 the other
160
00:14:57,730 --> 00:15:06,490
charge is at q 2 the charge q 1 is 200 micro
coulomb's micro is a prefix which means that
161
00:15:06,490 --> 00:15:08,870
10 to the power minus 6
162
00:15:08,870 --> 00:15:14,220
so this is 200 into 10 to the power minus
6 coulomb's or 200 micro coulomb's and this
163
00:15:14,220 --> 00:15:22,819
q 1 is located at point 1 0 3 see you can
go 1 unit along x so here in this particular
164
00:15:22,819 --> 00:15:28,220
figure i have rotated the axis to show you
everything hopefully in a clarified manner
165
00:15:28,220 --> 00:15:34,879
this is the x axis this is the y axis and
this height is the z axis so to locate point
166
00:15:34,879 --> 00:15:41,790
1 0 3 i have to move 1 unit along x there
is no movement along y so y is equal to 0
167
00:15:41,790 --> 00:15:50,120
now i have to move along z to z is equal to
3 to reach point q 1 or charge q 1 whose value
168
00:15:50,120 --> 00:15:56,309
is 200 micro coulomb please note that this
is the positively charged particle which i
169
00:15:56,309 --> 00:16:02,759
have put here okay similarly we have a charge
q 2 with a value of 30 micro coulomb's okay
170
00:16:02,759 --> 00:16:09,839
it is also positively charged and it is located
at 0 2 1 so 0 means x is equal to 0 so you
171
00:16:09,839 --> 00:16:15,579
are now lookaying only at the intersection
of y and z planes that is the point 0 2 1
172
00:16:15,579 --> 00:16:17,759
is in the y z plane okay
173
00:16:17,759 --> 00:16:23,500
so you can see here that i have to move 2
units to y and 1 unit to z in order to reach
174
00:16:23,500 --> 00:16:31,980
point q 2 now the question is q 1 is exerting
some force on q 2 find the force at q 2
175
00:16:31,980 --> 00:16:36,809
so coulomb's law will help us find this but
coulomb's law requires us to give both magnitude
176
00:16:36,809 --> 00:16:42,110
as well as direction magnitude is the easier
part okay direction is something that we want
177
00:16:42,110 --> 00:16:49,810
to understand if i denote the point where
the charge q 1 is placed by a vector r 1 and
178
00:16:49,810 --> 00:16:55,579
if i denote the point where the charge q 2
is placed by a vector r 2 then i know that
179
00:16:55,579 --> 00:17:00,480
because coulomb's law also tells me that the
force will act along the line that joins the
180
00:17:00,480 --> 00:17:05,890
2 charges q 1 and q 2 right it acts along
the line that joins the charges q 1 and q
181
00:17:05,890 --> 00:17:06,890
2
182
00:17:06,890 --> 00:17:14,270
i define r 1 2 as a vector which is starting
at q 1 and ending at q 2 okay in this direction
183
00:17:14,270 --> 00:17:22,190
is the force f 2 okay so i have r 1 position
vector for charge q 1 r 2 as the position
184
00:17:22,190 --> 00:17:29,100
vector for charge q 2 and r 1 2 is the vector
starting from q 1 to q 2 and i know clearly
185
00:17:29,100 --> 00:17:36,179
that r 1 2 is given by r 2 minus r 1 where
r 2 is the position vector of charge q 2 r
186
00:17:36,179 --> 00:17:42,520
1 is the position vector of charge q 1 why
because remember r i f i is the initial point
187
00:17:42,520 --> 00:17:47,610
f is the final point so the vector that would
be described by r i f is actually given by
188
00:17:47,610 --> 00:17:54,559
the position vector of f minus position vector
of i so r f minus r i okay
189
00:17:54,559 --> 00:18:00,300
so what is r 1 r 1 is the point which corresponds
to charge q 1 which is 1 0 3 this is in the
190
00:18:00,300 --> 00:18:05,100
x z plane so you can see that this is how
x will increase this is how z will increase
191
00:18:05,100 --> 00:18:13,299
so this plane is the x z plane and in this
x z plane r 1 is given by x hat plus 3 z hat
192
00:18:13,299 --> 00:18:18,590
similarly r 2 is in the y and z plane okay
193
00:18:18,590 --> 00:18:26,490
so r 2 is given by 2 y hat plus z hat and
r 1 2 is given by r 2 minus r 1 remember how
194
00:18:26,490 --> 00:18:31,710
to do r 2 minus r 1 i have to lookay out the
components the y component of r 2 subtract
195
00:18:31,710 --> 00:18:37,679
the y component of r 1 from the y component
of r 2 subtract the z component of r 1 with
196
00:18:37,679 --> 00:18:44,380
z component of r 2 subtract the x component
of r 1 with x component of r 2 okay you can
197
00:18:44,380 --> 00:18:49,630
do this subtraction and you will see that
r 1 2 is a vector that is in the three dimensional
198
00:18:49,630 --> 00:18:51,590
it has all the 3 nonzero components
199
00:18:51,590 --> 00:18:58,290
it is given by minus 1 x hat plus 2 y hat
minus 2 z hat this minus 1 is just for convenience
200
00:18:58,290 --> 00:19:04,789
i have written this is minus x hat plus 2
y hat minus 2 z hat and that is this vector
201
00:19:04,789 --> 00:19:11,750
r 1 2 now force acts along the line is fine
but what we want is not the force acting along
202
00:19:11,750 --> 00:19:16,290
that one what i want is not the statement
that force is acting along the line r 1 2
203
00:19:16,290 --> 00:19:19,360
what i want is a unit vector along the line
r 1 2
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00:19:19,360 --> 00:19:24,740
what is the unit vector along the line r 1
2 that will be the vector r 1 2 divided by
205
00:19:24,740 --> 00:19:30,360
its magnitude i have already found out what
is r 1 2 the magnitude of the vector r 1 2
206
00:19:30,360 --> 00:19:37,190
is its x component square plus y component
square plus z component square all under root
207
00:19:37,190 --> 00:19:42,470
okay and if you do that one you will get minus
1 square plus 2 square plus minus 2 square
208
00:19:42,470 --> 00:19:50,100
which will be 3 so the unit vector along r
1 2 that is along this vector r 1 2 is 1 by
209
00:19:50,100 --> 00:19:53,230
3 minus x hat plus 2 y hat minus 2 z hat
.
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00:19:53,230 --> 00:19:59,590
i hope that you derive this one for yourself
and convince yourself that this is correct
211
00:19:59,590 --> 00:20:06,190
okay so coming back to the force part i have
already found out the unit vector so now i
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00:20:06,190 --> 00:20:10,730
know that if i take the magnitude of the force
and then multiply it by the unit vector along
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00:20:10,730 --> 00:20:16,850
the line r 1 2 i will get the full vector
force f 2 right a vector force f 2 i will
214
00:20:16,850 --> 00:20:17,850
get
215
00:20:17,850 --> 00:20:22,929
what is the magnitude of the vector force
f 2 the magnitude of the vector is 1 by 4
216
00:20:22,929 --> 00:20:28,570
pi epsilon 0 r 1 2 square why epsilon 0 because
there was no specification on the medium so
217
00:20:28,570 --> 00:20:33,580
you just take the medium to be air or free
space and we know that for air or free space
218
00:20:33,580 --> 00:20:39,289
epsilon r is equal to 1 so i have 1 by 4 pi
epsilon 0 r 1 2 square
219
00:20:39,289 --> 00:20:45,070
now here is an approximation that you can
use frequently 1 by 4 pi epsilon not can be
220
00:20:45,070 --> 00:20:50,200
approximated as 9 into 10 to the power 9 because
epsilon not is in the order of 10 to the power
221
00:20:50,200 --> 00:20:56,670
minus 12 and when you multiply 4 pi and the
corresponding value for epsilon not this is
222
00:20:56,670 --> 00:21:01,649
an approximate value for 1 by 4 pi epsilon
0 we can put that approximation
223
00:21:01,649 --> 00:21:06,240
if you have a good calculator you do not have
to do this approximation but if you decide
224
00:21:06,240 --> 00:21:11,320
to do this approximation you can just write
it over here multiply the magnitude of the
225
00:21:11,320 --> 00:21:17,220
charges remember the force on a charge q 2
is proportional to the product of the charges
226
00:21:17,220 --> 00:21:23,980
charge magnitude q 1 and q 2 and inversionally
proportional to the square of the distance
227
00:21:23,980 --> 00:21:25,269
between the 2
228
00:21:25,269 --> 00:21:31,899
the distance between the 2 is 3 square that
is r 1 2 square so you are going to get 6
229
00:21:31,899 --> 00:21:37,320
newton’s okay this is the magnitude of the
force however in terms of the vector i need
230
00:21:37,320 --> 00:21:43,230
to multiply the magnitude by the unit vector
so this is the unit vector in brackets and
231
00:21:43,230 --> 00:21:49,669
i am multiplying that by the magnitude so
this is the force f 2 that is acting on charge
232
00:21:49,669 --> 00:21:51,850
q 2 because of charge q 1
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00:21:51,850 --> 00:21:57,440
now that you have seen this example re-lookay
at the example once more understand all the
234
00:21:57,440 --> 00:22:03,470
steps that are there in this example and then
try your hand at finding the force exerted
235
00:22:03,470 --> 00:22:09,019
by a charge q 1 which is placed at the origin
of three dimensional coordinate system 0 0
236
00:22:09,019 --> 00:22:16,740
0 and has a value of 100 nano coulomb’s
on charge q 2 which is placed at 4 3 and 0
237
00:22:16,740 --> 00:22:21,240
and with the value of 3 nano coulomb’s and
you can verify that the answer is approximately
238
00:22:21,240 --> 00:22:29,620
864 x hat plus 648 y hat nano newton it is
a very small value because the charge magnitudes
239
00:22:29,620 --> 00:22:35,390
are also quite small and the distance between
the 2 charges are quite large okay
240
00:22:35,390 --> 00:22:41,169
now before leaving this exercise you ask yourself
one question does it make sense does this
241
00:22:41,169 --> 00:22:46,320
answer make sense well the answer makes sense
because lookay at where the charge q 1 is
242
00:22:46,320 --> 00:22:53,720
placed it is at origin 0 0 0 okay now charge
q 2 is placed in x y plane only it is placed
243
00:22:53,720 --> 00:23:00,250
at 4 and 3 its z component is 0 so the force
that must be acting on q 2 must originate
244
00:23:00,250 --> 00:23:06,220
from that origin because the q 1 charges of
the origin and will be directed to q 2 and
245
00:23:06,220 --> 00:23:08,580
this force will only be in the x y plane
246
00:23:08,580 --> 00:23:14,000
there is no chance for the force to be in
the z plane for this problem okay so in this
247
00:23:14,000 --> 00:23:20,660
problem you expect only the x and y components
for the solution and that make sense you have
248
00:23:20,660 --> 00:23:25,700
to always do a sanity check on whatever the
answers that you find so that you are seeing
249
00:23:25,700 --> 00:23:30,730
that everything what you have done is correct
so this sanity check if you keep doing it
250
00:23:30,730 --> 00:23:35,950
you will not make mistakes while deriving
the answer okay
251
00:23:35,950 --> 00:23:41,539
i want to now introduce coulomb’s law and
change the notation that we use slightly so
252
00:23:41,539 --> 00:23:48,549
previously we had use a r 1 2 or a r as a
unit vector but when we go to different charge
253
00:23:48,549 --> 00:23:53,789
distributions it becomes easier to use a different
notation there is a reason why we go to other
254
00:23:53,789 --> 00:23:59,000
different notation i will tell you when we
go to that notation later we keep some source
255
00:23:59,000 --> 00:24:06,070
charges okay these charges i am still labeling
them as q 1 but i am also specifying the position
256
00:24:06,070 --> 00:24:07,529
of the source charge
.
257
00:24:07,529 --> 00:24:14,740
the source charge q 1 is at position r prime
r prime is a position vector from the origin
258
00:24:14,740 --> 00:24:20,370
to the source point and the test point okay
previously what we considered as charge q
259
00:24:20,370 --> 00:24:27,130
2 which is defined by the position vector
r is what i am calling as a test charge okay
260
00:24:27,130 --> 00:24:33,740
so this is known as the source point this
r without any prime is known as the field
261
00:24:33,740 --> 00:24:37,730
point this is the point where i am interested
in finding the force okay
262
00:24:37,730 --> 00:24:42,149
you will soon see why this is called field
points but for now just take the fact that
263
00:24:42,149 --> 00:24:47,870
the position vectors without a prime are called
as field points position vector with prime
264
00:24:47,870 --> 00:24:55,960
are called source points so this part has
not change all that we have done is to identify
265
00:24:55,960 --> 00:25:01,019
the position of the charges with q 1 is at
r prime q 2 is at r
266
00:25:01,019 --> 00:25:06,750
what is the unit vector unit vector is the
one that joins r prime and r so it is a vector
267
00:25:06,750 --> 00:25:11,380
that is directed from r prime that is from
the source point to the field point and we
268
00:25:11,380 --> 00:25:18,169
know that vector is given by r minus r prime
divided by its magnitude okay the magnitude
269
00:25:18,169 --> 00:25:25,179
of the vector is r minus r prime magnitude
the separation between the 2 vectors is magnitude
270
00:25:25,179 --> 00:25:31,159
square magnitude square part will make r minus
r prime square but gets multiplied by r minus
271
00:25:31,159 --> 00:25:37,840
r prime so as to give you r minus r prime
q okay so keep this in mind
272
00:25:37,840 --> 00:25:42,289
now you will frequently find as we go to the
next lecture you will frequently find that
273
00:25:42,289 --> 00:25:46,529
you are dealing with only one point charge
or it dealing with one charge distribution
274
00:25:46,529 --> 00:25:52,710
located at a point in space you will be dealing
with many charges or a continuous distribution
275
00:25:52,710 --> 00:25:58,070
of the charge so for that case we have know
how to obtain the force when you have more
276
00:25:58,070 --> 00:26:04,200
than one charge so let us i have 10 different
charges how do i obtain the force the answer
277
00:26:04,200 --> 00:26:06,940
comes in the form of superposition of forces
.
278
00:26:06,940 --> 00:26:13,710
what is this superposition of the forces fix
the test charge location and do not change
279
00:26:13,710 --> 00:26:19,529
that one so let the test charge be located
at the position r and we will call s q t just
280
00:26:19,529 --> 00:26:25,860
that subscript of t indicates to you that
this is a test charge so i have q t of r r
281
00:26:25,860 --> 00:26:30,899
standing for the position vector of the field
point where i am keeping the test charge now
282
00:26:30,899 --> 00:26:32,850
the source charges could be placed anywhere
283
00:26:32,850 --> 00:26:39,669
this placing of the charges anywhere i am
denoting this by r j prime where j corresponds
284
00:26:39,669 --> 00:26:47,750
to the j th charge of magnitude q j or sign
charge magnitude q j and what i want to find
285
00:26:47,750 --> 00:26:53,850
out is the force experience by the test charge
and this force will be given by the vector
286
00:26:53,850 --> 00:26:58,580
sum of this is vector sum this is not a scalar
addition this is the vector sum because individual
287
00:26:58,580 --> 00:27:01,260
forces are vectors
288
00:27:01,260 --> 00:27:07,260
what is this f 1 t mean it is the force because
of the first charge on the test charge force
289
00:27:07,260 --> 00:27:14,940
f 2 t is the force of the second source charge
on the test charge and so on and how are these
290
00:27:14,940 --> 00:27:20,019
forces obtained you simply break up the charge
or you simply find out the force because of
291
00:27:20,019 --> 00:27:25,280
each of these source charges and then sum
together and summing is happening only on
292
00:27:25,280 --> 00:27:31,220
the source charge distribution okay that is
why the summation sign with the j is applied
293
00:27:31,220 --> 00:27:34,110
to the source charge not to the field point
294
00:27:34,110 --> 00:27:39,779
so in fact you can pull this field point outside
and whatever that you are left with when you
295
00:27:39,779 --> 00:27:45,259
multiplied by q t of r is going to give you
the force by that j th charge on the test
296
00:27:45,259 --> 00:27:54,070
charge q t okay so this is the force of the
j th charge on the test charge and if you
297
00:27:54,070 --> 00:27:59,350
sum all these forces you are going to get
this particular expression okay we will not
298
00:27:59,350 --> 00:28:03,880
solve any problem with superposition of forces
we will in fact solve at slightly different
299
00:28:03,880 --> 00:28:06,379
problem of continuous charge distribution
later
300
00:28:06,379 --> 00:28:07,379
.
301
00:28:07,379 --> 00:28:12,059
here is where i want to talk about line surface
and volume elements and kind of finish this
302
00:28:12,059 --> 00:28:16,850
cartesian coordinate system for now we will
be requiring only cartesian coordinate system
303
00:28:16,850 --> 00:28:22,769
suppose i have a two dimensional plane a two
dimensional coordinate system this is the
304
00:28:22,769 --> 00:28:27,940
x axis think of the green board here or a
black board in your class as an example of
305
00:28:27,940 --> 00:28:34,090
a plane we have x axis and perpendicular line
which is y axis to point where these two meet
306
00:28:34,090 --> 00:28:36,309
is the origin o
.
307
00:28:36,309 --> 00:28:43,919
now if i move a certain distance say d x along
the x axis i will actually be generating a
308
00:28:43,919 --> 00:28:50,659
vector whose head or whose tail or the origin
lies at 0 and whose head lies at d x that
309
00:28:50,659 --> 00:28:59,159
is this particular vector that i have generated
by moving along x direction is d x x hat i
310
00:28:59,159 --> 00:29:06,710
could now move vertically to generate another
vector which is d y y hat of course this d
311
00:29:06,710 --> 00:29:11,580
y y hat is actually parallelly translated
from this particular vector right so you need
312
00:29:11,580 --> 00:29:13,559
to remember that one okay
313
00:29:13,559 --> 00:29:20,039
these vectors that i have generated are called
as the line elements this is called as the
314
00:29:20,039 --> 00:29:25,250
line element along x direction and this line
element along x direction is given by the
315
00:29:25,250 --> 00:29:32,659
distance that i have moved along x axis similarly
i have the line element along y which is the
316
00:29:32,659 --> 00:29:40,899
distance i have moved along the y axis this
is d y y hat okay if i now consider a general
317
00:29:40,899 --> 00:29:48,590
direction of movement i can write down this
as some d l okay and this vector d l which
318
00:29:48,590 --> 00:29:53,929
represents a vector from the origin to this
particular point call this as point p okay
319
00:29:53,929 --> 00:30:03,000
can be now decomposed into two components
d l x plus d l y which is nothing but d x
320
00:30:03,000 --> 00:30:11,909
x hat plus d y y hat what is the magnitude
of this d l vector the magnitude is d x square
321
00:30:11,909 --> 00:30:21,070
plus d y square under root the magnitude of
the d l vector is d x square plus d y square
322
00:30:21,070 --> 00:30:29,010
under root okay these are called as line elements
you will be seeing line elements and line
323
00:30:29,010 --> 00:30:35,830
integrals very soon so this is the simplest
line element that we have considered okay
324
00:30:35,830 --> 00:30:48,170
now we will consider surface elements for
this i will be going from two-dimension to
325
00:30:48,170 --> 00:30:53,639
three-dimension because with two-dimensions
i can only define one surface area with three-dimensions
326
00:30:53,639 --> 00:30:58,740
i can actually define three surface areas
okay so in accordance with the right hand
327
00:30:58,740 --> 00:31:04,850
rule i have this as the x axis this is as
the y axis and this becomes my z axis
328
00:31:04,850 --> 00:31:05,850
.
329
00:31:05,850 --> 00:31:10,419
you can imagine that this is the object that
i am considering the line that you are seeing
330
00:31:10,419 --> 00:31:14,870
along this one will be the x axis or maybe
we can consider this way the line that you
331
00:31:14,870 --> 00:31:21,029
are seeing here along this is the x axis this
would be the y axis and if you move from x
332
00:31:21,029 --> 00:31:25,389
to y you are going to see the direction along
z axis okay
333
00:31:25,389 --> 00:31:31,070
now if i ask you what is the area of this
particular phase the phase that you are seeing
334
00:31:31,070 --> 00:31:35,920
what is the area of this phase to get the
area of a phase i need to know its length
335
00:31:35,920 --> 00:31:42,230
and width what is this length some length
let us say but this if i am assuming that
336
00:31:42,230 --> 00:31:51,710
all these lengths are some d y d x and d z
so the phase has an area of d y so as you
337
00:31:51,710 --> 00:32:07,240
move along y you have d y and then d z right
so
338
00:32:07,240 --> 00:32:16,000
if i move a distance d z in this direction
along z and if i move a distance d y i will
339
00:32:16,000 --> 00:32:20,919
get the area d y d z that would be the area
d y d z
340
00:32:20,919 --> 00:32:30,919
so i have an area which is d s x i will call
this as d y d z this is the front surface
341
00:32:30,919 --> 00:32:37,710
area in which i have moved d y and i moved
d z now i want to associate a vector to this
342
00:32:37,710 --> 00:32:43,980
phase how do i associate a vector to this
phase i will simply multiply this by a vector
343
00:32:43,980 --> 00:32:51,629
x how is vector x pointing vector x is pointing
away from the surface right in fact this vector
344
00:32:51,629 --> 00:33:02,700
is perpendicular to this surface area so this
is how we associate a vector surface area
345
00:33:02,700 --> 00:33:03,700
okay
346
00:33:03,700 --> 00:33:10,750
we associate a vector surface area by giving
the area and then considering a direction
347
00:33:10,750 --> 00:33:17,120
that is perpendicular to that area or in mathematical
language we called this as normal to this
348
00:33:17,120 --> 00:33:25,750
surface area so this vector x is normal to
y z the surface area d y d z so this is the
349
00:33:25,750 --> 00:33:33,690
vector surface area along x similarly if i
go d x and then move d y along this one i
350
00:33:33,690 --> 00:33:40,029
am going to get the area of this top surface
or the bottom surface okay i will call that
351
00:33:40,029 --> 00:33:46,269
as d s z and say this is d x d y and z direction
352
00:33:46,269 --> 00:33:51,490
of course i could i have chosen the direction
of x for the front surface to be going inside
353
00:33:51,490 --> 00:33:57,190
which would be opposite to the x direction
but we always use the surface areas again
354
00:33:57,190 --> 00:34:03,960
as similar to the right-handed rule the right-handed
rule tells you that you go from x to y you
355
00:34:03,960 --> 00:34:10,950
will move along z similarly d s x will be
d y d z and move along positive x direction
356
00:34:10,950 --> 00:34:17,090
so d s z will be a vector surface area of
this top surface or the bottom surface pointing
357
00:34:17,090 --> 00:34:19,210
along the z direction okay
358
00:34:19,210 --> 00:34:25,800
so this particular bottom surface or the top
surface is the d s z here it would be pointing
359
00:34:25,800 --> 00:34:31,130
along z direction here it would be pointing
along x direction can you figure out what
360
00:34:31,130 --> 00:34:37,100
would be the vector area for the sides yes
you can easily figure this out this would
361
00:34:37,100 --> 00:34:46,520
be d s y and d s y will be d x so you move
now along like this d x and then you move
362
00:34:46,520 --> 00:34:57,360
d z so it would be d x d z along y direction
so these are the 3 surface elements or surface
363
00:34:57,360 --> 00:34:58,950
areas that you are going to see
364
00:34:58,950 --> 00:35:05,630
this is the magnitude of the surface area
and this is the vector that would be perpendicular
365
00:35:05,630 --> 00:35:11,840
to the plane that contains this 2 elements
okay d y is the direction along y d z is the
366
00:35:11,840 --> 00:35:16,040
direction along z in which you have moved
or distances that you have moved and a vector
367
00:35:16,040 --> 00:35:20,530
surface area will be something that will be
perpendicular to the 2
368
00:35:20,530 --> 00:35:25,920
what would be the volume element that is what
is the differential volume of this box while
369
00:35:25,920 --> 00:35:31,130
you have length you have width and then you
have height so this is nothing but d x d y
370
00:35:31,130 --> 00:35:38,310
d z is there area associated with that thankfully
no so the volume element is the scalar which
371
00:35:38,310 --> 00:35:49,520
is d x d y and d z and gives you the area
of this particular cuboid