0
1
00:00:01,733 --> 00:00:05,400
this is election No. 23 of a V232
1

2
00:00:05,666 --> 00:00:07,566
electromagnetic field theory
2

3
00:00:07,866 --> 00:00:10,866
today we will study matter of images
3

4
00:00:10,933 --> 00:00:14,666
and this will cover your section 6.6 of textbook
4

5
00:00:14,766 --> 00:00:17,700
this is Atomic week No. 14 of your semester
5

6
00:00:20,200 --> 00:00:22,766
the learning objectives of today's lecture would
6

7
00:00:22,766 --> 00:00:25,900
be to understand a method of images
7

8
00:00:26,200 --> 00:00:30,500
to find out the electric potential and electric field
8

9
00:00:30,866 --> 00:00:34,166
so basically in this uh lecture uh
9

10
00:00:34,166 --> 00:00:36,700
we will cover every way to find out the electric
10

11
00:00:36,700 --> 00:00:39,166
potential electric free landing uh
11

12
00:00:39,166 --> 00:00:42,500
forces uh being produced and active upon uh
12

13
00:00:42,500 --> 00:00:46,900
uh to charges and charge distribution on the other uh
13

14
00:00:46,900 --> 00:00:48,033
charge entities
14

15
00:00:50,333 --> 00:00:52,166
uh the matter of images uh
15

16
00:00:52,166 --> 00:00:56,266
was introduced by Lord Calvin in 18 uh 48 uh
16

17
00:00:56,266 --> 00:01:00,066
and is commonly used to determine the potential
17

18
00:01:00,333 --> 00:01:03,300
electric feed potential electrical
18

19
00:01:03,600 --> 00:01:04,933
electric flux density
19

20
00:01:04,933 --> 00:01:08,166
and corresponding lead surface channel density
20

21
00:01:08,566 --> 00:01:11,833
due to the charges in the presence of the conductors
21

22
00:01:11,966 --> 00:01:14,166
so always remember that
22

23
00:01:14,166 --> 00:01:17,700
the condition for the application of this method
23

24
00:01:17,700 --> 00:01:20,966
of images is only in the presence of conductors
24

25
00:01:20,966 --> 00:01:24,666
or frequently it can be the ground as well
25

26
00:01:24,666 --> 00:01:26,800
because at certain frequencies
26

27
00:01:26,900 --> 00:01:29,800
the grounds are also considered as the protect
27

28
00:01:29,800 --> 00:01:30,966
electric conductors
28

29
00:01:31,933 --> 00:01:33,000
but this method
29

30
00:01:33,533 --> 00:01:35,600
so I mean the the earlier
30

31
00:01:36,200 --> 00:01:38,133
earlier matters that we have already covered
31

32
00:01:38,133 --> 00:01:40,400
the presence or left mass equation
32

33
00:01:40,400 --> 00:01:41,433
can be avoided
33

34
00:01:41,466 --> 00:01:45,400
by realizing the fact that conducting surface in a
34

35
00:01:45,400 --> 00:01:47,500
is m equi potential body
35

36
00:01:47,500 --> 00:01:50,700
so we have already discussed that how this uh uh
36

37
00:01:50,700 --> 00:01:52,466
this uh connective service
37

38
00:01:52,466 --> 00:01:55,700
it is considered to be a equi potential body
38

39
00:01:56,100 --> 00:01:57,066
uh this matter
39

40
00:01:57,066 --> 00:01:59,600
uh doesn't apply to all electrostatic problems
40

41
00:01:59,600 --> 00:02:03,566
but it can reduce a formidable problem to a simple one
41

42
00:02:03,566 --> 00:02:04,966
so in in case
42

43
00:02:04,966 --> 00:02:06,366
uh the charges
43

44
00:02:06,366 --> 00:02:08,733
they are in the vicinity of your conductors
44

45
00:02:08,733 --> 00:02:10,666
or ground services
45

46
00:02:10,666 --> 00:02:14,066
then this method can be utilize effectively
46

47
00:02:14,066 --> 00:02:16,600
to find out the resultant potential electric
47

48
00:02:16,600 --> 00:02:18,900
Freeland electrical fractions
48

49
00:02:20,500 --> 00:02:22,800
let's start this method
49

50
00:02:23,500 --> 00:02:25,066
given charge distribution
50

51
00:02:25,066 --> 00:02:28,900
a congregation above an infinite grounded
51

52
00:02:29,066 --> 00:02:29,966
perfect
52

53
00:02:30,533 --> 00:02:31,766
conducting plane
53

54
00:02:31,966 --> 00:02:34,700
may be replaced by the child configuration
54

55
00:02:34,700 --> 00:02:36,633
configuration itself No. 1
55

56
00:02:37,166 --> 00:02:40,766
image No. 2 and an equal potential surface
56

57
00:02:40,766 --> 00:02:43,200
in place of the conducting plane
57

58
00:02:43,333 --> 00:02:44,800
let's see it one by one
58

59
00:02:45,400 --> 00:02:48,966
so the examples for these chart entities are the
59

60
00:02:48,966 --> 00:02:51,166
No. 1 is the is your point source
60

61
00:02:51,166 --> 00:02:53,866
okay so let's consider these point sources
61

62
00:02:53,866 --> 00:02:57,066
they are present in the vicinity of this
62

63
00:02:57,533 --> 00:02:59,133
perfectly conducting plane
63

64
00:02:59,133 --> 00:03:00,766
which is at potential zero
64

65
00:03:00,766 --> 00:03:02,966
that is also known as your grounded plane
65

66
00:03:03,000 --> 00:03:06,466
that is this one and it is of infinite
66

67
00:03:06,466 --> 00:03:09,366
remember infinite uh extent
67

68
00:03:10,066 --> 00:03:13,200
so in case if it is a point charge
68

69
00:03:13,200 --> 00:03:14,400
then it's equivalent
69

70
00:03:14,400 --> 00:03:17,666
image is replaced with the same magnitude
70

71
00:03:17,666 --> 00:03:20,833
but with d opposite polarity
71

72
00:03:21,300 --> 00:03:24,366
if it is a line charge distribution like this one
72

73
00:03:24,366 --> 00:03:27,200
then is image it is mirrored with the
73

74
00:03:27,366 --> 00:03:28,000
you can see here
74

75
00:03:28,000 --> 00:03:32,533
it is mirrored with the same magnitude of the line
75

76
00:03:32,533 --> 00:03:35,200
charge density but with opposite polarity
76

77
00:03:35,800 --> 00:03:39,166
and if it is the surface identity
77

78
00:03:39,166 --> 00:03:40,166
for example in this case
78

79
00:03:40,166 --> 00:03:41,966
it is given as a negative surface
79

80
00:03:41,966 --> 00:03:43,633
volume volume chart density
80

81
00:03:43,733 --> 00:03:45,166
if it is a volume chart density
81

82
00:03:45,166 --> 00:03:48,533
then in this case it will be again better as a
82

83
00:03:48,533 --> 00:03:52,033
you can see see here that it is equally be sphere
83

84
00:03:52,100 --> 00:03:52,800
mirror sphere
84

85
00:03:52,800 --> 00:03:56,400
but with the opposite quality volume chart density
85

86
00:03:56,933 --> 00:03:59,366
so the examples now we have seen that their pointline
86

87
00:03:59,366 --> 00:04:02,500
volume chart densities they can be surfaced as well
87

88
00:04:02,900 --> 00:04:05,833
about what effect conducting grounded pain
88

89
00:04:06,533 --> 00:04:07,566
given in this figure
89

90
00:04:07,566 --> 00:04:09,366
and correspondently image contributions
90

91
00:04:09,366 --> 00:04:12,633
you can see here that images have been replaced
91

92
00:04:13,066 --> 00:04:16,566
at the same distance at the same distance for example
92

93
00:04:16,566 --> 00:04:20,200
it is each about the this perfect conducting pain
93

94
00:04:20,200 --> 00:04:23,266
so it is at the same distance below this
94

95
00:04:24,533 --> 00:04:26,300
perfect land connecting pain
95

96
00:04:26,366 --> 00:04:29,533
and in place of this perfect connecting pain
96

97
00:04:29,533 --> 00:04:30,133
we have
97

98
00:04:30,133 --> 00:04:33,566
we have replaced that with equal potential surface
98

99
00:04:33,700 --> 00:04:34,900
and what is the charge of
99

100
00:04:34,900 --> 00:04:36,566
what is the potential of that surface
100

101
00:04:36,566 --> 00:04:39,266
for this particular case since it is a grounded one
101

102
00:04:39,266 --> 00:04:41,833
so it is taken as zero potential
102

103
00:04:41,866 --> 00:04:45,266
so that is the three steps we are going to do that
103

104
00:04:45,266 --> 00:04:48,066
first one that we are going to uh
104

105
00:04:48,066 --> 00:04:50,433
consider what is the actual child distribution
105

106
00:04:50,466 --> 00:04:52,533
then we are going to place uh
106

107
00:04:52,533 --> 00:04:55,366
a mirror and the opposite bloody charger uh
107

108
00:04:55,500 --> 00:04:58,966
distribution at the same distance below that uh
108

109
00:04:59,100 --> 00:05:00,700
uh perfect connectic plane
109

110
00:05:00,700 --> 00:05:03,300
and the third one is we're going to replace that
110

111
00:05:03,300 --> 00:05:05,766
perfect connectic plane with the equal potential
111

112
00:05:05,766 --> 00:05:08,500
surface of having same potential
112

113
00:05:10,566 --> 00:05:12,066
important considerations
113

114
00:05:12,066 --> 00:05:14,833
in order to apply this image matter is
114

115
00:05:15,100 --> 00:05:15,900
No. 1
115

116
00:05:15,933 --> 00:05:20,200
image charges must be located in the conductive region
116

117
00:05:20,333 --> 00:05:21,766
so this means that uh
117

118
00:05:21,900 --> 00:05:24,666
they must be in the vicinity of the conducting
118

119
00:05:24,666 --> 00:05:26,866
perfect conducting uh playing or reading
119

120
00:05:26,866 --> 00:05:29,966
something like this to satisfy me by John Secretion
120

121
00:05:30,466 --> 00:05:33,500
the second one is the image charges must be located
121

122
00:05:33,500 --> 00:05:35,866
such that on the conducting surfaces
122

123
00:05:35,866 --> 00:05:37,600
the potentially zero constant
123

124
00:05:37,600 --> 00:05:38,866
so this we have already seen
124

125
00:05:38,866 --> 00:05:40,833
seen that in the previous case
125

126
00:05:41,366 --> 00:05:42,300
the image charges
126

127
00:05:42,300 --> 00:05:44,800
they were placed in the vicinity above grounded
127

128
00:05:45,166 --> 00:05:46,133
connecting surface
128

129
00:05:46,133 --> 00:05:49,166
so it can be a constant potential as well
129

130
00:05:49,466 --> 00:05:52,100
to ensure that the boundary conditions are satisfied
130

131
00:05:53,900 --> 00:05:56,766
let's continue the first uh case case examples
131

132
00:05:56,766 --> 00:05:58,500
so the first case example is
132

133
00:05:58,500 --> 00:06:00,800
there's a point charge above our grounded
133

134
00:06:00,800 --> 00:06:02,466
conducting plane right
134

135
00:06:02,466 --> 00:06:05,333
so grounded conducting plane it is at zero potential
135

136
00:06:05,333 --> 00:06:09,900
so this is the uh your conducting plane with 0 printer
136

137
00:06:09,933 --> 00:06:12,700
and this is a point chart Q at a at a height
137

138
00:06:12,700 --> 00:06:17,200
so there's a point chart Q at a height H above this PC
138

139
00:06:18,700 --> 00:06:21,066
and this is considered to be our
139

140
00:06:21,133 --> 00:06:23,500
end of an infinite extent
140

141
00:06:23,533 --> 00:06:25,166
perfect conducting plane
141

142
00:06:25,966 --> 00:06:28,466
in which configuration is given in your figure B
142

143
00:06:28,466 --> 00:06:31,933
where the electric field at point P is given by
143

144
00:06:31,933 --> 00:06:35,033
so at point B we're going to
144

145
00:06:35,333 --> 00:06:37,766
in order to evaluate electric field at this point
145

146
00:06:37,766 --> 00:06:40,533
we're going to follow all these three steps
146

147
00:06:40,533 --> 00:06:42,800
No. 1 the original charge
147

148
00:06:42,933 --> 00:06:47,933
No. 2 the opposite charge with same uh magnitude
148

149
00:06:47,933 --> 00:06:53,066
but opposite polarity at same height but below that
149

150
00:06:54,566 --> 00:06:56,200
paint and the third thing is
150

151
00:06:56,200 --> 00:07:00,033
we're going to replace this with the zero potential
151

152
00:07:01,366 --> 00:07:02,666
this plan okay
152

153
00:07:02,666 --> 00:07:05,433
we could production plan so these are these three steps
153

154
00:07:05,733 --> 00:07:07,433
now what we are going to do that
154

155
00:07:07,566 --> 00:07:09,333
we're going to simply add up
155

156
00:07:09,333 --> 00:07:10,633
the electric field
156

157
00:07:10,700 --> 00:07:13,166
being produced by the original child distribution
157

158
00:07:13,600 --> 00:07:16,900
this one and this mirrored charge right
158

159
00:07:16,900 --> 00:07:20,066
so we're just going to apply the simple condoms law
159

160
00:07:20,066 --> 00:07:22,633
and the other laws by
160

161
00:07:22,766 --> 00:07:25,400
by using that super position principle
161

162
00:07:25,766 --> 00:07:28,533
so here at point P this uh
162

163
00:07:28,533 --> 00:07:29,466
the distance with
163

164
00:07:29,466 --> 00:07:32,466
with respect to this original charges R1 and for the
164

165
00:07:32,466 --> 00:07:35,166
this uh uh in the mirror
165

166
00:07:35,166 --> 00:07:38,566
mirror image it is R2 uh position back to
166

167
00:07:40,866 --> 00:07:42,866
we like to know our basic formulas
167

168
00:07:42,866 --> 00:07:46,066
to find out the electrical intensity
168

169
00:07:46,066 --> 00:07:49,100
that is you know that it is it is a
169

170
00:07:49,100 --> 00:07:51,400
it is charge it is a
170

171
00:07:51,800 --> 00:07:55,000
you remember that it is a force divided by the charge
171

172
00:07:55,000 --> 00:07:59,733
okay so from this particle of point it is a Q right
172

173
00:07:59,733 --> 00:08:03,200
q r one this is your R1 right
173

174
00:08:04,066 --> 00:08:06,366
this is your since you're using the
174

175
00:08:06,466 --> 00:08:08,933
you're replacing the unit tractor for example
175

176
00:08:08,933 --> 00:08:10,200
this is your unit tractor
176

177
00:08:10,566 --> 00:08:14,400
so you have to replace that with this rector
177

178
00:08:14,400 --> 00:08:18,600
divided magnitude of that rector okay
178

179
00:08:18,600 --> 00:08:21,366
so that's why this cube is coming in the
179

180
00:08:22,600 --> 00:08:24,566
denominator okay
180

181
00:08:24,566 --> 00:08:26,566
so this is the same formula that we have studied
181

182
00:08:26,566 --> 00:08:28,000
in chapter No. 3
182

183
00:08:28,533 --> 00:08:32,100
and the contribution of the electric field intensity
183

184
00:08:32,100 --> 00:08:33,766
due to this middle is
184

185
00:08:33,766 --> 00:08:34,700
is simply
185

186
00:08:34,700 --> 00:08:38,333
you have to replace this position factor with this R2
186

187
00:08:38,333 --> 00:08:42,000
and apply the way of charge with minus Q okay
187

188
00:08:42,000 --> 00:08:44,900
the next determine this disposition rector song
188

189
00:08:45,000 --> 00:08:46,966
which is back to this observation point
189

190
00:08:46,966 --> 00:08:49,333
which is placed at point P x
190

191
00:08:49,333 --> 00:08:52,966
y Z and you can see here that
191

192
00:08:54,733 --> 00:08:56,066
and the position right there
192

193
00:08:56,066 --> 00:08:58,566
it is your destination minus your source right
193

194
00:08:58,566 --> 00:08:59,766
destination
194

195
00:09:01,066 --> 00:09:03,766
minus your source right
195

196
00:09:05,000 --> 00:09:09,066
and this is first plus Q which is placed at 0
196

197
00:09:09,266 --> 00:09:11,233
0 h point
197

198
00:09:11,766 --> 00:09:15,266
whereas this mirror it is placed at 0
198

199
00:09:15,266 --> 00:09:19,633
0 minus h point along this ZX is right
199

200
00:09:20,566 --> 00:09:23,600
so you just need to um
200

201
00:09:23,700 --> 00:09:26,666
apply that distance formula like this one
201

202
00:09:27,066 --> 00:09:27,533
and then
202

203
00:09:27,533 --> 00:09:31,533
in order to find out the magnitude of this position
203

204
00:09:31,533 --> 00:09:36,466
rector you know that it is some of the scale of the uh
204

205
00:09:36,466 --> 00:09:39,333
each component and then you are too scared of that
205

206
00:09:39,333 --> 00:09:42,300
that whole quantity so it is the scale of the
206

207
00:09:42,466 --> 00:09:46,033
some of the scales of the each and each quantity okay
207

208
00:09:47,200 --> 00:09:52,166
and help the respective unit rectors have been replaced
208

209
00:09:52,333 --> 00:09:55,033
okay this one and this one
209

210
00:09:55,300 --> 00:09:58,466
so this is your final electric field at this
210

211
00:09:58,900 --> 00:10:02,700
which is being exerted at this point B
211

212
00:10:02,700 --> 00:10:06,666
due to charge Q in the presence of
212

213
00:10:06,733 --> 00:10:08,700
due to charge Q in this
213

214
00:10:08,700 --> 00:10:11,166
in the presence of this perfectly ground
214

215
00:10:11,166 --> 00:10:14,066
and perfectly grounded conducting pin
215

216
00:10:14,766 --> 00:10:16,966
and this is equivalent to the electric field
216

217
00:10:16,966 --> 00:10:18,100
due to this original charge
217

218
00:10:18,100 --> 00:10:19,433
plus the charge due to
218

219
00:10:19,700 --> 00:10:23,033
plus the electric field due to this mirror mirror
219

220
00:10:24,866 --> 00:10:26,400
charge that is minus Q
220

221
00:10:27,200 --> 00:10:31,300
okay so let's see once when Z is 0
221

222
00:10:31,400 --> 00:10:34,100
he has only Z component only
222

223
00:10:34,166 --> 00:10:36,633
let's see if you're Z this one
223

224
00:10:37,200 --> 00:10:38,000
the Z
224

225
00:10:38,400 --> 00:10:41,100
the Z is 0 this means your point is
225

226
00:10:41,533 --> 00:10:44,800
your point is coming on this plane right
226

227
00:10:45,166 --> 00:10:46,600
your part is coming on the screen
227

228
00:10:46,600 --> 00:10:48,566
for example right
228

229
00:10:49,000 --> 00:10:51,400
and so if this is the case
229

230
00:10:51,400 --> 00:10:54,766
if this is the case then you're going to only have this
230

231
00:10:54,766 --> 00:10:56,200
the component so
231

232
00:10:56,200 --> 00:10:57,666
this means that
232

233
00:10:57,700 --> 00:11:01,166
electric field is normal to the conducting surface
233

234
00:11:01,266 --> 00:11:03,466
so you only have the electric field
234

235
00:11:03,466 --> 00:11:08,333
and that is only normal to this conducted surface
235

236
00:11:08,333 --> 00:11:10,366
and that is your E
236

237
00:11:11,133 --> 00:11:12,266
Z right
237

238
00:11:12,300 --> 00:11:15,033
so if you are evaluating the electric field
238

239
00:11:15,366 --> 00:11:18,200
at this conduct on the surface of this conducting plane
239

240
00:11:18,200 --> 00:11:20,633
then it is only the z component
240

241
00:11:20,700 --> 00:11:21,600
and that is where
241

242
00:11:21,600 --> 00:11:24,533
satisfying your earlier boundary conditions as well
242

243
00:11:24,533 --> 00:11:24,866
so
243

244
00:11:24,866 --> 00:11:27,766
confirming that E is normal to the conducting surface
244

245
00:11:27,766 --> 00:11:29,566
and the potential at p is
245

246
00:11:29,566 --> 00:11:30,500
is the obtained
246

247
00:11:30,500 --> 00:11:32,833
from the equation that we have already discussed
247

248
00:11:33,066 --> 00:11:34,633
so let's see how it is Z
248

249
00:11:34,766 --> 00:11:38,633
let's see if you put this equal to zero okay
249

250
00:11:38,733 --> 00:11:39,666
this is zero
250

251
00:11:40,533 --> 00:11:41,766
this is it okay
251

252
00:11:41,866 --> 00:11:46,033
so this is going to cancel out with this one remember
252

253
00:11:46,100 --> 00:11:49,166
okay and then what else
253

254
00:11:49,166 --> 00:11:51,900
only these two things are left in this case that
254

255
00:11:51,900 --> 00:11:55,300
that is only your C component okay
255

256
00:11:55,566 --> 00:11:58,466
that is only your C component and
256

257
00:12:00,200 --> 00:12:01,000
right
257

258
00:12:03,533 --> 00:12:04,800
okay so
258

259
00:12:07,900 --> 00:12:08,800
uh now we need to
259

260
00:12:08,800 --> 00:12:13,166
find out the potential next uh find out this potential
260

261
00:12:14,100 --> 00:12:15,200
the potential it is again
261

262
00:12:15,200 --> 00:12:17,200
be simply the superposition of the potential
262

263
00:12:17,200 --> 00:12:19,066
contribution due to each uh
263

264
00:12:19,333 --> 00:12:21,200
charge entity that is uh
264

265
00:12:21,300 --> 00:12:24,466
your due to at work or perhaps or not into your
265

266
00:12:24,466 --> 00:12:26,433
this distance rector magnitude
266

267
00:12:27,766 --> 00:12:29,500
or the position vector in this case
267

268
00:12:30,366 --> 00:12:32,066
okay and uh
268

269
00:12:33,200 --> 00:12:34,666
so using equation
269

270
00:12:34,666 --> 00:12:38,466
um this uh 44 to find out the surface uh charge
270

271
00:12:38,466 --> 00:12:40,366
then 2 of the induced charge
271

272
00:12:40,366 --> 00:12:41,700
so in this case uh
272

273
00:12:41,700 --> 00:12:43,566
what is the case that you are evaluating
273

274
00:12:43,566 --> 00:12:46,200
at this Z is equal to 0 point
274

275
00:12:46,200 --> 00:12:48,000
so you are only going to have this
275

276
00:12:48,000 --> 00:12:50,100
normal component of the E field
276

277
00:12:50,100 --> 00:12:51,900
that is in the Z direction
277

278
00:12:51,900 --> 00:12:54,933
so you just solve this equation then you will
278

279
00:12:54,933 --> 00:12:58,166
you will find out that this is your resultant E field
279

280
00:12:58,200 --> 00:13:00,533
and that you will uh uh
280

281
00:13:00,533 --> 00:13:02,800
you just need to simplify the things
281

282
00:13:02,800 --> 00:13:04,733
multiply that is not your
282

283
00:13:04,733 --> 00:13:06,100
so this is the boundary condition
283

284
00:13:06,100 --> 00:13:07,400
that we have already studied
284

285
00:13:07,400 --> 00:13:10,433
that for the conducting plane
285

286
00:13:10,666 --> 00:13:14,533
the uh surface identity it is equal to the uh
286

287
00:13:14,533 --> 00:13:18,600
normal component of your electric cross density or e
287

288
00:13:18,600 --> 00:13:19,400
which is equal in V
288

289
00:13:19,400 --> 00:13:22,766
d optional not times b e n 4 free space medium
289

290
00:13:23,300 --> 00:13:23,700
so
290

291
00:13:23,700 --> 00:13:27,166
today in this charge on the connecting pen is now you
291

292
00:13:27,166 --> 00:13:29,533
you want to find out the charge on the pen
292

293
00:13:29,533 --> 00:13:31,100
you just need to integrate this
293

294
00:13:31,100 --> 00:13:35,200
with respect to surface differential surface element
294

295
00:13:35,500 --> 00:13:38,300
and uh since it is uh an infinite uh
295

296
00:13:38,300 --> 00:13:40,666
conducting plane so it's x and y components
296

297
00:13:40,666 --> 00:13:41,466
they are uh
297

298
00:13:41,600 --> 00:13:44,400
many from minus infinite two positive infinity
298

299
00:13:44,466 --> 00:13:46,733
now how to simplify this integration
299

300
00:13:46,733 --> 00:13:50,366
uh you just need to play around with the uh conversion
300

301
00:13:50,366 --> 00:13:53,300
uh into this vertical into the cynical corner system
301

302
00:13:53,300 --> 00:13:55,733
so use that uh simplification
302

303
00:13:55,733 --> 00:13:58,533
just use the change of variables that uh
303

304
00:13:58,533 --> 00:14:01,866
let's replace this uh with uh its respective
304

305
00:14:01,866 --> 00:14:03,900
uh raw component in the
305

306
00:14:04,700 --> 00:14:07,166
in your cynical coordinate system
306

307
00:14:07,166 --> 00:14:09,266
right and once you reduce this
307

308
00:14:09,266 --> 00:14:14,700
the replacement of this DX d y is Rodeo defy
308

309
00:14:15,066 --> 00:14:16,300
now replace this thing
309

310
00:14:16,300 --> 00:14:18,400
the equation will reduce to this form
310

311
00:14:19,500 --> 00:14:21,500
and accordingly the limits of your radius
311

312
00:14:21,500 --> 00:14:25,566
it will roll it is from 0 to infinity and for fire
312

313
00:14:25,566 --> 00:14:29,700
for fire it is from 0 2 to pi and no
313

314
00:14:29,700 --> 00:14:31,266
you need to integrate this right
314

315
00:14:31,266 --> 00:14:34,366
no let's use this substitution rule right
315

316
00:14:34,900 --> 00:14:36,400
and how you can integrate this
316

317
00:14:36,400 --> 00:14:38,766
just substitute this inner quantity
317

318
00:14:38,766 --> 00:14:40,700
that is in your denominator
318

319
00:14:40,700 --> 00:14:43,966
with you so the derivative with your 0
319

320
00:14:44,366 --> 00:14:49,033
your derivative of Jew is 2 row 0 okay
320

321
00:14:49,100 --> 00:14:52,466
and uh once you will do that then uh your
321

322
00:14:52,466 --> 00:14:53,900
this denominator for example
322

323
00:14:53,900 --> 00:14:56,466
it will come in your uh numinator
323

324
00:14:56,466 --> 00:15:02,166
you just apply this uh your uh respective uh power
324

325
00:15:02,200 --> 00:15:04,166
okay so once you will apply the power
325

326
00:15:04,166 --> 00:15:08,700
so it will become minus 3 by two plus 1 right
326

327
00:15:08,700 --> 00:15:11,200
so it will become your minus 1 by 2
327

328
00:15:11,200 --> 00:15:12,000
okay
328

329
00:15:12,500 --> 00:15:16,133
and once it will again go into the denominator then it
329

330
00:15:16,133 --> 00:15:18,000
will be a positive quantity
330

331
00:15:18,700 --> 00:15:24,733
alright and uh let's see what else uh you need to no
331

332
00:15:24,733 --> 00:15:27,933
you need to uh plug in in the previous uh slide
332

333
00:15:27,933 --> 00:15:30,300
you need to plug in that uh
333

334
00:15:30,300 --> 00:15:32,600
uh measure Linux
334

335
00:15:32,600 --> 00:15:36,400
and then this equation will reduce to the desired form
335

336
00:15:36,400 --> 00:15:37,500
that is your
336

337
00:15:38,133 --> 00:15:39,400
once you will er once you will er
337

338
00:15:39,400 --> 00:15:41,100
er put in the Linux
338

339
00:15:41,366 --> 00:15:43,766
anything divided by the infinity United
339

340
00:15:43,766 --> 00:15:46,033
it is zero right
340

341
00:15:46,133 --> 00:15:47,466
and for that case er
341

342
00:15:47,733 --> 00:15:49,600
you'll see that once you will put in
342

343
00:15:49,600 --> 00:15:52,300
then the dessert will be minus 2
343

344
00:15:52,333 --> 00:15:56,266
that is your er induced charge on the surface
344

345
00:15:56,266 --> 00:15:58,533
it is equal to minus Q okay
345

346
00:15:58,533 --> 00:16:01,666
and it is expected because all flex nines
346

347
00:16:01,700 --> 00:16:03,733
uh will terminate on the conductor
347

348
00:16:03,733 --> 00:16:07,166
would have been terminated on the image charge
348

349
00:16:07,166 --> 00:16:08,866
it is no conductor
349

350
00:16:08,866 --> 00:16:12,266
so since there's a conductor on the in this place
350

351
00:16:12,266 --> 00:16:12,533
so
351

352
00:16:12,533 --> 00:16:17,066
you can assume that there's a negative charge of equal
352

353
00:16:17,200 --> 00:16:18,666
equal magnitude level
353

354
00:16:18,733 --> 00:16:20,766
opposite quality on these circuits of charge
354

355
00:16:20,766 --> 00:16:23,900
so that is the takeaway from that by the first light
355

356
00:16:25,800 --> 00:16:26,400
now let's uh
356

357
00:16:26,400 --> 00:16:29,266
consider the another case that is your line charge
357

358
00:16:29,266 --> 00:16:29,466
uh
358

359
00:16:29,466 --> 00:16:32,966
distribution above our grounded plane connecting plane
359

360
00:16:33,100 --> 00:16:36,866
and uh consider our infinite line charge with uh
360

361
00:16:36,866 --> 00:16:39,666
Dansty Royale and uh
361

362
00:16:40,700 --> 00:16:42,800
uh located at a distance H
362

363
00:16:42,800 --> 00:16:45,500
again from the perfect ground connecting plane
363

364
00:16:45,533 --> 00:16:48,533
uh which is uh your at is 0
364

365
00:16:48,533 --> 00:16:51,066
0 plane this this plane okay
365

366
00:16:51,166 --> 00:16:53,266
and uh as you remember last figure
366

367
00:16:53,266 --> 00:16:56,066
we just need to repair this Q with d or L
367

368
00:16:57,866 --> 00:16:59,366
and uh uh
368

369
00:16:59,366 --> 00:17:02,300
you can uh you can uh just recall your previous uh uh
369

370
00:17:03,566 --> 00:17:07,266
previous uh uh concepts from your lecture number
370

371
00:17:07,266 --> 00:17:10,466
from your chapter number 3 okay
371

372
00:17:10,533 --> 00:17:14,833
and that infinite line charge maybe zoom in this case
372

373
00:17:14,966 --> 00:17:17,666
in this case X is equal to zero right
373

374
00:17:17,666 --> 00:17:18,600
so it is
374

375
00:17:18,600 --> 00:17:21,433
you can assume that if this is your plane right
375

376
00:17:21,766 --> 00:17:25,066
you can assume that it is your it is like this one okay
376

377
00:17:25,066 --> 00:17:27,633
like this one it is your role at okay
377

378
00:17:28,266 --> 00:17:34,233
it is at at height age from your this plane
378

379
00:17:34,700 --> 00:17:38,366
your grounded plane and it is
379

380
00:17:38,366 --> 00:17:45,333
it is oriented at this zero point on the x axis however
380

381
00:17:45,333 --> 00:17:48,400
it is changing from minus infinity to pause
381

382
00:17:48,400 --> 00:17:51,500
to infinite d and b by direction okay
382

383
00:17:52,100 --> 00:17:53,266
so that is the case
383

384
00:17:54,500 --> 00:17:57,200
and its image will be minus roll L right
384

385
00:17:57,200 --> 00:18:00,400
minus roll L like this one okay
385

386
00:18:01,366 --> 00:18:04,466
and uh back at the height minus uh
386

387
00:18:04,466 --> 00:18:06,300
minus 8 below that separate
387

388
00:18:06,300 --> 00:18:09,333
so that the two line chart densities are parallel
388

389
00:18:09,333 --> 00:18:13,300
to YXL y parallel dryaxies because x is equal to zero
389

390
00:18:13,866 --> 00:18:15,866
electrophilic point b is again
390

391
00:18:16,000 --> 00:18:18,266
given by the superposition of their individual
391

392
00:18:18,266 --> 00:18:19,466
electric fields
392

393
00:18:19,600 --> 00:18:22,700
and as you can see here the row l and what is row 1
393

394
00:18:22,700 --> 00:18:23,166
so it is
394

395
00:18:23,166 --> 00:18:24,366
the perpendicular distance
395

396
00:18:24,366 --> 00:18:26,400
between the evaluation point
396

397
00:18:26,400 --> 00:18:27,200
and b
397

398
00:18:27,666 --> 00:18:32,466
and that and your this line chart density okay
398

399
00:18:32,500 --> 00:18:33,866
this will be considered
399

400
00:18:33,866 --> 00:18:36,200
as suggested in our previous chapters
400

401
00:18:36,200 --> 00:18:38,333
and what is ERO it is the unit
401

402
00:18:38,333 --> 00:18:41,500
normal rector between the evolution point and the
402

403
00:18:42,100 --> 00:18:44,400
your line chart distribution
403

404
00:18:45,100 --> 00:18:47,700
and your only changes are your unit rector
404

405
00:18:47,700 --> 00:18:50,700
your this protector distance and your
405

406
00:18:51,300 --> 00:18:55,500
this quality is going to change for the second case for
406

407
00:18:55,500 --> 00:18:56,633
for the images
407

408
00:18:57,666 --> 00:19:00,000
but what are the perpendable rectors in this case
408

409
00:19:00,000 --> 00:19:01,900
we are just considering that
409

410
00:19:03,100 --> 00:19:05,066
we're just conceiving that we have partner
410

411
00:19:05,066 --> 00:19:08,466
but we have to graduate here right when we cheer the X
411

412
00:19:08,466 --> 00:19:12,833
y and Z so what you are going to do that you are
412

413
00:19:15,066 --> 00:19:18,166
no now you can find out this uh uh
413

414
00:19:18,166 --> 00:19:20,133
this perpencular distance between the uh
414

415
00:19:20,133 --> 00:19:22,533
evolution point or observation point and the line
415

416
00:19:22,533 --> 00:19:25,300
the uh identity using this again
416

417
00:19:25,300 --> 00:19:25,600
uh
417

418
00:19:25,600 --> 00:19:29,000
the distance from Mola that this is your destination
418

419
00:19:29,000 --> 00:19:33,766
right is your source y x is equal to 0
419

420
00:19:33,766 --> 00:19:37,700
and probably 12 because it is aligned with the y
420

421
00:19:37,700 --> 00:19:41,500
x is only okay and it is at height age
421

422
00:19:41,500 --> 00:19:43,566
and it's y coordinates raining
422

423
00:19:43,566 --> 00:19:46,633
because it is raining from minus infinity to two
423

424
00:19:47,600 --> 00:19:50,366
one stupid infinity so this is the partner
424

425
00:19:50,566 --> 00:19:54,233
you know this is this is you can say that
425

426
00:19:55,600 --> 00:19:59,066
this position director for this distance vector
426

427
00:19:59,066 --> 00:20:00,066
you can see in this case
427

428
00:20:00,066 --> 00:20:02,333
the distance vector between the evolution point
428

429
00:20:02,333 --> 00:20:03,700
and d source
429

430
00:20:04,000 --> 00:20:07,866
and likewise for the image only the differences
430

431
00:20:07,866 --> 00:20:09,766
this minus H
431

432
00:20:10,666 --> 00:20:13,966
okay now plugging back into this equation
432

433
00:20:14,133 --> 00:20:15,800
and accordingly you can
433

434
00:20:17,066 --> 00:20:19,566
you tell you can use this magnitude as well
434

435
00:20:19,566 --> 00:20:23,766
so that is your X square plus Z minus H square
435

436
00:20:25,400 --> 00:20:27,133
and uh so again
436

437
00:20:27,133 --> 00:20:30,400
and you can see over here that if you put Z equal to 0
437

438
00:20:30,400 --> 00:20:32,600
that is your Z equal zero over here
438

439
00:20:32,933 --> 00:20:36,233
so these two are going to cancel out right
439

440
00:20:36,766 --> 00:20:40,900
and uh what else right
440

441
00:20:40,900 --> 00:20:43,600
you can see it and you put this is equal to zero
441

442
00:20:43,600 --> 00:20:45,633
these two are going to cancel out
442

443
00:20:46,166 --> 00:20:48,900
and you're only left with Z compound again
443

444
00:20:48,900 --> 00:20:50,933
and then again it is confirming your statement
444

445
00:20:50,933 --> 00:20:55,900
that in the vicinity of the surface of your effect
445

446
00:20:55,900 --> 00:20:59,600
electric conductor plane your electric release
446

447
00:20:59,600 --> 00:21:02,166
always normal to that conducting surface
447

448
00:21:05,566 --> 00:21:08,600
the next uh find out the potential at point p
448

449
00:21:09,500 --> 00:21:11,166
uh by using the again
449

450
00:21:11,166 --> 00:21:13,700
the the same formulation that we know that it
450

451
00:21:13,766 --> 00:21:16,866
it would be the the resultant uh
451

452
00:21:16,933 --> 00:21:19,966
potential at devotion point p would be again
452

453
00:21:19,966 --> 00:21:22,233
this super position of the potentials
453

454
00:21:22,600 --> 00:21:24,400
of each individual lines
454

455
00:21:24,400 --> 00:21:26,666
line source that is your original line source
455

456
00:21:26,766 --> 00:21:29,300
and your image line source okay
456

457
00:21:29,900 --> 00:21:34,633
and now you have to take take care of this
457

458
00:21:35,966 --> 00:21:36,766
your
458

459
00:21:38,800 --> 00:21:40,166
see these these signs
459

460
00:21:40,166 --> 00:21:43,166
that is your minus coming out because of this thing
460

461
00:21:43,300 --> 00:21:45,300
and your
461

462
00:21:46,200 --> 00:21:49,200
this another simplification form that is your
462

463
00:21:50,000 --> 00:21:54,000
that the difference between the alarm terms
463

464
00:21:54,733 --> 00:21:57,633
you can replace with that division terms
464

465
00:21:59,333 --> 00:22:03,366
let's uh further simplify this thing that uh you know
465

466
00:22:03,366 --> 00:22:06,366
I need to uh put put in back that uh
466

467
00:22:07,366 --> 00:22:10,600
botanical distance between your line source
467

468
00:22:10,600 --> 00:22:13,066
and evaluation point that is this one okay
468

469
00:22:13,333 --> 00:22:14,266
so substituting
469

470
00:22:14,266 --> 00:22:16,166
the value of distance reactors
470

471
00:22:16,166 --> 00:22:17,966
from previous page equations
471

472
00:22:17,966 --> 00:22:21,200
this equation will reduce to this form
472

473
00:22:22,766 --> 00:22:24,400
so the surface chart density
473

474
00:22:24,400 --> 00:22:26,066
induced from the conducting pain
474

475
00:22:26,066 --> 00:22:29,466
is the pain given by using this boundary condition
475

476
00:22:29,466 --> 00:22:31,733
that is surface chart density
476

477
00:22:31,733 --> 00:22:34,800
it is equal to the normal component of the electric
477

478
00:22:34,800 --> 00:22:35,700
first density
478

479
00:22:36,566 --> 00:22:39,766
and again you just need to put in that Z equal to 0
479

480
00:22:39,766 --> 00:22:43,933
into this electrical equation that we have
480

481
00:22:43,933 --> 00:22:45,700
drive in our previous slide
481

482
00:22:45,700 --> 00:22:47,833
so this is your reduced form okay
482

483
00:22:48,366 --> 00:22:51,866
and not in order to find out the induced charge
483

484
00:22:52,500 --> 00:22:54,400
uh in this case on the conducting change
484

485
00:22:54,400 --> 00:22:56,300
you just need to simply this
485

486
00:22:56,300 --> 00:23:03,200
integrate this supercharger density along with this uh
486

487
00:23:03,200 --> 00:23:04,266
axis okay
487

488
00:23:04,333 --> 00:23:07,000
but in this particular case it is your x axis
488

489
00:23:07,133 --> 00:23:09,900
uh where the things are away okay
489

490
00:23:10,200 --> 00:23:14,066
and uh since it is uh along one line
490

491
00:23:14,200 --> 00:23:17,966
you can say that it is uh your uh
491

492
00:23:19,266 --> 00:23:22,900
so once you will uh integrate that uh uh you
492

493
00:23:22,900 --> 00:23:27,000
you you can see over here that uh it's uh dimensions
493

494
00:23:27,166 --> 00:23:28,400
you can you just uh
494

495
00:23:28,400 --> 00:23:30,300
play around with this equation and uh
495

496
00:23:30,666 --> 00:23:33,266
play with this uh integration right
496

497
00:23:33,266 --> 00:23:34,500
uh so these are the
497

498
00:23:34,500 --> 00:23:36,966
these are the construct with respect to the X right
498

499
00:23:37,766 --> 00:23:41,266
and uh what else the only changes okay
499

500
00:23:41,266 --> 00:23:42,000
this one okay
500

501
00:23:42,000 --> 00:23:44,866
here you have to again use this substitution rule
501

502
00:23:44,933 --> 00:23:47,600
and uh uh
502

503
00:23:47,733 --> 00:23:48,900
just just follow that
503

504
00:23:48,900 --> 00:23:50,500
it will be coming out in terms of
504

505
00:23:50,500 --> 00:23:52,000
once you will integrate that
505

506
00:23:52,000 --> 00:23:54,300
whether it's that to your eh
506

507
00:23:54,300 --> 00:23:57,100
this I'll find this case and put back this
507

508
00:23:57,100 --> 00:23:58,300
this limit standard to be
508

509
00:23:58,300 --> 00:24:00,100
it will be reduced to this form
509

510
00:24:00,100 --> 00:24:01,933
that is equal to minus or L
510

511
00:24:01,933 --> 00:24:04,300
again you can see here that it is uh
511

512
00:24:04,366 --> 00:24:07,666
for equal magnitude but with B opposite polarity
512

513
00:24:07,666 --> 00:24:12,200
so it is again uh confirming your previous case
513

514
00:24:12,200 --> 00:24:13,466
the department source case
514

515
00:24:13,733 --> 00:24:15,200
thing that deflect
515

516
00:24:15,200 --> 00:24:18,000
lines originating from the source
516

517
00:24:18,000 --> 00:24:20,966
will terminate on the image
517

518
00:24:20,966 --> 00:24:23,066
if the conductor is absent
518

519
00:24:23,100 --> 00:24:26,466
however this uh particle case and there's a conductor
519

520
00:24:26,466 --> 00:24:29,133
so the uh these uh lines
520

521
00:24:29,133 --> 00:24:32,066
the electric first lines originating from the source
521

522
00:24:32,066 --> 00:24:34,266
they will terminate on the conductor
522

523
00:24:36,066 --> 00:24:39,300
let's solve this example to understand further
523

524
00:24:39,300 --> 00:24:42,600
what is the significance of this matter of images
524

525
00:24:42,933 --> 00:24:47,766
so our point chart Q is located at point a
525

526
00:24:48,000 --> 00:24:51,700
0 and B so this means that your X component is
526

527
00:24:51,900 --> 00:24:54,000
your X component is a
527

528
00:24:54,000 --> 00:24:57,766
okay your y component it is for example X
528

529
00:24:57,766 --> 00:25:00,000
y and Z you can just take it
529

530
00:25:01,200 --> 00:25:04,433
in this direction okay this direction this is your your
530

531
00:25:05,900 --> 00:25:06,966
for example is your y
531

532
00:25:06,966 --> 00:25:09,600
Y1 kids Y0 in this part of the case
532

533
00:25:09,600 --> 00:25:14,300
okay and then Z then Z then Z component is your B
533

534
00:25:14,300 --> 00:25:16,700
okay this one right
534

535
00:25:17,366 --> 00:25:20,333
and uh between two semi infinite conducting pins
535

536
00:25:20,333 --> 00:25:21,466
so why semi infinite
536

537
00:25:21,466 --> 00:25:25,066
because they are fixed at this point zero
537

538
00:25:25,066 --> 00:25:27,866
but they are infinite in z direction
538

539
00:25:27,866 --> 00:25:30,100
and in this y direction
539

540
00:25:31,100 --> 00:25:32,600
they're intersecting at the right angle
540

541
00:25:32,600 --> 00:25:33,333
as you can see here
541

542
00:25:33,333 --> 00:25:34,733
they are prepared to look at each other
542

543
00:25:34,733 --> 00:25:36,200
90 degree orientation
543

544
00:25:36,666 --> 00:25:39,500
and we need to determine the potential at point P
544

545
00:25:39,700 --> 00:25:41,066
and the force and Q
545

546
00:25:42,266 --> 00:25:43,833
so let's try with one by one
546

547
00:25:44,333 --> 00:25:44,866
first of all
547

548
00:25:44,866 --> 00:25:48,000
since they are in the vicinity of the conducting pin
548

549
00:25:48,000 --> 00:25:50,100
which is grounded in this vertical case
549

550
00:25:50,566 --> 00:25:53,100
we need we need to create uh
550

551
00:25:53,100 --> 00:25:55,966
an image configuation for this thing right
551

552
00:25:55,966 --> 00:26:00,133
so in this case the three images are necessary
552

553
00:26:00,133 --> 00:26:01,766
to satisfy the conditions
553

554
00:26:01,766 --> 00:26:03,233
in the the
554

555
00:26:03,466 --> 00:26:07,300
that we have all discussed on previous slide that uh
555

556
00:26:07,766 --> 00:26:09,900
the poisons condition right
556

557
00:26:09,900 --> 00:26:11,266
poison condition only
557

558
00:26:11,266 --> 00:26:12,266
boundary condition that
558

559
00:26:12,266 --> 00:26:14,600
the images must be located in the vicinity of
559

560
00:26:14,600 --> 00:26:16,300
conducting plane the images must be located
560

561
00:26:16,300 --> 00:26:19,600
site that the conducting planes uh
561

562
00:26:19,600 --> 00:26:21,366
potentially zero as the Audi constant
562

563
00:26:21,366 --> 00:26:23,200
so these these are the conditions
563

564
00:26:23,400 --> 00:26:26,800
and then we will talk about something like
564

565
00:26:27,600 --> 00:26:30,500
the boundary conditions and the set it
565

566
00:26:30,500 --> 00:26:32,400
so yeah the first condition is to
566

567
00:26:32,400 --> 00:26:35,066
is necessary to satisfy the poison equation
567

568
00:26:35,133 --> 00:26:36,600
and the second condition that says
568

569
00:26:36,600 --> 00:26:39,766
to satisfy that boundary conditions okay
569

570
00:26:39,933 --> 00:26:42,366
and if you will specify those conditions
570

571
00:26:42,366 --> 00:26:45,600
then you need to put in that
571

572
00:26:45,600 --> 00:26:47,566
see images in this particular case
572

573
00:26:47,900 --> 00:26:48,966
so how to put that
573

574
00:26:48,966 --> 00:26:52,866
you can see here so the image of this one is here okay
574

575
00:26:52,866 --> 00:26:58,066
minus q at at this you can see that at the
575

576
00:26:58,133 --> 00:27:02,000
at the a distance that is your minus B okay
576

577
00:27:02,266 --> 00:27:03,066
right
577

578
00:27:03,933 --> 00:27:06,966
the second image of this one is on the other side okay
578

579
00:27:06,966 --> 00:27:11,800
on this left side that is that minus 8 distant
579

580
00:27:12,300 --> 00:27:15,066
now the third image it is indeed diagonal one
580

581
00:27:15,066 --> 00:27:15,900
right this one
581

582
00:27:15,900 --> 00:27:21,133
so it is that you can see that it just just label it
582

583
00:27:21,133 --> 00:27:25,133
for example this one and it is a distance you can see
583

584
00:27:25,133 --> 00:27:29,766
you can see that is clear plus B square
584

585
00:27:30,400 --> 00:27:33,033
okay because it is a diagonal
585

586
00:27:33,300 --> 00:27:35,466
we will see what is its position right
586

587
00:27:35,466 --> 00:27:37,966
in the next slide so this is the institution from here
587

588
00:27:38,000 --> 00:27:39,600
and then the provincial point P
588

589
00:27:39,600 --> 00:27:40,800
which is the direction point
589

590
00:27:40,800 --> 00:27:43,533
will be the stupid position or the provincial point B
590

591
00:27:43,533 --> 00:27:46,200
due to this individual point charges okay
591

592
00:27:46,533 --> 00:27:48,900
and accordingly the course at this queue
592

593
00:27:48,900 --> 00:27:50,300
due to these images would be
593

594
00:27:50,300 --> 00:27:51,566
these superposition
594

595
00:27:51,566 --> 00:27:55,133
of the forces being exerted by these images
595

596
00:27:55,133 --> 00:27:56,033
on this
596

597
00:27:56,666 --> 00:27:58,566
uh original charge
597

598
00:28:01,133 --> 00:28:03,466
first of all let's find out this potential
598

599
00:28:03,900 --> 00:28:05,666
so just
599

600
00:28:05,666 --> 00:28:09,566
you need to take care of the displities of the image
600

601
00:28:09,566 --> 00:28:10,433
and the source
601

602
00:28:10,500 --> 00:28:13,300
and then you have to take care of the corresponding
602

603
00:28:14,533 --> 00:28:15,900
distance directors okay
603

604
00:28:16,400 --> 00:28:19,900
for this first case the original original case
604

605
00:28:19,900 --> 00:28:22,866
your point is at a zero
605

606
00:28:23,733 --> 00:28:28,000
so this is your point to that a 0 n b
606

607
00:28:28,000 --> 00:28:30,100
so that is this one okay
607

608
00:28:30,333 --> 00:28:31,566
the second case is your image
608

609
00:28:31,566 --> 00:28:33,700
for the image you're on the left hand side
609

610
00:28:33,700 --> 00:28:37,666
it is minus a 0 n
610

611
00:28:39,300 --> 00:28:42,566
b for the image on the
611

612
00:28:44,133 --> 00:28:48,200
on the bottom okay on the on this
612

613
00:28:49,900 --> 00:28:50,700
yeah
613

614
00:28:51,800 --> 00:28:53,900
yes for the image on the diagonal
614

615
00:28:53,900 --> 00:28:58,700
right on the diagonal on the dial it is minus a
615

616
00:29:00,366 --> 00:29:03,533
0 n minus b point
616

617
00:29:03,533 --> 00:29:06,300
so the location of that to image with the
617

618
00:29:06,700 --> 00:29:09,600
with the charge of plus Q
618

619
00:29:09,933 --> 00:29:15,166
so this is your minus Q is your original source plus Q
619

620
00:29:15,300 --> 00:29:18,466
and this is your termimate that is minus Q
620

621
00:29:18,466 --> 00:29:21,166
and it is at point a
621

622
00:29:21,900 --> 00:29:25,500
0 minus B right
622

623
00:29:27,933 --> 00:29:29,600
and not find not
623

624
00:29:29,600 --> 00:29:31,600
find out the force being exerted
624

625
00:29:31,600 --> 00:29:33,200
on the original charge to you
625

626
00:29:33,200 --> 00:29:34,366
you just need to add up
626

627
00:29:34,366 --> 00:29:37,266
the forces being exerted by the new charge
627

628
00:29:37,266 --> 00:29:41,400
on this or by this images on the original charge okay
628

629
00:29:41,400 --> 00:29:45,700
ingestion while this uh chromosome formulation okay
629

630
00:29:45,866 --> 00:29:49,700
and uh for this case the distance
630

631
00:29:49,700 --> 00:29:54,700
the distance once you will uh solve this example okay
631

632
00:29:55,166 --> 00:29:58,100
now you need to find out the distance
632

633
00:29:58,100 --> 00:29:59,800
for example it is your Q
633

634
00:30:00,300 --> 00:30:05,233
it is your minus Q this is another Q and this is your
634

635
00:30:06,666 --> 00:30:10,166
sorry a plus 2 Q and your minus Q
635

636
00:30:11,933 --> 00:30:12,733
okay
636

637
00:30:13,800 --> 00:30:16,633
and uh in this case in this case
637

638
00:30:18,600 --> 00:30:20,166
uh you will see that
638

639
00:30:20,800 --> 00:30:21,966
you need to find the default
639

640
00:30:21,966 --> 00:30:24,566
being exerted by this minus Q
640

641
00:30:24,566 --> 00:30:25,800
image on this Q
641

642
00:30:25,900 --> 00:30:29,366
then you can see it the distance it is to be okay
642

643
00:30:29,366 --> 00:30:32,866
because it is at B and it is at minus B distance
643

644
00:30:33,500 --> 00:30:35,300
this is your access okay
644

645
00:30:36,100 --> 00:30:38,566
and likewise for this one
645

646
00:30:39,933 --> 00:30:42,633
so this is a sorry this is a
646

647
00:30:44,466 --> 00:30:48,033
a and for this one this is B n minus B
647

648
00:30:48,966 --> 00:30:51,866
okay and the distance would be to B
648

649
00:30:55,066 --> 00:30:58,000
K and what else
649

650
00:30:59,533 --> 00:31:01,900
for this Taiwan this one
650

651
00:31:01,900 --> 00:31:03,500
this one is pretty important
651

652
00:31:03,500 --> 00:31:05,000
okay for here
652

653
00:31:05,066 --> 00:31:09,600
you need to involve this because it is at a zero guess
653

654
00:31:09,600 --> 00:31:12,100
and B and this is at minus A
654

655
00:31:14,100 --> 00:31:18,200
0MINUSB you just enroll this distance formation
655

656
00:31:18,200 --> 00:31:20,666
to take this clear look of that
656

657
00:31:20,866 --> 00:31:23,966
and since you are enrolling this rectors over here
657

658
00:31:23,966 --> 00:31:25,133
unit rectors over here
658

659
00:31:25,133 --> 00:31:29,166
so that's where it is coming out to be 3 / 3
659

660
00:31:29,400 --> 00:31:33,566
the part of three because of this involving this
660

661
00:31:33,700 --> 00:31:36,966
this vector right R vector
661

662
00:31:38,600 --> 00:31:40,533
R q okay
662

663
00:31:40,533 --> 00:31:42,266
that is your basic column stop
663

664
00:31:42,466 --> 00:31:44,566
so once you will simplify the things
664

665
00:31:44,566 --> 00:31:46,866
you take the commons and the
665

666
00:31:47,266 --> 00:31:48,966
so this is your final choose form
666

667
00:31:48,966 --> 00:31:49,800
so electrically
667

668
00:31:49,800 --> 00:31:52,166
due to this system can be determined similarly
668

669
00:31:52,166 --> 00:31:54,700
and charge and use on the planes can be form
669

670
00:31:54,700 --> 00:31:57,400
okay and then if you want to find all the electrical
670

671
00:31:57,400 --> 00:31:58,833
so you know that it is
671

672
00:32:00,333 --> 00:32:04,166
e have to identify simply Q
672

673
00:32:07,333 --> 00:32:09,433
second example in general
673

674
00:32:09,600 --> 00:32:11,766
the method matter of images
674

675
00:32:11,766 --> 00:32:14,700
is used for a system consisting of a point
675

676
00:32:14,700 --> 00:32:19,000
charge between two semi infinite conducting plans
676

677
00:32:19,200 --> 00:32:21,466
so in this in this example
677

678
00:32:21,466 --> 00:32:23,866
we are only considering the simple cases
678

679
00:32:23,866 --> 00:32:26,166
that we only have two infinite
679

680
00:32:26,266 --> 00:32:27,733
semi infinite conducting plans
680

681
00:32:27,733 --> 00:32:32,466
and they can be inclined at any angle 5 number five
681

682
00:32:32,866 --> 00:32:34,000
then the number of images
682

683
00:32:34,000 --> 00:32:36,366
they can be determined in this formulation
683

684
00:32:36,366 --> 00:32:37,166
okay
684

685
00:32:37,400 --> 00:32:41,000
3 60 because you can you can consider that is complete
685

686
00:32:42,266 --> 00:32:45,233
sphere or circle or thing like this
686

687
00:32:45,733 --> 00:32:49,300
the angler distribution between these two planes
687

688
00:32:49,466 --> 00:32:50,500
and minus 1
688

689
00:32:51,700 --> 00:32:52,800
let's apply this
689

690
00:32:53,066 --> 00:32:55,333
given formula on all these cases that we have started
690

691
00:32:55,333 --> 00:32:56,133
okay
691

692
00:32:56,366 --> 00:33:00,800
because the charge and its image line on a circle right
692

693
00:33:00,800 --> 00:33:05,633
this extends your circle for example when the uh
693

694
00:33:05,866 --> 00:33:09,166
the angle distribution between the uh
694

695
00:33:09,166 --> 00:33:11,366
once the once the uh this plane
695

696
00:33:11,366 --> 00:33:13,200
connecting plane is at angle of uh
696

697
00:33:13,500 --> 00:33:15,566
this uh one 80 degree right
697

698
00:33:15,566 --> 00:33:17,500
that is your straight line right
698

699
00:33:17,500 --> 00:33:19,000
connecting plane is straight line
699

700
00:33:19,266 --> 00:33:21,700
uh then you're only going to have one image
700

701
00:33:21,700 --> 00:33:22,466
you can see here
701

702
00:33:22,466 --> 00:33:27,066
so this will be 1 80 degrees so 3 60 / 1 80 is 2
702

703
00:33:27,066 --> 00:33:28,200
2MINUS1 is 1
703

704
00:33:28,200 --> 00:33:30,466
so that's why we only had the one image here
704

705
00:33:31,366 --> 00:33:35,200
and if it is they are at the orientation of 90 degree
705

706
00:33:35,200 --> 00:33:37,633
that we studied in our previous example
706

707
00:33:37,800 --> 00:33:40,100
then uh you can see that 3
707

708
00:33:40,100 --> 00:33:43,866
60 divided by this so 5 it is 4
708

709
00:33:43,866 --> 00:33:45,866
4MINUS1 is your 3
709

710
00:33:45,866 --> 00:33:48,066
so that's why we had the three images
710

711
00:33:48,733 --> 00:33:50,566
and in case they are uh
711

712
00:33:50,566 --> 00:33:54,766
inclined at the angle of 60 degree for this case okay
712

713
00:33:55,400 --> 00:33:56,933
then these charges
713

714
00:33:56,933 --> 00:34:00,733
there will be these number of um images
714

715
00:34:00,733 --> 00:34:04,500
there will be you can see here 6 - 1 it is
715

716
00:34:05,800 --> 00:34:07,900
minus 1 it is five okay
716

717
00:34:09,400 --> 00:34:09,966
and you
717

718
00:34:09,966 --> 00:34:13,233
you need to place this five charges in this case right
718

719
00:34:13,333 --> 00:34:15,900
and what would be the angle distribution between them
719

720
00:34:15,900 --> 00:34:17,166
between them that would that
720

721
00:34:17,166 --> 00:34:18,633
that you know that it is your
721

722
00:34:19,566 --> 00:34:21,800
the angle that is your this one
722

723
00:34:21,800 --> 00:34:22,766
okay this one
723

724
00:34:22,766 --> 00:34:23,700
60 degree
724

725
00:34:24,866 --> 00:34:27,033
what is this 30 degree 1960
725

726
00:34:27,400 --> 00:34:28,266
yeah it is
726

727
00:34:28,266 --> 00:34:29,400
it is sorry
727

728
00:34:29,400 --> 00:34:30,466
it is 6 degree
728

729
00:34:31,366 --> 00:34:32,166
okay
729

730
00:34:35,566 --> 00:34:37,533
right and uh
730

731
00:34:37,533 --> 00:34:40,966
likewise they can be room treated like this okay this
731

732
00:34:41,300 --> 00:34:42,600
this okay
732

733
00:34:43,300 --> 00:34:45,866
uh but you need to take care of these priorities
733

734
00:34:46,400 --> 00:34:48,066
priorities okay
734

735
00:34:49,066 --> 00:34:50,100
priorities
735

736
00:34:50,800 --> 00:34:51,800
priorities
736

737
00:34:52,666 --> 00:34:54,500
and then we will solve the case
737

738
00:34:54,733 --> 00:34:56,166
the cases like
738

739
00:34:56,166 --> 00:34:56,900
for the potential
739

740
00:34:56,900 --> 00:34:58,733
and the problems for the presidential
740

741
00:34:58,733 --> 00:35:00,966
and the electric grid and electric forces
741

742
00:35:00,966 --> 00:35:03,466
just like the example that we solved
742

743
00:35:03,466 --> 00:35:05,066
in our previous slide
743

744
00:35:06,200 --> 00:35:07,933
so today we understood uh
744

745
00:35:07,933 --> 00:35:10,766
another matter that is known as the matter of images
745

746
00:35:10,766 --> 00:35:13,766
to find out the electric potential and electric feed
746

747
00:35:13,766 --> 00:35:16,766
due to point charge or the line charge uh
747

748
00:35:16,766 --> 00:35:19,533
or other charge distribution in the vicinity of
748

749
00:35:19,533 --> 00:35:22,300
remember in the vicinity of uh
749

750
00:35:22,300 --> 00:35:27,500
perfect conducting plane so that we can treat that uh
750

751
00:35:27,500 --> 00:35:29,766
equal to the equal potential surface
751

752
00:35:29,766 --> 00:35:33,466
and we replace that we add another image of uh
752

753
00:35:33,466 --> 00:35:36,066
the source with equal magnitude what
753

754
00:35:36,066 --> 00:35:37,866
but with the opposite planet
754

755
00:35:38,533 --> 00:35:41,800
in the next class we're going to have an oval here
755

756
00:35:41,800 --> 00:35:43,066
question number two
756

757
00:35:46,166 --> 00:35:47,300
alright thank you all
757

758
00:35:47,300 --> 00:35:48,900
if you have any questions
758

759
00:35:48,900 --> 00:35:50,666
uh they would be entertained uh
759

760
00:35:50,666 --> 00:35:52,966
through your emails or online
760

761
00:35:52,966 --> 00:35:55,466
synchronous sessions that we have arranged
761

762
00:35:55,466 --> 00:35:56,666
at the department