this is election No. 18 of baby 2 2
3 2 electromagnetic field theory 3
today uh 4
we will study electric field bound conditions 5
and this will cover your section 5.9 of textbook 6
the learning objectives of today's lecture would be 7
what are electric field boundary conditions 8
and uh what is the impact of uh 9
dialective to dialective boundary conditions 10
on the travelling of electric fee 11
from one medium to another 12
and likewise 13
the dependency upon this conductor to dielectric 14
boundary conditions 15
and conductor to free space of boundary conditions 16
upon the traveling of this electric field 17
from one medium to another 18
first of all the boundary conditions 19
once uh 20
the field exist in a medium 21
which is for the consisting of uh 22
two different media uh 23
and the properties of these two 24
dialectic properties of these two materials 25
they are uh different 26
then these are the conditions upon 27
the conditions are 28
the conditions that the electric feed must specify 29
at the interface of these two sexes 30
at the boundaries of these two sexes 31
and that's why they are known as 32
the boundary conditions so 33
these are the conditions that for this particular case 34
the electric feed must 35
satisfy and the interface between these two medium 36
so for electrostatic feed 37
uh we will consider following boundary conditions 38
which allow most importance 39
uh the dialector to dialectric interface 40
the conductor to dialectic interface 41
and the conductor to freespace interface 42
I know that they are always helpful in determining uh 43
the field on one side of the boundary 44
if we have the knowledge of electric field 45
on the other side of the boundary 46
so if we know the field on one side of the boundary 47
then we can the remain 48
the electric field on the other side of the boundary 49
so that is the significance of the boundary conditions 50
and uh in order to drive these boundary conditions 51
we make sure that the maximum equation uh 52
the maximum equation 1 and 2 that we drive for this uh 53
uh that is your closing tirular 54
electrostatic field around the closed part 55
it is uh equal to zero and that is your electric field 56
it is uh you know that it is uh conservated 57
yeah uh uh electroconturity 58
and the second one is your uh 59
is your uh that uh gaze is yeah gaze is gaze is clear 60
that the charge and closed uh is equal to the flex 61
uh passing to that uh close surface 62
and we will also make sure of this uh 63
decomposition of electric uh 64
feed and electric first density into two autobinal uh 65
components 66
uh that is your potential and your normal components 67
potential and your normal components 68
and we were also majors of this 69
constituted relationship between this electric feed 70
and the electric feed dynasty 71
let's have this first born recognition that is 72
medium 73
so consider the boundary between two dialectic 74
with permittivities ABS 1 75
1 and ABS 1 2 76
we know that uh 77
uh if they are other than the free space 78
they will be expressed in terms of their relative 79
permittivity multiplied with the 80
by mittility of the free space 81
so these are the two medium 82
your electric field 83
it is moving from medium 2 to medium 1 in this case 84
and uh or in the medium 1 we're having this uh 85
electrical 1 in the medium 2 86
we're having this electrical 2 87
which are further having the tenantial 88
and the normal components as express chair 89
when I solve this uh post uh maximum equation 90
uh that is your close in check it off line in 91
check it off electric field around disclosed park a B 92
C G 93
a in the given figure 94
and so in this case 95
we are assuming this close part that is a B 96
C G right 97
now let's solve this integration around this close part 98
right 99
and uh for this uh according to this medium 100
uh 101
we would take this uh financial and normal conference 102
so let's start from this this a to B 103
for this a to B we only have this in this medium one 104
we have this E1T 105
right it's initial component 106
and what is the length of this path 107
that is JTW right 108
okay for this segment that is your 109
uh B to uh B to that uh interface 110
you have this e 1 n but it is in B 111
it is in B opposite direction as to B 112
so it is in the upper direction right 113
even n it is in the upper direction and this uh your 114
direction of motion into this circulation part 115
it is in the clockwise direction 116
whereas the normal it is B 117
it is in the upper direction right 118
so that's why it is going to have this minus sign 119
because the data it is equal to 120
one 80 degree and we know that call of one 80 degree 121
it is equal to minus 1 122
okay for the second part uh uh that is 123
so for this part this 124
the differential land is equal to del 125
age divided by two 126
and also for this one this is dial age divided by 2 127
but the only difference is 128
you know 129
you are in the region 2 and you're having this E to 130
and normal component again 131
the direction is in the opposite direction 132
to this circulation part 133
that's why you are going to have this minus sign 134
and for this 135
length segment that is from C to D your turn is 136
you are in the second medium right 137
so your turn is your component 138
it is E2T 139
it is in the opposite direction to this circulation 140
pass so 141
that's why you are again going to have this minus sign 142
now for this normal component in this uh region two 143
uh the direction of this 144
normal is seen as your circulation pass 145
so that's where it is going to be positive 146
and again this uh first half 147
it is having this tell each divided by two 148
length and second half 149
it is also having this the age divider to land right 150
but the path is complete and it is equal to zero 151
because this circulation around 152
this complete path is equal to zero 153
okay now let's make the Assumption as you are 154
as you are closing towards this interface 155
so this means that your Bill age it is approaching zero 156
so very important Assumption 157
in order to determine this electric grid 158
on the surface 159
on the interface between these two different medium 160
so once you will involve this Assumption 161
that your dial age is approaching to zero 162
so this time is going to be zero 163
this time is going to be zero 164
this time is going to be zero 165
right so you are only left out with these two terms 166
you simplify these two terms 167
so this is the leftover term call this Del 168
a Del W divide is going to cancel out in the 169
in the common as a common 170
and then it will be cancelled out 171
on the right hand side 0 divided by that uh that W 172
you know that it is equal to 0 right 173
so this is the uh takeaway from these two sides 174
what is this what is the intuitiveness of this thing 175
this term 176
the intuitiveness 177
states that the tendential component of electric feed 178
we are equal at the boundary 179
or the interface of two different dialectics 180
this means that the potential component of electrical 181
undergoes no change on the boundary 182
and it will stay continuous 183
it will stay continuous across the boundary 184
let's see what happens to the uh normal component 185
so for this uh case for this case 186
now let's first see for this flextance tree 187
first of all flextance tree 188
let's see 189
what happens to this financial component of this 190
flextance tree 191
so since we know that uh this uh d plus 10 C 192
it is equal to absent on our times electric feed 193
so we involve this definition 194
the constituted relationship into this uh 195
first divided expression 196
from here we can see that your 197
d one t divided by abstinent 1 is equal to d 2 198
t divided by abstinent 2 199
and just simplify these two terms 200
you will get this thing 201
so from here you can into that potential component of d 202
it is undergoing some change 203
and that change across the boundary 204
it is due to this dialectic 205
materials are maturity 206
so this means that since these two are not equal 207
D1T is not equal to D2T 208
so that this means that d is discontinues 209
this continues across the boundary 210
so remember for electric field 211
dependential component of electrical 212
it is continuous 213
whereas the tangential component of this um 214
electric flux tensity it is not continuous 215
right because it is involving this permitted 216
let's uh see 217
what happens to the normal components of the trucks 218
density and the electric field 219
for this thing 220
uh we will uh make sure of this Boss's law 221
and as a boss can surface 222
we will take this uh 223
this example of a small pill box which is of 20% chair 224
uh so that we can uh 225
determine this uh 226
normal confidential boundary condition 227
so again we resolve this 228
we decompose this electricity to compromise 229
the financial and the confidence 230
and the surface value of this wasn't surface 231
it is DL s and the height of this surface it is DL H 232
again the two mediums 233
they are uh secreted by there 234
these electric permit duties 235
these are electric properties 236
let's involve this possible which changes the 237
the closed infigure of the dot 238
d s is equal to be charged and closed 239
that it reflects passing to a closed surface 240
it is equal to be charged and closed 241
okay so the left hand side here 242
uh we are assuming that since we are approaching this 243
uh 244
we're approaching this interface 245
so this means that this del H 246
it is approaching zero so this integral for this side 247
for this side 248
this integral is equal to zero for this side 249
and we only need to store this top and vitamin right 250
yes so for the 251
since it is a top product 252
uh and if we see that for D 1 n 253
for D 1 n the teacher of this E and the energies 254
it is normal to the surface 255
uh 256
this E N and this e N that is the normal to the surface 257
the last both are in the same direction that is your 258
right the first case in the first medium 259
if it is equal to zero 260
so the cause of uh uh that uh zero degree 261
it is equal to one and for the product specific 262
they are in the opposite directions 263
so that's why the angler displacement between them 264
it is 1 80 degrees so that's why because of 1 265
80 degree is for 2 - 1 266
and uh on the right hand side 267
what is in charge controls by this surface spins 268
we are only on this surface here 269
and let's assume that the surface identity 270
on this surface it is us 271
so discharged and closed by the surface 272
having this surface area of dial s 273
this simply is equal to the product of this war 274
s times dial s so it is from your basic definition 275
now this LS it is going to cancel outro 276
and this is your remaining term 277
so the relationship between the normal component 278
of the electrical first entity 279
is that the difference between the normal component 280
it is equal to the social chart s 281
E which is pleased 282
at the interface between these two mediums right 283
so this is the takeaway 284
and uh uh here 285
would you note that also that this is very important 286
that the integration with respect to this uh 287
side first date 288
which is go to 0 because this time 8 which is 0 2 0 289
no let's assume that this uh 290
there's no charge place at the boundaries 291
at the interface between these two dialects 292
that is your rule as is equal to 0 293
so for this case your d 1 n is equal duty to n 294
this means that the normal components of t 295
they are equal at the interface 296
at the boundaries between these two medium 297
and the normal component of electrical substance 298
t undergoes more change on the boundary 299
and it is continuous 300
across the boundary 301
now let's see what happens to the normal 302
comprehend of letrophy again 303
we enroll this constitute relationship 304
and you can see that 305
the e 1 n is not equal to 306
e 2 n 307
means that the electric field is this continental 308
across the boundary 309
so electric field normal component 310
it undergoes some changes which are related 311
using this electric permatuity of these two mediums to 312
that is the takeaway 313
right the normal component of electric fertility 314
they are continuous 315
if there's no surface charge density 316
at the interface of these two mediums 317
and uh electric uh normal components 318
normal components of electricity they are this one 319
yes across the bone 320
select uh 321
bring on these two reserves and further uh 322
uh drive some of the expressions indicates 323
uh if this is uh 324
in face in face between these two mediums right 325
and this is normal between these two mediums right 326
like this thing and you can straight like this thing 327
cream it is making the in the first medium 328
it is making the angle of Peter 1 329
like this one with respect to this one more 330
and this Peter 2 for this uh 331
you know 332
this cheetah tool with respect to this normal 333
for this electric field in this medium 2 334
now let's resolve this electric field in medium 1 335
on this normal like this way 336
like this way right so 337
you're making the projection of electric field 338
on this normal on this normal direction right 339
and so this normal even n it is equal to even 340
it is equal to e one cause of teacher 1 341
and light bulb d 1 n 342
it is equal to d 1 cause of teacher 1 and e 1 t 343
it is equal to e 1 sign of teacher 1 n d 1 t 344
it is equal to d 1 sign of teacher one and in medium 2 345
we we drop this 346
we drop this comprehend on this um 347
that is your normal number between these two such days 348
because here we have the information of Peter 2 349
which is the animal between E2 and the normal 350
between these two such days 351
and here we once we resolve these two components 352
then we will see that E2 353
it is equal to sign d tattoo and D two d 2 n 354
so e 2 t it is equal to E to sign Peter 2 n d 2 n 355
it is equal to d 2 cause Peter 2 back 356
we're just resolving this components 357
this this component on the normal 358
let's make sure realization of these two uh 359
things that we have drived on the previous uh slide 360
uh so uh divide this equation 1 with its question 2 361
use this uh 362
geometric identity of the anti physical 363
scientific web deposititor 364
and simplify this expression 365
so this is very famous uh 366
uh law that is known as the law of retraction right 367
and it is responsible for the bending of electric feed 368
towards the normal or away from the normal 369
once the dialectric property is 370
are different between two services 371
and here we are also having resumption 372
that be interfaces free of charge 373
there's no charge 374
that place between the boundaries of these two mediums 375
so that is only for this particular case 376
this law Protection is only vertical case 377
and depending upon the relative 378
permativity of these two mediums 379
you will see that this electric field 380
it is bending towards the media 381
towards the normal 382
between these two mediums or away from the media 383
personal discuss the second scenario 384
that is your one side of the medium 385
it is a dialectic and the other side of the medium 386
it is a conductor perfect conductor conductor 387
so this is the uh condition is the scenario 388
and we have to uh 389
drive the boundary conditions for the scenario that 390
what happens when the electric feed 391
it is going from uh 392
dialectic towards the conductor medium 393
so let's see what happens 394
so we know that inside the conductor 395
the electric field is zero 396
because your conductivity of that perfect medium 397
conductor it is infinity and this is um 398
Shin will maintain finite condensity 399
cause J Z Korro Sigma E we know that if uh 400
the uh 401
that you're from here you can see that is equal to 402
g to have a Sigma energy Sigma is very 403
very hard than anything 404
to have infinity is equal to zero 405
so that will say that inside the conductor 406
the electric release is zero 407
and we have already had the intention that uh 408
the electric turn it stays at the surface of uh 409
conductor and any induction of uh 410
electric uh 411
charge inside the inductor 412
will immediately travel toward the surface of that 413
conductor that's why the 414
uh electric field inside the conductor 415
it is equal to zero 416
so now let's um 417
uh solve these two things 418
uh the maxifications that we have 419
uh uh 420
solved for the previous case as well 421
so first of all 422
this close line will change well around this part 423
that is a B 424
C d a 425
the only difference in this case is 426
we only have electric field in this dialect medium 427
electric field in the second medium is going to be zero 428
right this thing 429
so the integrations with respect to this uh 430
uh normal 431
uh Indonesia components in the first medium 432
they will stay here 433
whereas the integrations in the second mediums 434
they will be equal to zero 435
and the just simply simplified case 436
we are we are taking that e 1 t it is equal to t n 437
likewise e 1 n which is equal to e n right 438
because we only have the electric green 439
in the plus medium that is the dialect medium 440
okay now again 441
involved Assumption that day 442
you are approaching this interface 443
so once you will approach this interface 444
then this will also equal to 0 right 445
and this will also equal to 0 446
so what you are left with is that is this thing 447
and from here you can 448
into that the potential component of electric field 449
it is equal to zero at the interface 450
R boundaries of these two uh 451
these two mediums 452
that is the uh connector dielectric mediums 453
and likewise your uh furniture component of your uh 454
this uh electric first energy it is actually Puerto Co 455
now let's see 456
what happens 457
to the normal component of the electric grid 458
and let's see first for the electric personality 459
involving involving this causes law 460
and using this cause in surface 461
now again in the second medium 462
this electric with the zero and we are left with this 463
uh DNL s is equal to the charge and closed 464
and again the same Assumption 465
that you're approaching this interface 466
because you're Dell H it is approaching zero 467
again inside this uh conductor electrical is 4:00 468
and uh we know that uh 469
if Ruas is placed at the interface between these two 470
such days 471
then the charge and close will be is equal to rosdls 472
and then Ldls will be cancelled out 473
and you can see that 474
your normal component of lexicregency 475
it is simply is equal to the surface charge density 476
which is placed at the surface 477
at the interface between these two mediums 478
that is your conductor and the electric medium 479
and likewise 480
you can find out the electric feed in this case 481
the normal component of the electric feed in this case 482
this and aesthetic conditions 483
the following conclusions can be drawn 484
for a protect conductor 485
that no electric feed may exist within a conductor 486
because electric feed is equal to zero 487
inside a connector and the volume 488
charge density inside the connectors 489
is also equal to zero because the electric charge 490
it will immediately travel towards the surface 491
of that conductor due to the high conductivity 492
and if we use this relationship 493
and the relationship between the potential 494
and electric field 495
if your electric field is equal to zero 496
so the grade level is equal to zero 497
so this means that there's no chain in the potential 498
on the surface of this electric conductor 499
and it can be said that it can be said 500
that no potential difference between any two points 501
inside this conductor 502
and that's why we see that a conductor 503
it is an equiferential body 504
and the tyranese 505
and electric field must be external to the conductor 506
and once it is like this 507
so it must be normal to each surface 508
that is we don't have the 509
we don't have the financial components 510
inside this connector we only have the 511
we only have the normal components 512
we only have the electric field and that too 513
it's normal component at the normal 514
today this conductor 515
and there is no financial component even at the 516
at the 517
at the surface at the interface of this conductor 518
we only have the normal component of electric feed 519
or electrical agency if we have the superchargency 520
at the interface between these two mediums 521
that is the conductor and dialectic 522
and uh that is relationship between these two 523
election wall and study 524
this very important property of this conductor 525
dialectic boundary conditions 526
there is an important application of the fact that 527
we know that electric field inside the conductor is 0 528
so it can act as an electrostatic screen material 529
so if a connector a get x 0 potential like this way 530
and it is sounding a connector B 531
connector B as shown in this case 532
then we can say that 533
B is set to be electrically screened by a 534
electrically screened by a form 535
other electric circuit such as conductor C 536
which is outside a because this conductor a 537
it will not allow any of the 538
electric field to pass through it 539
because the e electric field inside this conductor a 540
it is equal to zero 541
similarly we can see that the connector C 542
outside of a is screened by a 543
from this connector B 544
thus here in this particular example 545
this a the connector a 546
it is acting as a screen 547
screening or shielding material and it is uh 548
very useful in your electric uh 549
uh suckers that uh you can uh you can screen the uh 550
it is also used 551
uh to reduce the electromagnetic interferences 552
interferences and the uh 553
interferences due to the external electric fields 554
uh on the electric secretary of your electric circuits 555
so electrical conditions inside and outside 556
the screen are completely independent of each other 557
due to the involvement of this screening or shielding 558
conductor a 559
the next uh consider our third and last scenario 560
that is a special case of your conductor 561
dialect conditions that here for the dialect 562
you are simply having the free space medium 563
and we know we know that what's the 564
it is a free space medium 565
then it's like the relative permittivity 566
it is equal to one 567
and uh in this case your simply your reserves 568
it will reduce to that 569
your normal component of this electrical density 570
it is equal to do s 571
and your normal component of electric uh field 572
it is equal to do s t y P epslon not 573
and this financial components last 574
they will stay same because it was equal to zero 575
so if you must approach a conducting surface 576
normally as the financial component is equal to zero 577
so as some people know 578
that 579
we don't have the potential component in this case 580
as well 581
and we only have the normal components in this case 582
and that is only available 583
components of electric field 584
and the electric first chance 585
and they are related with the surface charges 586
like this relationship 587
so this is a complete summary of this lecture No. 18 588
for different boundary conditions 589
that how the normal and genitial components the vary 590
at the interface between these two dialective machine 591
mediums 592
and if we have the information of the lactophile on 593
in the in the one medium 594
then we can utilize these boundary conditions 595
to find all the lactophile 596
and the electric first density in the other medium 597
using these relationships 598
and likewise it is the summary for the conductor 599
dialector Bondi conditions 600
and the conductor free space body conditions 601
so I'm here today's lectures 602
so today uh 603
we study uh what is electrically bounded conditions 604
what is the utilization and what are the the 605
the three different scenarios uh 606
for which we need to study the boundary conditions 607
and which are utilized to determine the electric field 608
and electric first density 609
from one medium to another medium 610
if we have the information of electric field 611
or electric prestency in one medium 612
then we can determine the electric field 613
and the electric prostency in other medium 614
utilizing these three boundary conditions 615
that we have derived 616
next time 617
we will again review these electric bonded conditions 618
and we will further clarify these bonded conditions 619
by solving some case examples 620
so this is very very important lecture 621
here I thank you all if you have any questions 622
uh they will be entertained uh 623
to your emails or online synchronous sessions 624
that we have arranged at the department