this is election No. 20 of a B 2 2
3 2 3
Today 4
we will understand what is electrostatic boundary value 5
problems and their significance 6
this will color your sections of a chapter No. 6 7
that is a 6.12 6.4 8
so the learning objectives of uh 9
today's election would be uh 10
to understand electrostatic boundary problems 11
to derive by zones and that last equations 12
to understand what is uniqueness here 13
and thereafter to solve the case examples 14
uh in previous chapters uh 15
we have uh determined this uh electric uh field 16
yeah either using this uh columns law or causes law 17
uh 18
when the charge distribution is known are this uh uh 19
this relationship uh 20
that is 21
the relationship between electric field and the uh 22
electric potential 23
once this electric potential is not in more common 24
most practical applications 25
neither of these charge distribution nor the potential 26
electorponentialist distribution is known 27
so what is the realt 28
so that is the the realt is given in this chapter 29
uh uh 30
that's a practical 31
electrostatic problems are considered 32
when only electrostatic conditions 33
lactostatic conditions at some boundaries 34
electrostatic conditions at some boundaries 35
so these conditions can be charges or the potentials 36
or the electric fields at some boundaries are given 37
and from there when we can find out the electric field 38
and the potentials um 39
in other mediums or in other uh in other locations 40
and such problems uh 41
as well is using this uh 42
poisons or lap class equations are the metro 43
images that we will discuss 44
in our next coming lectures 45
very important 46
indications are from your electrostatic point of view 47
that is prysons and laplassecutions 48
and let's see how to drive them 49
so they can be easily drive from the causes law 50
so what was the causes law 51
it was a delta de v quatu robe the volume chart density 52
and 53
if we use the constitutive relationship between this d 54
and the electric field and also 55
the relationship between this electric field 56
and elective potential and plug in this relationship 57
this relationship between this discretion 58
then this is your uh 59
relationship for in homogeneous medium 60
so why in homogeneous 61
uh because it is not constant over anti medium 62
that's why we cannot take this outside 63
we cannot take this outside 64
and treat this as a constant 65
we cannot do this thing however 66
if uh the medium is homogeneous then we we 67
we can drive a poison equation 68
this means that the ABS non electric permitivity 69
it is the cost trying to cost 70
the medium in which we need to find out the potential 71
and the electric fee 72
so and alternate this in particular case 73
you can in this particular case but you can do that 74
no you can treat this as a question it 75
and it can come outside of the integration 76
and once you solve this 77
take this on the right hand side and 78
you treat this Dal dot d 79
you know that it is equal to your Dal square okay 80
Dal square 81
that is your lapation operator as well 82
okay so in case there's 83
there's no charges to present in the medium 84
this means that the it is free of the charge region 85
then it can simply convert to a homogeneous yes 86
see that 87
it is homogeneous second order differential equation 88
it is non homogeneous okay 89
because this right hand side is not equal to zero 90
this is homogeneous secondary differential equations 91
and this aggression is known as the leplassequision 92
for a charge free region 93
so the class 94
and thousands equations are useful in solving 95
electrostatic field problems 96
and various other problems as well 97
and we what are those problems as well 98
and where these equations can be utilized 99
this is only by simply replacing the variables 100
in these two standard equations 101
so if we treat this v as a magnet 102
this electric potential as a magnetic potential 103
then we can 104
we can utilize the same equation in v magnetic statics 105
we'll see in that coming textures as well 106
and also this so we 107
if it 108
if it is treated as treated as a temperature variable 109
then it can be utilizing the heat conduction problems 110
and if it is treated as a stress function 111
then it can be utilizing the fluid dynamics problem 112
this food food uh food flow dynamics problem 113
and uh you must remember from your chapter No. 2 114
what are these uh uh 115
second order uh 116
differentiate equations or the the liplash 117
uh liplashing operators in the second order 118
uh form and uh you must uh 119
use them and you must remember them 120
according to their respective card 121
uh cardination uh this uh coordinate system 122
and similarly for deposits equation 123
you can see that the right anxiety 124
it is not equal to 0 so 125
this means it is in homogeneous second order equation 126
in homogeneous 127
second order 128
differential 129
equation right RPD partial differential equation 130
second order but it is homogeneous 131
second order 132
Piti right so you must remember that 133
what equation you need to involve 134
according to its cardigan or cardinal 135
is the cardinal system either Cartigian 136
cynticol or strical system and heavy 137
it is known as the potential according to its 138
uh the coordinate system 139
remember these multipliers 140
always remember these uh cautions 141
these multipliers these cautions okay 142
these scope multipliers the cautions 143
either they are outside of the uh 144
this differential or they are inside the differential 145
no no 146
no be threatening what is unique testing them 147
there are several matters of solving a given problem 148
so we know that either it is an analytical problem 149
or it can be solved using some graphical approach 150
or it can be solved using some grammarical method 151
or it can be solved uh to your experimental uh studies 152
so solving Le class equation in different ways 153
will give you different answers 154
but the question is if a solution to Le class equation 155
two different ways 156
satisfy a given set of boundary condition 157
a given set of boundary condition 158
is this the only possible solution 159
either we solve this equation 160
for this given boundary condition 161
to this 3 form matters so 162
the solution that we are going to find out using this 163
for matter 164
is it going to be the same for all this matters or not 165
yes it is seen 166
because it is going to satisfy a given set of 167
boundary conditions so 168
either resolve that Lepause equation using analytical 169
medical graphical or experimental way 170
but if if if 171
the solution is going to satisfy 172
with given set of boundary condition 173
then the solution will be the same 174
for all these four different matters 175
and that's why we say that there's only one solution 176
and that solution is unique 177
if it is going to satisfy a given set of 178
boundary conditions 179
so unique mesteinum uh states that any solution of uh 180
leclase equation which satisfies uh 181
steam boundary conditions must be the only solution 182
regardless of the method used 183
premium applies to any solution of poisons 184
or lip gloss equation in a given region 185
or closed surface 186
so it is valid for both lip gloss and poison equation 187
so pre acquisition 188
for solving any of the body valley problems 189
you must remember all these three uh pre acts 190
these three steps as well 191
before proceeding or solving these uh uh 192
these problems so a rare solution is uniquely becomes 193
so rare solution is uniquely 194
describes a problem 195
that we have only a single solution 196
to these types of the problems 197
so first of all appropriate differentiate equation 198
so either it will be a class equation 199
or it will be a presence equation 200
a solution vision where you need to solve the question 201
prescribe boundary conditions 202
require bound boundary conditions or boundary problems 203
a problem does not have a unique solution 204
and cannot be solved completely 205
if any of these three items is missing 206
that either is nor 207
it cannot be described in a differential equation form 208
or either 209
you don't have some particular boundary conditions 210
then you will have uh many possible equation 211
that in finitely many possible equation 212
solutions that we see in our Mathematica terminology 213
however 214
if it is going to satisfy all these three period acts 215
that that the problem 216
it is going to have a particular 217
differentiate equation 218
it is being solved in a particular region 219
given region and it is going to satisfy the particular 220
bond conditions then you will have a unique solution 221
so general procedure for swarming poisons of 222
for our laplassecrations and these are that for laplas 223
you know that the wall of chart density 224
it is equal to 0 225
for the poisons it is not equal to 0 and if uh 226
they say there's only one variable involved 227
then 228
you can solve this using a direct integration method 229
so direct integration when we 230
is a functional only single variable 231
and um if one of them are able 232
and they can be seperable from each other then 233
then then you can solve for this uh 234
variable b using this separation of a even matter 235
we will see these matters in the upcoming uh 236
uh examples as well 237
solution at this point is not unique 238
but expressed in terms of unknown integration costumes 239
so 240
this means that as far as there are multiple costumes 241
and solutions are going to be multiple in finite 242
many solutions so how to find out a unique solution 243
you need to involve the boundary conditions 244
apply the boundary conditions 245
to determine a unique solution 246
for V that is your electric connection 247
in this electrostatic case 248
boundary problems imposingly given boundary conditions 249
makes the solution unique 250
and having obtained the potential 251
you can find out the electricity 252
by using this relationship 253
having found out this electrical 254
you can find out the electrical business 255
using this relationship 256
and if desired you can find out this charge as well 257
inducing on a conductor using this relationship 258
that charges the 259
uh is the surface integral of the uh circuit identity 260
where uh for this conductors 261
we know that uh 262
this rule as is equal to normal electrotroxancy 263
from this uh 264
boundary conditions that we discussed in check 265
in our chapter number five 266
which is the normal component of the electrical stansty 267
it is equal to this sufficial dansty 268
and there on if we integrate this ruas 269
ruas is equal to bn 270
d and can be found out using this relationship 271
and if we were 272
E can be found out this using this relationship 273
and the V can be found out using this 274
uh based on this boundary problem 275
and again uh 276
you can find out this capacity as well between those 277
between these two conductors 278
even if 279
if you know that this charge coming out from here right 280
and this potential coming out from this 281
over boundary problem 282
so you can also proceed in this way as well 283
the next show were some of the examples 284
to further clarify the concept of body valley problems 285
that how we can enroll these poisons and labour 286
secretions to find out the electric potentials 287
and electric fields 288
so here very important example in your 289
this is your practical example 290
the current 291
carrying components in high voltage power equipment 292
must be cooled 293
so cooling phenomena is to be exist 294
uh to clean because the heat is uh you know uh 295
it is eminating from these uh 296
high voltage components 297
and this company must be cooled down 298
so that they are not uh 299
uh hampered or there's no uh 300
you know fire hazards or there's no damages to the uh 301
overheated components 302
so you must uh uh have the mechanism 303
mechanism to cool down these electric components 304
uh because uh due to the again 305
uh because they are crank and compromise 306
so due to the only clouds 307
uh they are going to eminate or dissipate the energy 308
uh or dissipate the heat in their surroundings 309
and the surrounding has to be cooled down 310
so there must be some uh phenomena 311
and to carry out of any heat caused by the army classes 312
okay 313
physical to ice 314
so 315
a means of pumping is based on the force transmitted 316
to the cooling fluids to the cooling fluids 317
so a cooling fluids so inside this tube a fluid 318
a cooling a coolant 319
you can see as well a coolant or cooling fluid 320
it is uh uh residing inside this um 321
and inside the you know and 322
and you can you can see that it is residing 323
residing inside this are circulating as 324
as well inside this coolant two 325
okay so uh 326
transmit to calling uh 327
so a means of pumping you must must have a some 328
some procedure of pumping of this uh 329
this electric 330
this coolant in inside this tube so that the 331
it can cool down the sounding okay 332
so electro so this phenomenon is known as the electro 333
hydrodynamic pumping and it is more 334
it is more than this uh 335
figure and that the region between electro electrodes 336
that is electrodes 337
contains a uniform chart and steel or not 338
which is generated by the left electrode and collector 339
Gatby right electrode 340
and it is going to pumping 341
pumping this uh or you can uh 342
you can see that it is going to 343
circulate this uh uh this coolant in its uh in its uh 344
ultra parameter or inside this so that uh 345
the sounding is cooldown so a region between electrodes 346
the region 347
between these electrodes contains a uniform Cha chaunsy 348
so here you have this unit 349
this means it is going to work 350
going to be a Byzance problem 351
boundary problems and the left uh left 352
left toad and then eclectic on the right 353
left toad and accordingly 354
the things are going to be sinus idles 355
so that the coolant it can be uh it can be circulate 356
uh in this uh complete uh tube and what else 357
so the left side 358
we can assume that the potential at the left side 359
it is be not 360
and on the right side it is going to collect it down 361
so the potential on the right side 362
it is assumed as the ground or zero vote 363
so on both boundaries we know that 364
we know the potentials inside 365
we know that the there's a chart density inside this 366
and it is constant in this case 367
so unlike the traditional 368
Thermal cooling technologies 369
that rely on mechanical pumps 370
ESD cooling 371
uses electric fields to pump the coolant to the tube 372
so that is the electric field 373
which is generated by this 374
applying of this external voltages 375
electro supply 376
the voltage that pushes the coolant to the tube 377
ESTB systems what is the advantage 378
they are lightweight they consume a little bar 379
because they are involved with this 380
electrostatic phenomena 381
now 382
let's solve this example to find out this differential 383
uh since uh Robbie is not equal to 0 384
so we can treat this problem as a present equation 385
and this is a positive question okay 386
and this is a Slender Girl farm right this is a slender 387
so we can involve this slendrical coordinates 388
systems form of this presents equation okay 389
to body conditions we are given at z equal to zero 390
at the start of this tube 391
at these equal to 0 the potential is v 392
not at the end of this tube 393
at these equal t the potential is 0 where the 394
the coolant is going to collect it 395
and then the cycle will reverse 396
and the coolant will circulate 397
in the opposite direction 398
and the phenomenon will repeat 399
and uh show that uh 400
the V depends only on C okay 401
there's a Z 402
only depended upon the distance from the electrodes 403
so you plug in uh these two values 404
the only variable the only variable 405
on which these changes in the potential 406
they are depending is only in the z direction 407
right in the z direction 408
because these changes are only with respect to this Z 409
so this is going to be zero 410
the changes are for potential 411
with respect to this fi company is going to be zero 412
the changes with respect to this raw company 413
is going to be zero 414
and the only changes that we have in this uh 415
vertical phenomena that is with respect to just Z 416
because these changes are with respect to Z coordinate 417
so this is the second order differential equation 418
that you need to solve 419
okay 420
so uh 421
it only involves one variable that is your potential 422
so we integrated once right 423
uh this is the integration constant 424
the right hand side it is only having one constant 425
right 426
so integration of constant with respect to Z is Z right 427
so integrated twice to find out this v 428
you can see that 429
first integration eliminates the first 430
uh the second elevator of this differential 431
the second integration will eliminate 432
the first elevator of uh 433
this differential so this way if you twice 434
do twice integrate this equation 435
you are going to eliminate this 436
second order differential 437
so that it is reverse phenomena right 438
so integrate it twice to eliminate this 439
second order differential equation 440
so this will cancel out okay 441
and right hand size you need 442
you need to solve the equations accordingly 443
since it is uh us 444
uh second uh tricing integration 445
so you are going to have two 446
two constants here in the second um 447
integration since there's no one 448
one along with this 449
so it's integration is going to be Z 450
for this first integration 451
we have this Z here so 452
each integration is going to be Z squared divided by 2 453
okay and this is Rona divided by absmont 454
it is going to be constant okay 455
so we need to find out this a and B integration 456
questions so let's apply the boundary conditions 457
at these equal to zero we have the potential 458
these will be not so put in this equation okay 459
so first time is going to be 0 460
second time is going to be 0 461
and we are left on living B for your B constant 462
it is equal to B not now plug in this um 463
be not into this okay 464
and I'll put in this second differential 465
second boundary condition that your is equal to 0 466
at this d is equal to 467
as is equal to 3 that is your this end 468
this end and the first case does this end 469
now her plug in this says Z is equal to d 470
Z is equal to d and v is equal to 0 471
and solve the equation to find out this 472
second question that is your is equal to this one 473
and now you plug in this a and B into this 474
uh into this uh 475
standard equation of this uh 476
what age in terms of the question 477
then you can find out this that plug in this e 478
I'm being 479
today's to find out the final equation of this 480
uh voltage for this particular boundary problem 481
so what is that uh 482
final equation 483
that you can simply replace this thing with this 484
and this thing with this okay 485
and the next thing you need to find out is 486
the electrical 487
so that is the minus gradient of differential 488
and since it is clinical partner system 489
and the dependencies are with respect to Z only 490
so these two are going to be the zero and you 491
need to find out this thing 492
and this thing 493
you have already found out in your previous slide 494
as well or you can again integrate 495
you can again difference 496
differentiate this equation with respect to Z 497
to find out this thing 498
and if you differentiate this with this 499
with respect to Z then this is going to be zero 500
this is going to be one and this is going to be 2 z 501
right 502
2 2 503
2 2 is going to be cancel out 504
and you are going to laptop with Ronald Z 505
and you have a option on 506
and since there is a negative over here 507
so it is going to be positive 508
and it is going to be negative 509
and I plug in this constant E over here 510
so this is your electrical equation 511
okay now we need to find out in that force 512
F force we know that it is equal to q e and what is q 513
it is d volume and trigger larger row V 514
and what is row V row V is constant in this case right 515
okay so it would come outside right 516
and the volume and trigger 517
we can divide this into the surface integral 518
and into the third varying dimension 519
that is your d dimension 520
and this uh surface integral 521
we can simply treat this at the 522
treated as the surface area of this conductor right 523
okay and we are now left with only one integration 524
that is with the spectrum z divering 525
divering waver 526
and e e 527
e e 528
you can involve from here right 529
so the first term integration with the start of Z is 530
since it is on the constant 531
so it is going to be Z 532
second time it is going to be Z square divided by two 533
return time time 534
time it is the constant 535
so it is going to be Z so that is your force 536
you further simplify this for this integration 537
from 0 to D 538
and you will see that for the zero 539
all terms are going to be 0 for the Z equal to d 540
so this is going to be d square 541
this is going to be d square 542
these two are going to be cancelled out 543
and you are left out with this first term 544
okay 545
and then 546
this day is going to be cancelled out over here 547
and this is the only thing that you are left out 548
after the intergation no 549
we need to find a deforsed pressure 550
that is your force by Unitarian 551
so activate the s as is going to be cancelled out 552
and now you plug in your zonaut 553
the given or not and v not 554
so this is your final pressure 555
which is applied by the external potential 556
or accordingly the electric feel 557
the applied electric feel within this tube 558
so that is the pressure 559
which is applied on the coolant for pumping in 560
into this tube 561
let's follow this 562
another example to find out this what is an electrical 563
so what is this semi infinite conducting pain at 564
5 equal to 0 and 5 0 5 by 2 565
so I start driving geometric semi infinite plane 566
connecting plane at 5 0 0 567
so this is your x y z X is uh plane 568
so add X is equal to of 5 569
we're not at 5 equal to 0 at this point 570
so it is semi infinite plane like this one semi y 571
it is a half plane 572
but it is infinite in this extra action 573
second manager of 5 is equal to 574
second plane is at 5 is equal to 5 by 6 575
that is your 576
I guess 30 degree angle by by 6 wanted to divide yeah 577
it is 30 degree angle 30 degree angle okay by by six 578
okay so that is another plane 579
second infinite plane in this direction 580
an infinite in this direction 581
the boundary conditions that we are given is that at 5 582
5 0 we have the voltage zero right 583
so this means it is a ground this thing 584
okay the 585
and the second bomb definition that at 586
at second infinite plane it is visical to v not 587
and in between these two in between these two plane 588
it is 589
an insulating gap here 590
you can can see that the gap is over here right 591
so this is the gap between these two planes 592
so that they are not shorted 593
and we need to find out the 594
potential and electrical 595
in the region between these players that is your 596
this region you can see okay between this region 597
so 598
let's go 599
what is this particular case 600
uh in this case 601
uh you can see that you're not provided with the 602
any of the information of uh density 603
so this means that row v is equal to 0 right 604
so it will be equal to zero that it is going to be a 605
the class equation problem 606
and since the things are given in a row 5 z 607
so it is a splendidal partner system problem 608
so variations are being specified 609
so rest of the variations are zero 610
because we have seen that 611
the boundary conditions are given with respect to 612
by corner system 613
so we'll only need to solve this 614
homogeneous differential equation 615
and you can sense at sense rose equal 616
zeros excluded due to the insulating gap right 617
there's a gap 618
so at rose equal to 0 and we don't have any potential 619
or you can simply eliminate this factor as well 620
because using this elementary mat 621
that you can take it on the right hand side 622
so it would be eliminated 623
and you are only left out with this equation to solve 624
so again 625
you need to intigate it twice to get rid of this 626
second order differentiation 627
to solve this potential 628
first integration is going to provide you this 629
a only 630
the second generation is going to provide this y and B 631
okay so it's a second dimension 632
applying boundary conditional 633
to find out these integration constants 634
so at 5 is equal to 0 b is equal to 0 635
so this will give you b is equal to zero 636
and plug in this over here 637
and not integrate this second 638
uh not 639
plug in this second boundary condition 640
that at 5 is equal to 5 not 641
which is equal to v not so v not is equal to this one 642
and this is your is equal to v not to verify not 643
now plug in these two constants into the this one 644
b is equal 0 is equal to this one 645
and this is your final form of uh potential in the uh 646
in terms of this varying uh variable 647
that is your fine 648
okay now when we need to find out the electrical field 649
second part that is minus gradient of b 650
so there are two ways there are two ways either uh 651
you in uh 652
you differentiate this equation with the spectify 653
or you can also uh bring in 654
bring in this d read by right 655
defire from this first integration start 656
and then uh 657
uh know what would you need to find out substituting 658
you need to evaluate 659
basically you need to evaluate this uh uh 660
electrical at this uh point and where uh and this uh 661
vertical boundary conditions 662
so what are these boundary conditions that uh 663
at this uh 5 equal final 664
that is your 5 by 60 second boundary condition 665
your B is equal to V not okay 666
so this is your final form of potentials 667
and electric field reading with respect to this 5 okay 668
and what else 669
and here you can see that 670
the variations are with respect to your 671
for electrically 672
the variations are with respect to your look 673
k and for potential degraations 674
that will structure your fine 675
so let's um find out whether these two equations 676
they satisfy our boundary values 677
uh boundary conditions that that we have 678
we have been provided earlier 679
you can see that at 5 equal to 0 680
this is evaluating to be equal to 0 681
and at 5 equal 5 by 6 682
this is evaluating it equal to hundred 683
so plug in this 5 equal 5 by 6 right 684
5 by six so this will cancel out 685
this will cancel out 686
and the voltage is going to be hundred 687
so this means that for this poison equation 688
boundary weight problem we have satisfied the 689
these two conditions boundary boundary conditions 690
let's all disturb the example 691
this is given in your uh central partner system 692
uh cervical partner system okay 693
uh two connecting cones 694
so first connecting cone it is your inner cone right 695
it is at uh 5 by 10 there 696
this one 697
and the second connecting cone it is your outer cone 698
uh both are uh infinite in extent yeah 699
infinite in this uh in this in this 700
in this extent and separated by uh 701
infinite infine infinite infinite decimal 702
infinite decimal disabilities 703
infinite decimal gap that is this gap 704
with the insulation right so that they are not shorted 705
and uh we have given the boundary condition that 706
for the first corn it is at zero potential 707
for the second corn it is at 50 potential 708
that is where we not okay 709
and again we need to find out the electrical 710
and the electrical potential in this area 711
in between these two planes here 712
let's solve this uh by treating it uh as a uh 713
boundary belly problem so again 714
in this case we don't have the information of uh 715
1 inch identity we don't have it right 716
so it is a left last equation case 717
and since it is in uh your cervical corner system 718
so this is the 719
and the changes that we respect to treat only 720
so we are left with only this equation of this uh 721
second order uh differential equation 722
and again the changes at your data is equal to 0 723
and as since I since r is equal to 0 724
and at the data is equal to 0 and pi are excluded 725
we can um 726
multiply these two terms on both sides 727
and also mathematically it can be excluded 728
because it is treating as a constant 729
because the variations are with respect to this 730
the teacher 731
okay so since R is equal to 0 and r is equal to 0 732
you're uh it is both conditions of cheetah 733
Cheetah is vating from uh zero to pi 734
they can be excluded uh 735
because uh the mortgages are going to be zero 736
at these two particle points 737
so magnetically can be eliminated as well 738
by bringing them on the right hand side 739
so this is the thing that we need to fall 740
integrated ones we are left out with this thing right 741
on the right hand side we have the constant 742
we know that integration of the constant uh 0 743
it is a constant and the differentiation of the 744
with the special tea diet is very important concept 745
it is constant 746
and the differentiation of a constant is 747
zero because it is constant 748
so you must remember 749
this differentiation and integration of the constant 750
that the integration of 0 751
it is some constant and the integration of some 752
differentiation of constant it is zero 753
okay so bring this on the right hand side 754
or we are left out with this thing 755
but we have to take these second integration 756
so that we can eliminate this thing 757
and we are left out with the prevention only 758
okay so in this case 759
we need to enroll this uh geometric identity 760
your uh that uh sign of teacher 761
it is equal to sign of teacher 762
it is equal to 2 times cause of teacher by 2 763
and sign of teacher by two 764
uh what is this property that is your sign of 765
A plus b 766
it is equal to 2 767
times cause affair n 768
sign off 769
B essence a and B both are Peter by two 770
right 771
and here you can also treat it like A's t W 2 772
b is t W2 and you can write this in terms of this 773
okay so we already have this sign of Tita 774
so we have converted it like this okay 775
so multiply and divide this by this Cobra Peter by two 776
so that we can have the uh 777
we can have what we can convert this thing into a 778
into a this thing 779
this thing so divide it over here 780
so this is your 10 Peter by two 781
and this thing is your core square Peter right 782
and you take it in the numerator 783
so it is your second square Peter by two 784
and now 785
involve this very important property in English 786
and property 787
if you have the databasive of the denominator 788
in the numerator 789
we know that deteriorated of the stand theater 790
it is equal to Sequence Square Theater 791
so we can treat it like that 792
this is this integration is equal to Alan of the 793
this denominator 794
so here the denominator is tantita by 2 795
so integration of this is 796
it is Allen of tantita by 2+ the constrat 797
this e constrat of integration 798
okay and also you know that one of our X 799
it is equal to alanoff 800
integration is equal to alanoff x plus 801
construct okay 802
so you can also use this chain rule as well 803
this this substitution rule as well 804
so these are the techniques of integration 805
that's why I constantly say that 806
you must know 807
what are the differentiation and integration rules 808
and how to find out the differentiation 809
and integrations 810
to have success in this particular course 811
but the things are simple 812
over here we have we have found out this voltage 813
no we need to find out these costumes 814
and let's solve this equation 815
at this given boundary conditions 816
so for the first boundary condition 817
at Peter's equal to Peter 1 818
we have this water is equal to 0 819
so from we plug in this to this boundary condition 820
so this is your constant b 821
okay now plugging this b over here 822
what does uh 823
Peter is equal to Peter 2 right 824
Peter is equal to Peter 2 and reads equal to 825
we not and uh 826
what we are now going to find out 827
we are going to find out simply this think that 828
that in the first step in the previous step 829
you know that Eleanor 830
a divided by B 831
it is equal to Ellenoff a minus Ellenoff B 832
so we simplified in this previous stat that 833
I don't know if uh 834
tta by 2 minus E I don't know tta uh 835
10 tta 1 / 2 we have simplified after putting in this 836
uh 837
uh this found out uh 838
question of integration that was your b 839
uh you can simplify this subtraction of Alan 840
Alan terms in the form of this 841
division of these two in a terms 842
and then take this Eleanor 843
Eleanor outside of this division term 844
so remember this identity 845
the logic media identity 846
that the difference of Ellenoff two terms 847
it is equal to the Ellenoff ratio of these two terms 848
so the division of the two terms 849
uh Allen of two terms 850
it can be written as the uh 851
difference between the Allen of these two terms 852
in the other way 853
okay so once we have done this thing 854
you can plug in this second boundary condition 855
and you can simplify this thing 856
to find out the second question right 857
uh E and not plug in this into this 858
okay into this 859
to find out this standard potential 860
in terms of your varying constant 861
that is your variable variable in your 862
in your variable or variable coordinate 863
that is your teacher in this case so this is your final 864
equation in terms of your data 865
only for this vertical vulnerable problem 866
we found out in London London 867
we need to find out this 868
uh electrify and this electrify 869
it is minus period of fee 870
so this is your diss track right 871
uh since changes are over the special teacher 872
so that's 873
that's part of this gradient terms are going to be 0 874
that is your changes with spectify 875
and changes with respect to your R there 0 876
and your you need to solve 877
well solve this 878
only tell the differentiate part of this gradient term 879
so either you can 880
take its differentiation okay 881
or you can basically uh juicy 882
uh first integration 883
uh that we found out uh in our previous life 884
the first integration 885
once we done the first integration step 886
you can take this uh DVR data from data as well 887
but remember to multiply that with this minus 1 or R 888
so this is your electric field and it is in terms of 889
you can see that it is in terms of your boat 890
teatime car 891
let's uh verify 892
let's find out these evaluate these to uh 893
uh potential electric Protection 894
electric field at the given uh boundary conditions 895
so at the time teacher 1 896
we are given that it is power 2 897
power 10 teacher 2 898
it is power 6 and we not it is equal to 50 899
so this is your final expression 900
for the electric potential 901
and the electric fee and you can find out this 902
you can verify this 903
electric potential equation 904
for this boundary value conditions as well 905
that at t ties equal 5 or 10 906
it it should evaluate equal to 0 because uh 907
you will see that you 908
you can value it at your own times 909
and then the t tie is equal to uh 910
by the 6 it will value equal to be not that is your 50 911
okay and then it will also satisfy your 912
what is on the question this lapation equation as well 913
so you double differentiate this thing 914
with respect to this data 915
then you will will 916
you will find out that it is going to be equal to zero 917
message from your lecture 918
for this electrostatic boundary problems 919
we understood what are the 920
what is the significance of these 921
electrostatic boundary problems 922
and how to find out this poisons 923
and the class questions 924
and what is the geneticness theorem 925
and then uh how 926
how to involve this uh 927
electrostatic boundary value problems and uh 928
these poisons and laprasecrations 929
to find out the electric uh 930
field and electric potentials 931
and by solving these boundary value problems 932
based upon these by zones and the class equations 933
next time we will determine this 934
the resistance and the capacity 935
and by solving these Bondi value problems 936
here I thank you all 937
if you have any questions 938
uh they will be entertained 939
uh with your emails or online 940
synchronous sessions that we have arranged 941
at the department