this is election No. 22 of a V232 2
electromagnetic field theory today 3
uh we just build on our uh previous lecture 4
uh that was how to determine this uh 5
resistance and capacity of uh 6
registers and capacitors using the boundary problem 7
and in this in this lecture 8
we are going to solve the examples of that 9
previous case and this will again cover your section 10
as remaining part of your Section 6.5 11
so we will uh will solve the 12
some of the case examples 13
remember this uh slide that we discussed in details 14
in our previous lecture that if uh 15
if you assume uh for the capacitor 16
if you assume that discharge is constant 17
and you find out the V in terms of Q 18
okay then you can find O d capestinence 19
and how to find O d v uh 20
this is a potential 21
uh across that uh parallel capestial that is 22
uh for that we know that the relationship is uh 23
the line integral of your electric field 24
and how to determine this electric field 25
you can take help of either causes 26
law or decoulents law 27
okay and then uh what was the second uh 28
second uh method that you 29
you assume that the voltage is constant across the 30
by the plate capacitors uh 31
and then you find out Q in terms of 32
q in terms of your E 33
in terms of your V okay 34
in terms of your V not okay 35
and how to find out the eh eh 36
this eh queue in terms of v not 37
you again need to take out of this eh 38
eh flux equation 39
and for this you need to find out electric field 40
electric field is related to the 41
there by this gradient 42
relationship with the elective potential 43
and how to powerfinder this potential 44
then based then according to these scenarios 45
either as you can apply this lapse equation 46
or the portions equation 47
so you need to solve this boundary problem 48
uh for that uh for these particular cases for 49
for which you have this volume circumstancety 50
or you don't have this volume circumstancety 51
and how to find the resistance of the capacitor 52
for this case the first matter 53
presumed that your vote is across that resistor 54
it is constant 55
you need to determine this current using the 56
this grand density and we're using this homeslaw 57
you can find out the tendency for this 58
again you need to find out this electric field 59
and it is related to the elective potential 60
by using this relationship 61
and again you need to solve this 62
the class and pause 63
and equation to find out this potential 64
uh this again uh 65
recap of your previous structure that uh 66
these are the capacitances uh 67
uh of different types of uh capacitors that we have uh 68
drived for the generalized cases 69
okay and then accordingly 70
you can find out 71
the legal distance between the players 72
that is uh 73
your uh due to your medium 74
uh uh that uh that can be 75
that can be found out using this relationship 76
that was your relaxation time relationship 77
that is the ratio of your primitivity 78
and the connectivity of your medium 79
and using this relationship 80
and using this capacity over here 81
you can find out the resistance 82
like like this 83
like this in this case but remember 84
that this resistance is the leakage resistant 85
it is not the distance of the capacity it played 86
but it is the leakage resistance between the plates 87
first of all this example 6.8 of your textbook 88
so a metal bar of conductivity 89
Sigma is bent of form of flat 90
90 degree sector 91
so it is 1/4 of the 1/4 of your complete circle and 92
its inner radius is a this is your inner radius 93
its outer outer radius is B 94
that is this one this one outer radius 95
inner radius 96
and its thickness is thickness is t okay 97
and we need to find out the resistance of the bar 98
between the vertical curve surfaces 99
at r is equal to a n r is equal to B 100
so that is the 101
you need to find out the resistance between these two 102
okay between these two surfaces okay 103
this sensor inner surface and the outer surface 104
and what is the second thing 105
this is chance between two horizontal surfaces 106
that is your this one and this one 107
top one okay 108
between your top and the bottom surface 109
so that is your second thing 110
our bar 111
our bar and our SO2 resistances you need to find out 112
um between a vertical curved 113
and located at this first radius 114
roads equal to a and roads equal to B 115
the bar has a non uniform cross section area 116
because you can as you can see here that uh uh 117
that uh this area is going to change right 118
this area is going to change between these two uh 119
between these two vertical 120
these two vertical these these are the verticals okay 121
inner inner surfaces okay inner surfaces interceptors 122
so between vertical curved ends 123
located at rose equal to a and rose equal to B 124
the bar has non uniform cross section area okay 125
because it is a see here it is a 126
it is sector of row is equal to B 127
and sector of row is equal to a 128
so let's assume the potential difference 129
be not constant 130
potential difference between these two curve 131
that row is equal to a and 132
row is equal to B 133
that we assume that at rose equal to a 134
and rose equal to B so 135
that we just take the Assumption that 136
if this potential difference 137
we not is constant between them 138
then add inner radius you potentially 0 okay 139
at this your potentially 0 140
and at the outer radius 141
you are taking the potential as we not 142
okay 143
and now we need to find a 144
since we don't we don't have the uh wallet identity 145
uh so uh 146
we need to solve this labrass equation 147
in this Slendrical Partners okay 148
why Slender Partners you can see this material here 149
this Dmitry is your slendricals material here okay 150
the thing is so 151
so the things are changing with this back to your row 152
the variable is row okay 153
so you uh you take its first integral 154
so you you just need to recall your uh 155
this uh laplash integration in terms of your uh 156
slander regarding assistant from your earlier chapters 157
okay 158
and uh then integrate it once take this rubber here 159
integrate it twice 160
you know that the integral of this uh 161
one or row it is and enough 162
and then off row like this one OK 163
with respect to this G row 164
okay so as you're having two constants of integration 165
you need to find out 166
these two constructs of integration 167
so you need two boundary conditions 168
always remember the number of unknowns 169
for the unknown under and for the 170
for same 171
for the number of boundary conditions required 172
to determine the number of unknowns 173
must be same of 174
same numbers two unknowns 175
so two boundary conditions 176
so the first boundary condition is at inner radius 177
your what is 0 178
and the second boundary condition is your 179
at your outer radius your what 180
radius constant that is v not so plug in these two 181
um boundary condition so for 182
for the first one you will find out this B constant 183
which you use this uh you plug in this into your hair 184
and then you plug in this second 185
boundary condition to find out this a 186
okay so in the 187
at this stage 188
we have found out these two construct of integration 189
uh no 190
uh no 191
you you uh you need to uh just rearrange the things 192
okay just uh rearrange the things 193
plug in these constant values and uh 194
you you also know this uh 195
rule of your love with me 196
rule of subtraction that 197
subtraction of love 198
of two terms is equal to the love of their ratios okay 199
like this one okay 200
okay so now you know the potential 201
you just take its gradient to find out electric feed 202
so this is your electric feed 203
and did you have found out this 204
so I did there there are two things 205
either you just take the gradient of this thing 206
or you can just use this uh 207
uh first 208
the things that we have found out 209
after taking the first interior of blood 210
cash and equation on the previous life 211
you can directly plug in that as well 212
but take care of these symbols and things like this 213
so it is always better it is always better to 214
take the gradient of this this thing okay 215
to avoid any confusion of the corner system 216
no involve this homes bar that your surface condense t 217
it is equal to the stigma e 218
and here see that your d this 219
surface surface area differential surface area 220
it is having minus sign so y minus sign 221
y minus sign because this is your positive 222
and the inner is your 0 0 negative 223
so electric field it is moving from outward to inward 224
so that's why is your 225
this radial direction of the electric field 226
is in minus a row direction 227
so this is very very important point over here 228
that's why it is minus sign over here 229
because your electric field 230
it is against this radial direction 231
okay so plug in this in this relationship and what is 232
what is your left with 233
everything is constant over here okay 234
and uh 235
you just need to integrate it with respect to defy 236
so you will have power 2 okay 237
why power 2 because it is a 90 degree sector okay 238
so here 5 it is 5 by 2 239
the second thing that it is raining is your DZ okay 240
the DZ in this direction okay 241
in this direction 242
because your changes are in the radio direction 243
so your surface idiot is differential surface idiot 244
it is equal to this just recall 245
recall your chapter number two 246
and the second variable is yours changes are to two 247
2 variables 248
the first one is your uh this uh English changing 249
the second one is your length is changing 250
like this one okay 251
and uh what else 252
the integration with respect to Z is Z only 253
and then this will replace your TR okay 254
the thing is already constant 255
and this is your private dessert 256
this in integration with respect to your fine 257
so you have found out this I 258
in terms of your we not that was 259
which was your and objective 260
so this we not and this we not will be cancelling out 261
and this is your resistance that you need to follow 262
for this between these two vertical plates 263
between these two vertical plates in and out of plate 264
let's not solve the second part of this example 265
where you need to find out the resistance 266
between these two 267
horizontal horizontal surfaces okay 268
and here we assume that at the bottom surface 269
there is your Z equal to 0 270
we are assuming that is potential is zero okay 271
and at the top surface 272
we are assuming that it is having constant potential 273
that is v not okay 274
and this and this surface it is v not 275
and this surface it is the bottom surface 276
it is zero okay 277
and uh no the things are changing 278
the things are changing with respect to your Z axis 279
okay Z axis 280
so that's why you have 281
we are having variations with respect to your Z 282
Z and the lip class equation 283
again we will use this to find out the potential 284
and according to the electric fee 285
and uh again this very uh 286
this is slendical uh 287
uh slendical uh corner systems 288
lap class equation and then integrating it twice 289
then this is uh uh 290
v in terms of potential in terms of your uh 291
uh constant uh in integration 292
conscience of integrations 293
and now you need to apply this 294
simply the boundary conditions to find out the 295
these constants that is your b equal to 0 and a 0 296
b not divided thickness and uh 297
this is your potential 298
there's your potential in terms of here 299
you're evading parameter that year Z 300
and as you take the gradient of this and uh 301
that is your electric field 302
now if you know the electric field 303
then you can find out the current density 304
using this Homs law and not again 305
to find out its surface integral with respect to 306
to find out this current you can 307
you see your head 308
and you have taken the bottom surface at zero 309
the top surface at me not 310
so you electric feel 311
electric feel it is 312
it is from top to bottom 313
that is your minus easy direction 314
okay so the normal here 315
it is minus easy okay 316
and so the things are wearing with this 317
so what is changing over here 318
what is changing over here 319
your radius is changing your fire is changing 320
and your indeed this normally is your 321
easy okay 322
normal to these two separate surfaces 323
and the wide way of taking minus easy 324
because your electric feel 325
it is from top to bottom okay 326
no the things are very simple here 327
again you need to uh 328
integrate this with respect to fire 329
that is your fiber tube the sector is your fiber tube 330
and uh 331
then integration with respect to row it is your row 332
scale on 2 333
the image they are from a to B and uh 334
just plug in these two things and uh 335
not use this current over here 336
and then you can simplify the things 337
so this this is the 338
and this is the uh expression that you 339
you were required to dry 340
now alternate alternatively 341
for this thing is using just basic formulation 342
because for the top surface 343
the cross section area 344
it is uniform both top and bottom surfaces 345
they are having the same same 346
same surface area you see overhead this 347
this is the same surface area 348
the same as as I'm sure bottom one 349
so you can apply the basic formula as well okay 350
the only difference is that here the surface area 351
so as here over here the thickness 352
what is the suppression between them 353
the distance between them 354
it is your t okay 355
and what is your uh what is your uh this uh 356
what is your uh this uh 357
uh that is uh 358
surface area since it is out one fourth of your 359
circle so that is your 1/4 of your 360
one fourth of your by a square okay 361
and what is over here that is your B minus E 362
the effective the effective 363
this your effective radius 364
so that is the another method 365
let's solve another example 366
6.9 a coexial cable 367
contains an insulating material of conductor desigma 368
if the radius of the central y is a 369
and that of the sheet that is your most 370
it is a ground and that is a b 371
so we need to uh 372
show that the conductors of cable per unit land 373
is this one okay 374
conductors per unit land for the dialectic 375
it is denoted by G and it is the recipe proclub here 376
this leakage resistant so actually 377
we need to find out this are here in this case 378
okay so again we uh 379
we use this uh uh 380
boundary problem case that for this particular uh 381
cylindrical quaxia case that uh 382
the inner radius the inner 383
the inner inner one is at your potential uh 0 384
as you have seen in your 385
in your previous slide that the inner 386
inner thing at it is at inner radius conductor 387
it is at differential 0 388
and the outer one is at potential v 389
not nothing's are changing with respect to your role 390
for this slendical partner system 391
no as you involve again delapasin equation 392
take its double integral okay 393
and so it is just like your example 394
6.8 where we found all the voltages 395
and the current and electric wheel 396
as per these expressions 397
not involve this electric field over here 398
not here again again this is your zero okay 399
this is your positive 400
we not this is in your minus 0 direction okay 401
that's why it is negative sign is coming over here okay 402
so what you need to do that 403
you just take the drop product of this 404
with this differential surface area 405
take its integration 406
so the fire it is changing from 0 to 2 pie the Z 407
your this Z direction you can take it as a C direction 408
okay 409
it is changing from zero to L and 0 to L okay 410
and you just intigrate this thing okay 411
so just intigrate this 412
so you're going to have this to pie 413
you're going to have this L over here 414
and the other restings at your constant okay 415
either constant okay 416
okay so this Roy is going to cancel out with this one 417
so resistance per unit land resistance per unit land 418
you can just take it like this that resistance 419
what was the resistance it was we not where I okay 420
just plug in over here and for portion of land 421
you just need to divide the distance with the R 422
divided by this complete length okay 423
so this will provide you the distance portion of plan 424
and just take your stress approval to find out 425
deep conductor conductor plan 426
yeah another is useful example 427
the connecting sphere steady 428
connecting sperical shell with your dye 429
is equal to 10 centimeter 430
and b is equal to 30 centimeter 431
are maintained 432
at a potential difference of hundred volt 433
such that your alter radius 434
in this case your alter radius 435
see alter radius see that 0 degree and 0% 0 volt 436
and the innerry is at your this hundred volt 437
so electric creates in Europe this a r direction 438
pause to be our direction okay 439
you need to determine this voltage 440
and the electric wheel 441
in the region between the shells 442
and where this homogeneous dialect medium is 443
having this relative permitivity 444
you need to determine the total charge 445
induction of the shells 446
and the capacity of the capacitors 447
let's uh 448
proceed one by one 449
since the things are changing in a videos 450
back to your radio directions for this uh spherical 451
um capacitor 452
uh 453
so you need to involve this lip Russian equation again 454
um because there's Ruby's equal to 0 0 here okay 455
Ruby equal to 0 456
just take your stimulant trigger and so this is 457
this is your potential 458
in terms of the integration constant 459
now 460
apply this boundary conditions that at inner radius 461
your voltage is 462
we not at the outer radius your voltage is 0 463
and once you will find out 464
the construct of integration 465
just plug in into your this main equation 466
so this is your potential for this 467
this concentrate uh spherical uh spherical capacture 468
triplet capacture 469
but you have found out the potential 470
take its gradient to find out the electric field 471
once you have found out this electric field 472
put this electric field into this equation 473
now you to determine this capacity 474
we need to find our BV on either of the shell okay 475
uh q discharge on either of the shell okay 476
so here you have a zoom d 477
we not as a constant 478
remember we not as a constant you have a zoomed it 479
and so you need to find out the 480
this queue in terms of 481
we not remember this relationship that the charge 482
it is the surface integral of your electric feed no 483
what things are changing over here 484
your tita tita for this 485
you know that in this vertical coordinate system 486
it changes from 0 to pi 487
your this Fi it is changing from 0 to pi 488
and they change that with respect to your defined data 489
and this is in your surface 490
the normal was your er okay 491
the normal unitractor was your er 492
so what you need to interview 493
okay so as you 494
so what we have assumed that being with constant 495
and we need to determine q in terms of v 496
involving this solving this Lapras is equation 497
no just simply put this into your hair 498
so you're not really canceled out 499
so this is your capacitors for this radical capacitor 500
no you need to evaluate this 501
find out this capacity for this particular case 502
that radius are given and even notice 503
given just all these voltages and electric read 504
and the charge on these uh 505
inner and outer shells 506
and accordingly deep capacity between these two shells 507
okay another example 508
uh 6.11 509
uh in section 6.5 we uh 510
we solve the same example but there we assume that uh 511
Q was uh we 512
so what was that previous guess 513
uh previous guess was that uh 514
there are true matters uh 515
contestants can be founded 516
by either as human Q and finding out v or by either uh 517
assuming V as a constant and finding out Q 518
the formal approach was used in Section 6.5 519
where we assume the q as a constant 520
but here in this case 521
we're going to assume that your V not is constant 522
and we will find out the q in terms of your V not 523
so this is the little matter 524
and here again the boundary conditions are given 525
that at your X is equal to 0 526
that is your changing variable 527
your X is equal to 0 528
and your potential is taken to be 0 529
and X is X is equal to d 530
the suppression between these two players 531
the potential is we not okay 532
so that's why you're this electric wheel 533
it is in your minus a X direction 534
okay so again solve this lapation equation 535
take the double in trigger 536
put in this boundary conditions 537
to find out the constants of integrations 538
and once you have found out 539
so this is your 540
and voltage in terms of your changing variable acts 541
is to find the questions 542
first to find the charge on either of the plate 543
so here you remember that your charge 544
it is equal to your 545
that uh the charge it is equal junior 546
and this uh circus integral of here 547
uh this uh density this is circuit chart density 548
and you remember that from your boundary condition 549
that your surface chart density 550
your surface identity it is the drop product of here 551
the electric proxanity 552
with the normal genitomimal factor 553
so that's why you can you can also treat it like that 554
it is equal to d dot DS okay 555
and here's your D uh 556
here's your d is uh 557
your judges is target in terms of your uh 558
in terms of your basic definitions that your charge 559
it is a surface integral of the charge dynasty 560
and it planned it destroys the protigue detour 561
you can equently you can 562
you can equently treat it treated like this that uh 563
the using the boundary conditions for the conductors 564
your surface density is equal to the uh 565
top product of your unitomal and e lactic uh 566
flux tensity 567
okay now simplify this thing okay 568
and what you need to do 569
you need to just take the gradient of this thing 570
so this is your electric field 571
okay and uh for low place 572
uh you know that your uh uh a normal 573
it is in the post 2 direction for the upper plate 574
your normal is in B minus the X direction 575
so uh your raw s your raw s is equal to uh minus aplon 576
we not do advantage okay 577
and uh so if this is the case 578
if this is the case you see here 579
so how you're playing around that 580
your rule as it is equal to appslon right 581
appslon e dot a n okay 582
and where's your E it is equal to this one okay 583
so that's why your 584
this race is coming out to be minus ABS 1 585
b not d for this bottom plate 586
and if we are solving this equation 587
this equation okay 588
this equation so what 589
what is using a simplification 590
that this integral s is equal to your simply s okay 591
so that is another simplification over here 592
and so you are just just playing with this equation 593
so this equation is no reduced to 594
your Q is equal to raw s and s 595
that is the thing okay 596
where your raw ass it is coming out from here 597
and this raw ass is coming out from here 598
and this E is coming out from here 599
so everything is interrelated 600
so this is the charge on 1 601
1 of the plate and if you find one 602
need to find out the capacitys 603
between these two players 604
just take the magnitude of this charge 605
divided with the constant 606
potential that we have resumed 607
so this is the same thing that we have derived 608
using our first method 609
where we assume the charge constant 610
and found out the potential 611
let's solve another example that is 6.12 612
here 613
just need to determine the capacity of the capacitors 614
and here there are two mediums okay 615
X1 R 1 and X1 616
2 and the separation between them 617
these two plates the capacitors 618
the plates 619
the only thing is there are two conductors okay 620
and there are two 621
two mediums to instead of single medium 622
there are two mediums between them 623
okay that is the difference 624
and these two mediums are 625
they can be changing in this 626
in this manner or they can be changing in this manner 627
okay 628
next next 629
solve this two cases one by one 630
that what is the impact of this configation 631
on the capacitors of this capacitor 632
so in the first case your 633
this electric field 634
they are normal to the dialectic interfaces okay 635
normal you can see here 636
it is originating from here 637
and it is terminating from here 638
it is normal okay 639
it is normal to the dialectic interface 640
what is dialecting interface 641
that is your dialectic interface 642
okay that is this one 643
or on the other hand on the other hand 644
you can assume that the charge is constant 645
across this capacitor okay 646
and the capacitor can be treated 647
so in this case 648
the capacitor can be treated like two capacitors 649
in theory okay 650
because this charges 651
the charge is constant 652
the charge is constant across these two capacitors 653
whereas in this second case 654
your voltage is constant across these two 655
across these two mediums or the inner 656
where the capacitors 657
okay so the capacity 658
and Figure 1 can be treated as two capacitors 659
C1 and C2 this is your C1 660
this is your C22 capacitances 661
and they are treated to be in series 662
because the charge is constant 663
remember q is constant 664
in this configuration and also 665
the electric feed is normal to the dialectic interface 666
so if this is normal than for C1C1 667
you see here the surface area 668
the surface area seem s okay 669
however the depth the distance between these two 670
the plate and the interface 671
it is d by 2 and likewise 4C2 672
now you just need to involve the formulation 673
uh 674
for the Cds capacitors to find out the total capacitor 675
and solve this thing 676
and we'll wait our deep particular given values 677
and if the electrical 678
they are parallel to the dialecting interest of 679
what did that is your 680
this case was to play it not to play it 681
and in this case 682
the voltage is same across these capacitors 683
so if the voted is same across these two capacitors 684
then they will be treated as battle capacitors okay 685
C1C2 686
and in this case the only difference is 687
another difference is 688
the distance is d between these two plates 689
but the surface area of these two plates 690
it is passed by to 691
really last example 692
uh a slink like a pastor has a red eye of uh 693
inner red eye of uh 1 cm or 2.5 cm 694
if the space between plate is filled with a homogeneous 695
in homogeneous remember this is no 696
this is different case in homogeneous dialect media 697
so it it is going to exhibit the uh 698
variations in its relative permittivity 699
with respect to this variable 700
that is your role the radial radial direction 701
now we need to find out the capacitys per metre land 702
in this for this capacity again 703
Jews this guardians involved the squads law 704
so we 705
we already studied that the electric field for these 706
these types of the electric field 707
electrically in between these two 708
concentric shells for this aesthetic slendical pasture 709
it is varying in this manner okay 710
and if we know this 711
we can find out the voltage using this 712
e dot g relationship okay 713
and here you just need to take care of this thing 714
that absline is not absline 715
not in this case 716
rather it is actually not time digital permittuity 717
and where the relative permittuity 718
in this particular case 719
it is our function of very well 720
that is your role okay 721
so here is going to cancel out this thing 722
this thing and your left out with this thing 723
and the integration of this thing 724
integration of this thing 725
it is equal to Eleanor 726
this 10 + 2 because you remember that one of a row 727
it is equal to Eleanor Row 728
because it's derivative is 729
it is available in the numerator 730
and again it's derivative which is one 731
it is available in the numerator 732
now just put put put put my put in this uh 733
uh radius to find out this uh voltage 734
and then we'll wait it at the given 735
at the given values of the radius 736
and since we need to find out the capacity per metre 737
so just put this allergy for two 1 bit 738
so this is this is the assignment for those from the 739
this chapter No. 6 740
so all these three and these practice 741
practice exercises of your chapter No. 6 742
for this second edition book 743
so all campus students can be conserved 744
with your own campus students 745
for the exact problems that you need to solve 746
so I'm here today's lectures 747
so today we just build on our previous uh lecture 748
uh where we determine the resistance and capacitys 749
for standard uh standard types of the capacitys 750
and we solved here uh some examples 751
some case the case examples to uh 752
determine the capacitys and the resistances 753
using either the causes law or the 754
by solving these labrasians equations 755
and then 756
based upon these different matters that there are 757
if we if we keep the voltage cost 758
and then we need to find out the charge 759
in terms of the water is and if 760
and vice versa now we if we need 761
if we keep the uh this uh current cost 762
and then we need to find out the voltage 763
in terms of year 764
current to find out this uh 765
resistance for the conductors 766
in the next class we're going to study another method 767
to find out the electric potential 768
the electric field for the charges 769
and that is known as the method of images 770
here I thank you all 771
if you have any questions 772
uh they will be entertained 773
uh through your emails or online 774
synchronous sessions that we have arranged 775
at the department