this is election No. 10 of a V232 2
electromagnetic field theory 3
today we will solve examples from your previous trophy 4
that is 5
electric feed due to continuous charge distributions 6
and electric first density 7
this is again 8
from your section 4.3 and 4.4 of your textbook 9
the learning objectives of uh 10
today's lecture are to solve 11
the examples regarding degeneration of electric fields 12
a due to continuous 13
a chart distribution and find out the electric 14
electricity due to these distributions 15
earlier we studied point charge 16
and electric vehicle intensity 17
generated by this point charge 18
at some elevation point using the coolant slot 19
then we covered how to uh 20
how to evaluate the electrified intensity 21
at a given point due to a line charge distribution 22
having symmetrical charge distribution on its surface 23
and we uh we 24
we studied that if there is a small 25
differential element on this aligned charge density 26
then the charge on the smaller differential 27
length element would be equal to the charge density 28
times e 29
differential element length element 30
and 31
if you want to find out the total charge on this line 32
charge density then 33
we have to simply integrate that 34
differential charge element 35
on that 36
complete complete line 37
complete line chart dances and length like this one 38
the second case the third case that we studied 39
was regarding the surface chart dynasty 40
that 41
if the chart distribution is uniformly distributed 42
on a surface 43
then then 44
then on on a differential surface element DS 45
the charge would be raw as times d DS and again 46
in order to find out the total charge on the surface 47
of that surface job density 48
we have to simply integrate that role as times DS 49
or overly complete surface limits of that 50
that surface job density 51
the third one was the 3D volume charge density Roe v 52
and for that the differential element 53
charge element is equal to all times GV 54
the differential volume element and again 55
we if we want to find out the total charge 56
present on the surface of this volume 57
chart density then 58
we have to simply integrate over the entire volume 59
for that given volume charge density 60
so that is your 61
that will give you total charge on that 62
residing on that volume charge density surface 63
there on uh we uh 64
we study that the electric intensity due to a point 65
charge is given by couloms law 66
that is queued about the call by ABS law 67
not our sphere times 68
and the unit rector and the unit distance rector 69
and if you want to find out the elective intensity 70
due to these different containers 71
charge distribution 72
we have to simply replace this Q by the total charge 73
available on this line charge density 74
or the total charge available on the circuit 75
charge density 76
or the total charge available on this wallet 77
charge density 78
and these are the three 79
these are the three cases that we studied 80
in our previous lectures 81
and later on we studied that if you're aligned charge 82
if your line charge is a finite finite land 83
then this is this line charge equation 84
it can be simplified to this equation 85
and here we straight the ID alpha 1 86
alpha 2 R 87
d 88
angular distributions 89
are depended on the peniplear distance 90
between the uh 91
between the revelation point and the uh 92
and this uh headline charge uh planning conductor 93
and 94
alpha 1 is from the starting point of that conductor 95
and the alpha 2 is from the terminating point of that 96
conductor with respect to that uh uh uh 97
uh with respect to that uh perpendicular distance 98
and in case our line chart distribution 99
it is off in finite line 100
then it will be extending in finitely in the positive 101
z and negative z direction 102
and again and in that case we 103
we studied that the alpha 1 and alpha 2 would be plus 104
minus pi by 2 105
and electric create 106
intensity would be simplified into this form 107
then we discussed what happens if we 108
we want to find out the electricity 109
due to this surface identity 110
then it is simplified to this rule 111
as divided by two astronaut 112
which is independent of the 113
independent of the distance from the revolution point 114
but that particular surface identity 115
and for the volume charge dynasty 116
we found out that it is equal to Q 117
divide report by astronaut RSP AR 118
where Q is d 119
put a charge residing on that 120
that volume charge dynasty 121
then we studied what is electric plugs 122
so it is a number of lines passing through a surface 123
and if it is a junior surface 124
then it is known as electric class density 125
and with a studied relationship between the electric 126
intensity and the electric prostency 127
how we can find out electric prostency 128
by simply multiplying that electric frequency 129
with that permittivity characteristics of that medium 130
for this free space is abstract 131
which is the 8.85 inch tenders 4 minus twelve 132
okay 133
let's solve uh this uh first example of your uh book 134
that is 4.5 so uh here uh uh finite sheet 135
so it is uh truly problem 136
and the final sheet is given which is having the 137
so it is the final sheet 138
which is having the dimensions of 0 139
2 1 in x direction and 0 2 140
1 in y direction and it is on Z plane 141
so it is x y plane so that is Z plane so use the 0 142
0 and it is having a surface chart density 143
and here note that the surface chart density 144
it is in terms of XY variables 145
and what we need to find out 146
we need to find out the total charge on the sheet 147
that is Q we need to find out the electricity 148
at this elevation point and then 149
we have to find out the force being experienced by 150
are given charge at this regulation point 151
so 152
these are the three things that we have to determine 153
first of all the total charge residing on this warning 154
charge density 155
is formed by simply integrating that surface 156
chart density over that complete surface 157
so here it is very simple 158
that you just need to put in that role 159
as then since it is a cardigan partner system problem 160
so your DS is equal to d x d y 161
and further you can uh now 162
you can simplify this integration 163
by taking half of this substitution rule 164
so you can uh replace the x square by zu 165
you just take its differentiation 166
your you know 167
that extra x square differentiation is the 2 x 168
and then along with it's a variable offer integration 169
a differentiation and then from here 170
you can read in to find out this X times d 171
X is equal to the U2 172
so if you use all these two these two uh replacements 173
substitutions 174
then you can simplify the equation into this like that 175
you replace the X square by shoe 176
you replace this X times d X with this U by 2 177
like this one 178
and now this this integration is very simple 179
now you have to simply integrate with this 180
back to this U and then you can 181
since this thing is inside this 182
so it's uh uh involved this uh pyro 183
and it is the integration is equal to the 184
whatever is the inside uh is part at 3 by 2 + 1 185
that will come out to be 5 by 2 divided by N+ 1 186
so again it is 5 by two and then the differentiation 187
you know what it is again the inside integral 188
the differentiation of inside integral is 1 189
4 / 1 is the same thing 190
so from here you put the uh 191
the integration image that were 0 to 1 192
for this again particular is you variable 193
so for again you have to play around with the Inter 194
with this integration limits as well 195
so far 0 as for a value of a 0 for given x 196
the g is again is equal to 0 197
if the value of a parliament of x is 1 198
then g is equal to 1 199
because it is at x square function 200
now you simply uh 201
you simplify these uh uh these limits 202
plug in these values and simplify this uh this thing 203
then it will the first the upper limit will come out 204
comes out to be 26 with all because it is 1 205
and then plus 25 is 26 206
and then the lower limit it is 0 207
so it will comes out comes out to be as 0 + 25 208
again it is while still plus 25 209
now the things are left out in terms of variable why 210
next uh against all this uh 211
uh uh 212
this uh integration with the spectrum y 213
by taking help of this substitution again 214
we replace this y square by variable v 215
again we take it to differentiation 216
again we find out this uh product y 217
d y and what we need to do 218
we need to replace this y divide with this d V about 2 219
and this y square with this v 220
no this is the integration that you have to perform 221
and once you do the integration 222
with respect to this thing 223
then it will come 224
come out to be this 5 by 2 will come out to be 7 by 2 225
this will be 7 by 226
2 divided by this is 7 by 2 227
now you again put in the limits 228
it will comes out to be 27 like this thing 229
and the other thing it will it will come out to be 25 230
again 231
you can consult the book for the explanation as well 232
and once you will simplify the things it will 233
it will give you a total charge on this volume 234
as a 32.15 235
Newton coolants 236
well let's uh 237
find out the electric fuel intensity 238
at the evolution point 239
so here the evolution point is 0 0 240
5 right and your source is a a fine at sheet 241
so since the axis varying varying in this sheet 242
in this sheet y is varying in this sheet 243
and these constant that is equal to zero 244
so involving the distance vector 245
so this is your distance erector minus 246
minus x minus y n 5 247
and now you want to find out your magnitude 248
so you can find out by this 249
x square plus y square plus 25 square root no 250
plug in this value in this given formulation 251
you know that a r is equal to r vector 252
divided by magnitude of r 253
so it will be a r cube and due to the spare root 254
it will be a 3 by two 255
right so now you can simplify this thing 256
that this thing will be cancelled 257
this thing and what we are left with is 258
this thing 259
after the translation we are left with this thing 260
now you can open up this product 261
so the first product is x square right 262
minus x square y 263
then second product is minus y square x 264
and the third product is x y 265
now you simplify uh 266
no use uh no 267
solve this integration one by one using 268
using this each term for the extra 269
so since it can be seperable 270
so you can solve this integration 271
this integration separately 272
and put in their respective limits 273
so we know that the integration of extras 274
with respect to a d x 275
s x 276
Q by 3 so put in this uh limit 277
so it is 1 or 3 for this y 278
it is y squared by 2 put in limits again 279
uh for the second part it is uh x squared by 2 and y 280
y q by 3 and put in limits simplified 281
for this thing it is x squared by 2 and y squared by 2 282
put in limits and simplify 283
so this is the electric field at 284
the given elevation 285
point due to this finite surface identity 286
now you want to find out the force being exacted 287
at this evaluation point 288
if some point charge of minus 1 million 289
8 minus one 290
mini column is placed at the same evolution point 291
let simply involve this columns law 292
and then it is equal to q and 2 e 293
and then you can simplify this thing 294
the second example of your book is uh 295
that you have uh two planes 296
that is X is equal to y and y is equal minus 3 297
and intercaring uh the charge uh 298
that's equal 299
is equal to 10 nanoquents per meter square 300
and 15 nanoquents per meter square 301
and this 1/3 charge distribution 302
that is a 9 charge distribution 303
and here it is as you can assume that it is 304
it is along y axis because it is it is an 305
it is due to the intersection of two planes 306
that is actually equal to 0 and is equal to two 307
and it is carrying this uh 308
10 pine and coolant square metre charge distribution 309
and what we need to do 310
we need to find out the lack of intensity 311
at this given point 312
little these three different charge distributions 313
now you can assume like this 314
that the electric field at this given point 315
it is a solution of three electric fields due to this 316
two two out of them are due to this plane 317
these two planes 318
and one is due to this crank and conductor 319
so first of all the electric intensity 320
and this evaluation point due to this sheet 1 321
so it is actually equal to two 322
so what you need to do is your access 323
so it is your uh uh uh 324
so you are currently focus on this uh 325
pictorial diagram that your magical representation 326
that it is your X y and Z and eh 327
this first name 328
it is a question distance of x is equal to 2 329
so it is in finite sheet 330
which is varying in the z direction 331
and in the y direction and it's x coordinates constant 332
so how to find out the normal 333
towards the evaluation point 334
so what is the valuation point 335
it is 1 1MINUS1 so you can consider like this it is one 336
see it is minus 1 right 1 - 1 and 1 337
1MINUS1 so so it is you can take it like sorry 338
if you can 339
you can achieve it like this thing that it is 340
uh like uh this is one right 341
okay y 342
X is one right like this one 343
so this is your x y plane point and then it is minus 1 344
so it is in the down direction 345
so you can take it like it is the 1 346
1 minus one regulation point and what you need to do 347
you would need to find out the normal vector 348
pointing from the this uh 349
sufficient identity towards the evaluation point 350
so since it's x x of the relation point is 1 351
so it is it is towards the 352
towards this negative E extraction 353
the normals the normal from this 354
explain towards the revolution point 355
is towards the minus extraction 356
so a n is equal to minus e x 357
and since it is independent of the distance 358
from the surface identity 359
towards the revolution point 360
you just need to put in this surface identity 361
and divide by beam free space permittivity 362
since northern medium is mentioned over here 363
and then you can simplify like this 364
the second uh second sheet is a y is equal to minus 3 365
so this is falling over here 366
right is your y is equal minus 3 however 367
in this case your Z is in finite 368
and your axis in finite in both directions 369
like this like this 370
and now again you need to find out this normal 371
so here is your elevation point 372
so this is your elevation point 373
so in this case 374
your normal is pointing the words this direction right 375
so this is equal to a Y like this 376
so this normally is why again 377
put in this uh 378
official identity and you can evaluate this E2 379
let's find find out this third electric intensity 380
and due to this line distribution 381
and here first of all 382
you remember that this hero and this rule 383
they are due to the perpendicular distance vector 384
the perpendicular distance vector from the 385
from your source towards the destination 386
and uh from the distance formula 387
if you want to find out the perpendicular distance 388
sector what it is that your line 389
your line was comprising of x is equal to 0 390
and z is equal to 2 10 intersection 391
so here your uh 209+ your 9 392
your X is equal to 0 right 393
your y is varying and your Z is 2 like this one 394
all the points are here 395
only your y is varying from minus infinity to infinity 396
so if you want to find out the botanical distance 397
between this elevation point 398
and the line what you need to do that 399
you need to keep this variable y 400
same as the 401
same as the y coordinator of this evaluation point 402
that is equal to one right 403
so for this line 404
you know that your x is equal to 0 constant 405
your is equal to constant and since your y is variable 406
no you 407
if you keep the y is equal to by steam coordinate 408
that is of uh 409
your evaluation point in this case it is 1 410
then 411
you can find a different pendicular distance factor 412
between these two uh uh this uh line uh line charger 413
uh line charger uh density and this uh evolution point 414
and if you see 415
basically you are moving towards this point 416
like this one and this is going to be your 417
are in this case like this one 418
and it is representing your perpendicular distance 419
the magnitude is representing your 420
a tangler distance between this line chart dance 421
G and B evaluation point 422
and if you want to drive in the 2 423
deprecatory representation 424
you can further explain it like this that on this 425
on this 426
on this X Z on the Z 427
explain your rise equal to rise up to one constant 428
and you can find out from here 429
that revelation point is on the studio is 1 1 430
1 common minus 1 that is your accent Z 431
and here you just need to find out the distance 432
rector between these two things 433
and so this was your distance rector 434
coming out from there in the 435
in the form of your unit factors 436
you take its magnitude 437
it is the perpendicular distance 438
between this linechardency 439
and this revolution point 440
you know that the unit factor is equal to factor 441
divided by the magitude of the lecture 442
and from here now you plug in these 443
these values in this given formula and a like this 444
a row is like this your row is the magnitude of per R 445
and you can simplify this like this 446
okay and 447
since you have evaluated all the electric intensity 448
now the resultant electric free 449
in the intensity at this human evaluation point 450
is the superposition 451
or the summation of all these three 452
electric free intensity 453
and you can simplify like this one 454
now let's solve this example 455
that is regarding the determination of electric flux 456
density at this given evaluation point 457
and it is due to our No. 1 point charge 458
point charge which is placed at this point four zero 459
zero so if this is X 460
y and d so this is your point charge location right 461
4 0 0 0 your x coordinate is 4 and y and and y and 0 462
so it is a longer x x is line the 9 chart distribution 463
it is along your y axis right 464
again your y is going to change 465
your y is going to change right 466
and since since it is on y axis 467
so it's X is going to zero 468
it's y is going to be variable 469
and it's disease again going to be zero 470
now your evolution point is 4 0 3 471
so here again you see that it is your 4 right 472
this was again before 473
this is your uh rise again 0 and this is your 3 474
so you are over here 475
so this is your evolution point four zero and three 476
okay now you first of all 477
you need to find out the electrically 478
due to this point charge which is at this 4 479
0 0 point location 480
just involve this coolum floor electrical intensity 481
multiply by this absolute not 482
you remember the formula it is equal to d absolute not 483
E right 484
okay this uh this decision rector 485
you know that it is the uh 486
position rector of this destination 487
minus the position rector of this force 488
and we know that this the position rector of this uh 489
destination with respect to origin is 4 0 490
3 and for this source point point charge it is 4 0 491
0 492
you can replace this 493
this is unit factor by this distance vector 494
divided by the magnitude of distance vector 495
and it will simplify to this 496
and distress vector 497
divided by the magnitude tube of this a distance vector 498
now you can uh take this magnitude 499
so and uh you can put in this uh uh genetractor 500
it will be the subject over here and you can simplify 501
you can take take the cube of this magnitude 502
and which will comes out to be 27 right 503
and as you can see 504
your book quality for the calculations as well and 505
now this is your minus point one three 506
eight in B a Z direction 507
and you can see over here 508
it is also in B pointing in B negative z direction 509
like this thing 510
okay 511
no we need to find out the electrical identity 512
due to this line charge distribution 513
and again we need to find out this patentical distance 514
between this line charge identity 515
and this regulation point 516
again your why is the rating in this thing 517
and what is the why of this relation point 518
that is zero may you keep 519
may you need to keep this varying y of this 9 520
chart density equal to this y of the regulation point 521
so these two must be same 522
and since it is a y aligned line chart density 523
so it x and z r already equal to zero 524
because it is being created in intersection of this x 525
and z plan 526
now phone this uh distance vector fourth minus this uh 527
minus the uh destination minus the fourth 528
divide with the magnitude of this uh 529
distance vector so it will come out to be 5 530
and then you uh need to plug in this uh 531
this is unitractor the perpendicular distance 532
distance factor 533
unitractor divided by the magnitude of this unitractor 534
so you can simplify this in terms of this again since 535
again since it is a 536
it is a superposition or a summation problem 537
that you have to uh uh 538
you have to add up the effect of this electric flex 539
flexes due to this uh uh 540
due to this uh 541
point charge and the 9 chart distribution 542
in order to find out the overall electric trucks uh 543
uh at the electrical density 544
had be given evaluation point 545
due to these two different charge distributions 546
and that is your resultant 547
after dissermation of these two individual uh 548
electric trucks density 549
uh this is a very useful uh tutorial to understand uh 550
columns law and uh 551
it is uh developed by 552
and the simulator is developed by a CP12 553
exploration series 554
and you can find all the simulator uh 555
a link from this uh source 556
uh and I 557
I've also copied the simulation uh 558
uh video on your uh shared folder 559
and so what is the explain to do that in this tutorial 560
and in the simulations 561
you can create number of charges 562
you can assign the platies you can find out there 563
you can you can place them at any of these locations 564
you can find you can adjust the good separation 565
that scale on this front panel 566
and then you can find out that 567
how the electrical lines are generated from pulse 568
to charge and electric charge 569
and how the uh 570
ecurrential lines are generated 571
in the surrounding of an uh uh 572
uh charge and uh 573
how to find out the electricity in the uh 574
at some particular uh 575
point and give vicinity of these to uh 576
be different types of the charges 577
and you can also find out the elective potential 578
which is a scale of quantity at the given charges 579
so this complete tutorial 580
will provide your brief overview 581
and you can also play around with this 582
uh to further uh 583
get an understanding of the coolants now and how the uh 584
things that uh are there in the area life once uh more 585
more than uh 586
more than one number of charges are present 587
uh individuality of the another charge 588
this is your assignment for week No. 5 589
and you have to solve these two questions 590
from your chapter No. 591
4 of textbook and kindly follow this second edition 592
if you don't have the second edition 593
then you can mash the questions from your colleagues 594
which are on campus 595
and take the snapshot of those questions 596
which are waiting in your 3rd edition 597
with those factories 2nd edition 598
so I'm here today's lecture 599
so we uh after covering the 600
a brief for introduction on this electric field 601
due to this uh 602
continuous charge distributions and the electric flux 603
tensity due to this uh 604
charge distributions uh 605
we sold uh some of the examples of your book and uh 606
the 607
and we saw that how to find out the electric trucks 608
intensity and electricals tensities due to uh 609
due to the presence of uh 610
uh more than one uh electric uh 611
uh good uh 612
presence of more than one uh 613
electric charge distributions 614
either it is in the form of point charge 615
or it is in the form of line charge distribution 616
or it is in the form of circus charge distributions 617
so if they are more than one 618
then we need to find out them individually 619
for that evaluation point 620
and then we'll have to sum them up 621
to find out the oral impact of that 622
child distribution 623
in the next class uh 624
we will cover that uh cause Islam or we can uh 625
find out the electrical tensities uh 626
due to this uh recharge distribution 627
if they are exhibiting some sort of this smetry 628
either they are aligned with some some access 629
they are they are aligned with some surface 630
or they are aligned with some given point 631
including swear lightings 632
then we we can involve to 633
we will involve this Maxwell education to drive this 634
uh gazes law and then we will utilize the uh 635
gazes law uh 636
where we can uh in 637
in in different uh 638
applications that how we can uh 639
apply this uh 640
gazes law to simplify the uh 641
this uh 642
finding out of uh 643
electric trucks tensities and electric intensities 644
uh with the help of this uh 645
gazada in case 646
if they are exhibiting this 647
symmetrical distribution properties 648
here I thank you all 649
if you have any questions 650
they will be entertained to your emails or online 651
synchron 652
a session that will have arranged at the department