this is lecture No. 11 of AV 2 3 2 electromagnetic 15 2
Today we will study Gaza's Law based on that 3
we will drive max notification 4
application of Gauze's Law 5
and this will color section 4.5 6
and 4.6 of your textbook 7
this is a demonstrate No. 6 of your semester 8
the learning objectives of today's lecture 9
would be to understand causes 10
law for determining the electrical 11
then on to derive the Maxel's first equation 12
and then 13
we'll discuss the basic applications of causes law 14
Gaza's law 15
states that 16
the total electric flux to any closed surface 17
remember 18
it is total electric flux to any closed surface 19
is equal to the total charge 20
and closed by that surface 21
so 22
the total number of lines passing to a closed surface 23
for example it is your closed surface 24
then total number of electric 25
the net electric number of electric uh 26
flux lines passing through surface 27
would be equal to be charged 28
which is and closed by this closed surface 29
so the net flats 30
Netflix out of this surface 31
that is size is equal to the total charge 32
which is enclosed by that surface 33
so that's why it is size equal to Q and closed 34
so if you remember that 35
the total number of electric clocks lines 36
it is equal to a closed surface 37
it is equal to the dock 38
product of your electric clock density 39
with that surface area 40
and uh also by the the basic definition of that uh 41
uh charge if we know we 42
we know the one in charge density of uh 43
of that uh at the geomaxi 44
then we can find out the total charge 45
and closed by that complete 46
uh volume by simply integrating uh 47
integating that uh 48
volume charge density over that volume 49
uh volume uh volume surface uh volume uh volume 50
charge against the object okay 51
so let's not utilize this basic definition of Moses 52
that equation 53
so we just need to equate these two equations 54
let's create these two questions over here 55
right beside the that is 56
your flux is equal to be total charge 57
and caused by that surface 58
so this is stated as the first 59
first equation of Maxwell 60
so this is Maxwell's first equation in integral form 61
and it has been dried 62
by using the definition of causes law 63
okay now 64
let's apply the divergent serum that will study that 65
we can we get 66
convert a close surface in chiguel into it 67
and draw volume and draw 68
and draw open volume integral 69
by choosing the definition of this divergous over 70
of that same factor field 71
so now if you involve this definition 72
you create these two equations now right 73
you plug in first of all you plug in this over here 74
then you 75
then it will be a equation of these three questions 76
right and then they are on 77
you can see that these two onions 78
they are going to cancel out and then what will be 79
we will be left over 80
with that is row V is equal to del dot d 81
and this is known as D 82
Maxwell's first equation in differential form 83
so always remember these two equations 84
from your electrostatic point of view 85
toward this course labels 86
and it is a first of four Maxwell's equation 87
okay volume charge density is the same 88
what does it state 89
that the volume charge density will be 90
is the same as the divergent of electric trucks density 91
to that close surface 92
Goss law is an alternative statement of Cooms law 93
this you can see in your 94
and problems that how to drive it 95
drive the Cooms law from the from this or Goss law 96
this picture shows that 97
what is the Netflix out of this close close object 98
close surface 99
so for example this uh closed surface uh 100
it is uh 101
having the charge 10 nanoculum and minus 5 nanoculum 102
and also in its exterior 103
there are 20 nanoculum and 15 nanoculum however 104
the Netflix the Netflix out of this complete surface 105
would be equal to the 106
would be equal to discharge and closed by this surface 107
so for this first uh closed surface 108
the net plugs leaving this V1 is equal to 10 - 5 109
that is equal to 5 nano column 110
and for the second surface 111
there is nothing and closed by this surface 112
so that's why this total tax 113
out of the surface is equal to zero 114
so why this is equal to zero 115
because the number of 116
lines entering due to this surface 117
for example due to this charge 118
are equal to the number of lines leaving 119
this close surface and likewise 120
the number of line and change 121
the surface due to this plus 15 coolant 122
are equal to the number of lines 123
leaving this close surface to 124
so at the end there's no Netflix out of this complete 125
complete closed surface and likewise for this 126
the first example uh 127
it is in material or independent offers 128
uh exterior charges that are 20 nanocolor man 129
15 nanocolor man and the total flex 130
out of that surface is only dependent upon the charge 131
which is enclosed by this surface 132
that is 10 nanocolor man minus 5 nanocolor 133
let's not discuss what is the significance of this 134
God's law 135
God's law provides and easy means of finding electric 136
or electric field electric field density 137
or electrical density 138
for symmetrical charge distribution 139
so it is only applicable to these types of 140
the charge distribution such as point charge 141
an infinite um inline charge 142
Australian distribution 143
or it may be an infinite charge distribution as well 144
so it is a 145
applicable if there is a symmetric charge distribution 146
and for that we construct a mathematical 147
closed surface of unknown geometric 148
and this surface would be known as Gaussian surface 149
and the cavities of this guardian surface are 150
the d 151
should be normal or peninsure to that guardian surface 152
and if it is a normal to that surface 153
then the maximum flux will be passing to that 154
because for example if this is your surface right 155
so what is the the unit vector of this surface 156
that is in this section right 157
you remember the unit vector of that surface 158
it isn't this direction now 159
if your d your electric flux 160
your electric field is passing to the surface 161
in a direction which is normal to the surface 162
so for that case 163
the angle between the normal to that surface 164
and this d would be 165
this angle in between these two will be 0 degree right 166
but in in other direction 167
it will be a 1 or 2 degree can it can be 1 or 2 degree 168
so if it is a 0 degree you know that cause of 0 0 to 1 169
so this this means that 170
the maximum flux would be passing to this 171
so surface 172
and in case it is a potential like your for example 173
let's change the color so if it is like this 174
your d is in this direction 175
then in this case 176
the angle would be between them would be 90 degree 177
and your cause of teacher would be quarter 0 178
and no flux will be positive to that play 179
that close surface 180
so guys in surface is chosen 181
such that d is constant in the parts of the surface 182
for which d is normal to the surface 183
because for the financial 184
the Netflix would be equal to 0 185
so this will see in our coming uh 186
slides as well 187
that how this d is constant on this complete surface 188
that we have assumed for the calculation of this uh 189
uh this uh 190
flags are the total charging closed 191
and therefore 192
we can bring d outside the integration sign and uh 193
and the solution is uh easier no 194
so if we treat this complete d as a constant term 195
then we can definitely create treat it uh 196
uh we can definitely bring it out of the constant 197
that integration so that are uh 198
our mathematical solution is easier 199
so guys this now we will apply this causes now 200
that is the electric substance to 201
is equal to the total charge 202
and closed by that close surface 203
and we will utilize this equation right 204
so for a point charge per point charge 205
it would be enclosed in a Gaussian surface 206
then this enclosed the charge would be equal to q 207
and if it is a inline charge distribution 208
infinite line charge distribution 209
and the total charge enclosed by that Guardian surface 210
if the line charge distribution 211
has the density of a royale 212
would be simply equal to the integration of that royale 213
over the entire land of that Guardian part 214
and if it is a sheet of infinite 215
a sheet of charge distribution withdraw 216
then they with charge and see the portable s 217
then 218
we will integrate that to find out the total charge 219
and caused by that 220
goes in surface that we consider an outcome 221
and slide and if it is a walling 222
charge distribution 223
so first of all 224
appoint charge Q is located at the origin 225
so this is your charge 226
okay it is located at the region 227
we have to determine the uh d at uh at point P 228
for example point P 229
and in order 230
in order to media the electric create or uh uh 231
this d at this given point 232
we need to consider our Guardian surface 233
and for this point if we consider in a threed geometry 234
then we can see that it can be a very good surface 235
that can contain this evaluation 236
point b and 237
this will also sacrifice the symmetry of this 238
complete problem right 239
so you because uh it's very good symmetric object 240
so this is everywhere normal to the surface 241
since this is uh moving away from the electrical lines 242
and moving away from the charge 243
so if you see over here 244
so at every point on the surface 245
this electric field or your in other words 246
the electric pregnancy is normal to this surface 247
like this 248
okay 249
so if it is a normal to that surface 250
surface 251
so that you know that it is in the radiant direction 252
it is also in the radiant direction 253
so both are having the TikTok 254
this is your AR direction right 255
this is your Dr 256
so both are this is your surface 257
like this surface and surface normal will be like this 258
the end so 259
both are having 260
the angular distribution of displacement of 0 degree 261
so call of 0 will be one 262
now you know who we apply that to Gaussian's Law 263
that is the electric clock is equal to the charging 264
close by that surface so first of all 265
did I said uh 266
first of all this uh 267
that your side we know that side 268
you could close and check lovely dot BS 269
since we are here we are here 270
you can see that on since it is a symmetric geometric 271
so at all points at all points of the surface 272
the value of this 273
the value of this electric flux density 274
it is going to be staying 275
is going to stay the same 276
okay so if we solve this complete integration 277
we can treat this electric looks 10C as a constant 278
we can take it out of this integration 279
and then we can solve this integration 280
now observe that 281
since it is as radical partner system problem 282
so DS is equal to Oscars and pwpw five 283
and since in the Sperical Partner System 284
the feature approaches from 0 to pi 285
and this 5 is 286
the emperor is approaching from 0 to 2 pi 287
and since Dr is constant 288
it can be taken out of the integration 289
and we know that this this 290
if you either 291
you call this integration or from the institution 292
since it is a complete sphere 293
so it's area the surface area of this complete space 294
report to 4 by R square or also 295
you can find this in the same answer 296
the same for by R square 297
by swallowing this integration in your own time 298
now the total charge and close by this guardian circles 299
not all guardian circles for the sphere 300
now the total charge 301
and close by this guardian circle is equal to Q 302
right now you can create this LED 303
this left right this right hand side of this equation 304
and you just rearrange this thing 305
you can find out your d 306
R is equal to q divided by 4 by r square 307
and in vector notation 308
since it is in the radio direction 309
so it is equal to q divided by 4 by r square 310
into a unichrector a r 311
and from there you can find out the electrical field 312
by using this constituted relationship 313
that is B is equal to optional not E 314
so E is equal to divide B divided by optional 315
not for free space medium 316
and your electric field is going to be 2 / 4 317
by astronaut Harshcare er 318
and this is the scene that we found out using your uh 319
uh in our previous section as well 320
that the electric field generated by this uh uh 321
by this point charge with the help of Coolum's Law 322
we found out that it is equal to 323
to divide therefore 324
perhaps we're not asking er 325
now the next uh 326
symmetric charge distribution is infinite line charge 327
of uniform charge distribution 328
roll L and it 329
is in this case we have taken a simple case 330
that it is oriented along the zxs 331
like this 332
so this is your roll L 333
Auckland L and note to determine a d at point P 334
for example it is your evaluation point here here 335
here here 336
here or here 337
we have considered a cylindrical surface containing B 338
to satisfy this symmetry condition 339
so here this symmetry is being observed over here right 340
and this what is that symmetry 341
that symmetry is of cylindrical shape 342
so your line chart 343
it is also exhibiting that person intical shape 344
your Gaussian surface it is also exhibiting that 345
persometry of symmetrical shape 346
and uh we know here we here we 347
we can see that d is constant 348
and normal to this cynical surface 349
so again the electric plugs density 350
it is imminating from this line chart density in this 351
in this area direction like this one 352
and here if this is your a N right 353
this is your a roll then 354
you can see that these two are parallel to each other 355
and also these 356
these perpendicular disk to this normal to disk in god 357
in surface 358
so Q and closed is equal to uh 359
so this is your uh sign so sign is equal to B dot DS 360
and here d is in the row direction 361
this is the radial direction 362
for this clinical partner system 363
and these constant on 364
and normal on these clinical partnering surface 365
and for the rest of the since it is a closed surface 366
so we have to take care of all these uh three surfaces 367
that is your top bottom end 368
this side surfaces so if you see this geometry 369
then you can observe that 370
for the top and bottom surface 371
your a n is a n is in this direction right 372
your this arrow is in this direction right 373
so the angle between them is 90 degree 374
right 375
so for this case you are um that uh electric prestency 376
it is a tendential to that surface so for that case 377
the Peter between the normal to that surface 378
and this electric prestency 379
it is going to be zero 380
so these two things are going to be nullify 381
so it is zero no 382
clock signs are passing to this top and bottom surface 383
and for the side surface 384
for the side surface no 385
we have to solve this integration 386
50 has no component along the a Z 387
also you can see that there's no component along a Z 388
so this integration is going to be the zero as well 389
so d is a tenantial to the top and bottom 390
so that's why 391
these two innovations are also coming out to be zero 392
the d is only having the radial component 393
so that is another factor of these two terms to be 394
coming out to be zero 395
and now we have to solve this on the integration 396
and again since it is constant 397
to be constant on the anti causing surface 398
it can be taken out of this integration 399
and if we solve this integration 400
this integration for this 401
symmetrical object 402
you know that the DS is equal to road defy 403
defy D 404
so you're uh this this one is equal to D5D Z right 405
because your aero is a normal 406
okay so if you solve this complete integration you get 407
so Royce taken as constant 408
integration of this D5 is equal to 2 409
5 for this complete slender 410
and what is the integration of will start to d Z 411
so that is your length that is your L complete plan 412
remember that so the so the integration over this 413
the surface area you can also take that 414
the surface area of this complete slender 415
it is equal to to pyro at 416
okay now you can uh uh 417
you can uh use the basic definition of the charge 418
so the charge uh 419
which is enclosed or which is possessed by this uh 420
infinite uh uh line charge distribution 421
it is equal to 422
it is equal to row L times B complete length of this 423
infinite line charge distribution so plug in over here 424
use a take out of this basic equation 425
that your size equivalent closed 426
use uh simplify this equation 427
then you can find out as your DSP to do 428
and do a bit to Pyro 0 429
and it is the same thing that will derive 430
in our section 431
now section that is your 432
uh 4.3 and there we took help of that uh 433
uh your uh 434
uh that basic definitions of uh your coolance law 435
okay now let's uh 436
take the case of this infinite uh 437
sheet of charge switch your card distribution 438
and it is for this case we have considered that 439
it is aligned on your extra plane 440
so that is easy for zero plane to determine 441
a d at a point P 442
we have consider a rectangular box as a garden surface 443
in this case so this is your 444
this box is 3D boxes here 445
and causing circus in this case 446
and it is uh cutting uh this uh 447
sheet of charge magically into two halves and has uh 448
two of its faces parallel to this sheet 449
so these two faces are apparently to this sheet 450
okay now d is normal to the sheet like right 451
if this d 452
if electrical is being generated by that sheet 453
for it would be generated either in this direction 454
or in this direction so it is the normal to the sheet 455
and it is in the d Z direction 456
so d has no component along this a y and a X 457
so it has no component along these two things 458
these two unitractors 459
now we if 460
if we apply the closing to the definition of this 461
electric plugs 462
so these are BS for the anti guardian surface 463
and we see that for this side for this side 464
that is these four sides that is this one 465
this one the back one 466
the left one so turn back right and back side 467
for this side the angle between the normal to this side 468
that is for example a N 469
right this you can treat it like a X or a y 470
so the angle between a X and a y 471
you can treat it like this one 472
the dot product of this a X and a C 473
and the dot product of this a y and a Z 474
it is equal to 0 because they are normal to each other 475
and for this these four services 476
be normal to the circuits without the accent e 477
y and the d it is only having the Z component 478
so that's why these these four integrals 479
they are going to be zero 480
and now 481
we only have to solve this top and bottom integral 482
okay for this 483
for this top for this top 484
we know that it is a n is equal to its a n is equal to 485
easy right 486
and for this bottom this a N is equal to 487
minus easy 488
right 489
and for this DZ 490
this direction will be positive AZ for the top one 491
and this negative AZ and d for this 492
bottom surface so this 19 active day 493
they will be basically cancelled out with each other 494
and at the end it would be simply a two times the 495
better solution of integral on of one side 496
okay so if we simplify this integration 497
we simplify the other side 498
what is the other side of the causes 499
well 500
that is the charge and close by this complete circle 501
now you see that the charge and close by the surface is 502
let's change this marker 503
charge and prove that his surface is this one right 504
and this surface area is equal to a 505
that is this right this thing 506
sector failure the integral of this thing 507
you can assume over here 508
the integral of this DS equal to a right 509
and also the uh you can solve the estate 510
solve the side and side using the same Assumption 511
that this integral is equal to a 512
this integral is equal to a 513
so it is easy taken as common 514
so it is two times a right two times 2 times d Z 515
a and I use just a real for this a 516
it is going to cancel out 517
you just rearrange that 518
so that your d is equal to Royce 519
to Abbot to a Z 520
and from there you can find out the electric grid 521
that is your d to Abbot Astronaut 522
and for the free space medium 523
and it is equal to Royce to Abbot to astronaut a Z 524
and it is the same that we drive in at section 4.3 525
let's consider the last case 526
that is the uniformly chart sphere 527
consider sphere of radius a 528
so this is your sphere a 529
the uniform charger distribution or not 530
and we have to find out the electric 531
field inside and outside the outside 532
the outside of that sphere 533
okay and for that 534
we have to take help of two Gaussian circles 535
for the first one 536
the Gaussian surface will be having a radius less than 537
or equal to this radius of this Jas 538
medical chart distribution 539
and for the second one 540
this r will be greater or equal to this E 541
now from the first case uh 542
we take that uh uh 543
for this uh left hand side that is your cue enclosed 544
we think cue enclosed is equal to Rovi 545
and Rovi is constant 546
so that's why it is taken out of the integration 547
and now we have to solve this integration right 548
and here uh the inside we 549
if we saw this inside the uh 550
that uh inside Guardian Surface integration 551
so we can see that the total charge imposed by this in 552
by this Guardian Surface uh would be equal 553
the radius of that god in surface 554
will be equal to the same um 555
uh same radius that that is the of the god in surface 556
so the total charge 557
total charge is radical as radical distributions 558
radius would be equal to the radius of that 559
causing surface so that is the catch in this case 560
so your so that your your radius of integration 561
it is equal to going to stay the same 562
that is of your Guardian surface 563
in this particular case 564
because everything closed by this surface 565
the total charge closed 566
by this surface will be equal to that 567
to the same radical charge distribution 568
radius that is off Guardian surface 569
okay now you can simply write this integration 570
this integration is equal to the same as 571
the volume of your spherical object 572
with radius R also 573
you can solve this integration 574
and you will come out that it is equal to 575
the volume of a sphere with radius 576
is equal to R 577
unless all the fluxide of this guardian law 578
so that is a close interglow 579
d dot BS is equal to again this electrical 580
it is eminating in the radiant directions 581
that is the er so d R can be taken out as a constant 582
so this integration is same as the 583
as these surfaces 584
you are that here you can also follow this integration 585
now you can equate these two sides 586
these two sides right 587
these two pins you can equate using the causes law 588
you can rearrange 589
then you will find that is equal to r divided by c rona 590
p r where r is the radius of your Gaussian surface 591
so remember that here the integration 592
to find out the total charge enclosed by that cost 593
Gaussian surface would be the r 594
now let's uh 595
consider the second case where the causing surface 596
we need to find out the elector field outside of this 597
uh assertical chart distribution 598
for this case again 599
we are going to integrate the one in charge density 600
run out his constant however in this case 601
your radius 602
your radius it is your radius 603
it is not going to be the radius of that 604
your god in surface rather rather it is US mission 605
that's mission from 0 to A+ 606
a to R so it is that mission of 2 integration 607
and for this a to R the d is equal to 608
it is charged and closed by this uh 609
by this surface is equal to 0 610
so the total charge enclosed by this guardian surface 611
which is having this radius greater than the uh 612
square cervical charge distribution 613
square radius uh which is the a is the same 614
is the same as the uh as the radius of this 615
complete vertical charge distribution 616
so that is that is the difference in this case 617
so here 618
in order to determine the charge enclosed by the 619
causing surface 620
which is having the radius greater than that 621
uh charge distribution 622
uh will be having the radius equal to that 623
uh that offer charge distributions here 624
so in this case it is equal to a 625
now going to solve the integration 626
under the change is your a 627
that the radius is equal to the radius of your 628
charge distribution 629
radical object 630
now for the uh side side 631
you just again need to solve this integration 632
here there's no change 633
because the golden circle is having the radius of r 634
you again simplify this equation 635
now you equate these two equation 636
you can find out that 637
your d is equal to a cube divided by your squared 638
or not the r 639
and here is the radius of your critical job 640
distribution and RD Radio Secure goes in circles 641
now let's summarize this history question 642
in the graphical presentation 643
in your if you plug this to these two equations 644
now you can see that if you plot either R is equal to 645
so boundary condition is R is equal to a 646
apply this R is equal to a at this to the 647
at this boundary condition 648
your uh d is equal to a divided by C are not also 649
you can apply this put this article to error here 650
and you can find out this 651
that is pure so this will come out to be the same 652
divide by cheat or not so if you observe 653
overheard that 654
if you are measuring the electric field in an 655
in a 656
at a point inside this radical charge distribution 657
with a radius 658
less than the radius of this vertical chart 659
distribution then 660
these are direct relationship 661
between the electric field 662
and the radius of that Guardian Star 663
Guardian surface so um 664
as bigger as the radius of that garden surface 665
would be 666
so the electricals and the density would be greater 667
because here 668
you can see that it will come closer to that surface 669
of that giant distribution 670
so it is that in your relationship for the second one 671
if you move this is your inverse relation 672
that is you're moving away from that surface 673
that's vertical chart distribution 674
and this relation is inverse square 675
that these 676
inversely proportional to the square of this 677
are squared and this you 678
you know that the investigation of this are squared 679
it is it exhibits some sort of here quadratic equation 680
so that is your parabolic decaying in this case 681
until parabolic quadratic equation 682
so that's why it is decaying in a 683
in this parabolic quadratic equation manner 684
and again since you are moving away from this surface 685
of charge distribution 686
that's why your deeds going to decrease 687
as you move away from this charge distribution 688
let's solve these two examples of your book 689
so in the first case your DS given 690
it is having dependency on the Z direction 691
and we have to calculate the chart density 692
at this reduation point and we have 693
we have to also can be the total charge 694
and caused by the this cylinder of radius uh uh 695
1 meter and uh 696
the uh that uh z direction uh dimensions minus the 2 2 697
2 698
so that is your guardian surface you can see over here 699
okay 700
so first of all your Roby that is your child Dansky 701
the one in Child Dansky it is equal to that 702
don't be so 703
only 704
you have to simply take the divergence of this thing 705
and it is only having the dependency on this easy 706
so 707
only we have to take the Basha derivative of this DZ 708
with respect to Z and there's only Z present over here 709
and whose differentiation with respect to Z 710
is equal to one 711
so at the end your row is equal to this one 712
and simply put this row 713
y and Z and this equation 714
and you can simplify this chart 715
and then see at this resolution point 716
so remember this is your Maxwell's first equation 717
differential form 718
so to find out the total charge and fold by this number 719
there are two ways so the first one is the basic one 720
that is 721
using the direct definition of your total volume 722
volume that chooses your total volume chart 723
against the integration and it is in this case 724
you know that your row v 725
your row v it is equal to draw cause 726
square 5 727
and your DS as your DV in this clinical corners system 728
it is r o d 5 d 0 d Z 729
and your limits for the integration these are from 730
since it is a slender so far 5 it is from 0 2 731
2 5 732
4 your uh Z 733
it is from minus 2 to 2 734
and uh for this road we will be scared with it 735
it is so scared now you just simply integrate this 736
so it's integration with the Z 737
so upper minus 4 to 2 minus minus 2 738
so it will be a 4 739
and your integration of this thing can be 740
find out using this half an angle identity 741
so solve this in your own time 742
so the result will come out to be bye 743
because this is going to be zero 744
its integration would be going to zero 745
and this would be equal to 2 pie 746
right this would be equal cancer why this is zero 747
because sign of pie 748
sign sign of 2 pie and sign of 0 is equal to Z 749
and for this thing 750
it is the rocky divided tree and put this up 751
and also it will come out to be one of a tree 752
so that is your answer 753
what is the second way 754
second way the way the help of your boss is long 755
and for that 756
your cues equal to size with the colonies to close 757
and take a lot of C dot DS here 758
since it is a sylintrical object 759
so again 760
there are three surfaces surfaces side top and bottom 761
since your d it is in d d direction 762
so for this size for this size 763
there's no zero component is d so for this size 764
your this integration is going to be zero right 765
because again 766
the angle between them will be equal to ninety 767
okay and cause of 90 is equal to zero 768
remember this thing that we have discussed earlier 769
access as well so for this top and bottom surface 770
it was yellow here for the first of all 771
for the top surface your DS is equal to again 772
in the slanderal quality system 773
is equal to row divided row and Z for the top 774
it is equal to 2 right 775
and your radius it is the range from 0 to 1 776
your 5 equal to 0 to 5 777
and you can solve this integration 778
know how to solve this integration 779
your Z is constant so simply put it equal to 2 780
your deal uh so it is a Roku by 3 781
so solution is 1 by 3 782
and again this uh uh cost square five solution 783
you know it is 5 from your previous life 784
likewise for the for this bottom 785
the only difference is your normal 786
to that surface element it is minus easy 787
and also use easy for 2 - 2 at this bottom surface 788
solve this integration again 789
you will find out the same answer 790
that is because minus 791
minus been going out to be canceled 792
so answer is 2 5 by 3 some all the 3 intergation 793
so the answer is the same that is 4 794
5 by 3 that we found out in the previous life 795
let's solve this uh problem 796
uh for your uh surical charge distribution 797
so the case is for this 798
uh uh for this charge distribution with videos r 799
R and inside that your dependencies on a row 800
not r divided by r 801
and outside this radius of this radical chart 802
distribution here 803
it will be equal to zero 804
so the first thing we have to find out 805
the electric field inside that sperical charges 806
solution so for this case 807
your radius of integration 808
to determine this Rovine integration 809
would be the same as the radius of your garden surface 810
that is your arm now you solve this integration 811
you can take this doughnut outside the integration 812
this R is also constant this R is going to be Q right 813
and the integration of the science data 814
respect to data is minus false data 815
put this uh uh put this uh uh 816
that uh integration limits are 817
in this vertical coordinate system 818
of the data that we know that it is from 0 819
2 5 820
4 d complete square and for 5 it is from 0 to 2 pi 821
now solve this integration 822
it is very simple 823
so sign the integration is minus fast data 824
then this R Q intervision is R 4 / 4 825
put this limit inside this and you can 826
for the left hand side for this 827
for this close integral side that is your flag 828
you can use the definition of that 829
secretary of that sphere so that is equal to 4 bioscare 830
you know that these equal to astronaut e 831
electrical intensity 832
and then you can simply simplify this equation 833
to find out the e 834
inside that um that's radical object 835
so for that outside of that circle object 836
so the only difference is as I mentioned earlier 837
that your Gaussian surface 838
it will be a combination of two Gaussian surface 839
two spherical limits intergation limits 840
so the first one is approaching till the radius of that 841
till the radius of that cerical charge distribution 842
and the second one is outside of that 843
cerical charge distribution 844
because if this is your are out of the 845
all of your uh cerical charge distribution 846
this is your causing circle 847
so obviously from here to here 848
you are going to have this volume chart dancing 849
and also this 850
this volume chart dancing is going to be zero 851
so you don't only simply have to fall the integration 852
which is up to there up to this 853
uh up to this uh 854
uh radius of this uh cervical charge distribution 855
and here you can again relate the same 856
uh case No. 4 857
that the total charge enclosed by this uh 858
by this causing surface is equal to D 859
complete charge which is uh 860
uh which is uh 861
which is on this sperical charge distribution 862
having the radius of this are 863
now 864
you again simplify this integration as of a last slide 865
and what is so left hand side is going to be 866
stay the same because 867
it is not having the dependencies on this integration 868
and only 869
we have to take the integration of that surface 870
at your factor gaussian surface 871
however 872
for the right hand side so your robot is constant 873
your eyes going to be constant 874
uh right and you can solve this integration 875
and this integration is going to be R cube 876
R cube integration is going to be R 4 by 4 877
so this R will be cancelled by this R4 878
and R cube will be staying over here 879
and then you can simplify this integration 880
to find out this user 881
so I'm here today's lecture 882
so we understood what is causes law 883
how to find out the electric field 884
intensity 885
and electric first density using this causes law there 886
and we divide the max size of first equation 887
in the integral and in the differential dot notation 888
and finally 889
we found out some useful applications of this 890
Gaza's law that is applicable to that 891
symmetrical charge distribution 892
to find out the electrical instance and intensity 893
and electrical agency using this Gaza's law 894
next time we are going to study 895
or understand this electric potential 896
and determine the potential difference 897
due to this electric charge distributions 898
here I thank you all if you have any questions 899
that will be entertained to your emails or online 900
central 901
a session that will have arranged at the department