this is election No. 19 of a B 2 3 2 2
Today we will have an overview of electric field 3
boundary conditions 4
and thereafter we will solve some of 5
some of case examples 6
this will cover your rest of uh section 5.9 7
this is academic week No. 11 of your semester 8
so today we will uh review uh 9
electronic conditions 10
and thereafter we will solve multiple examples 11
first of all the boundary conditions 12
uh run the field exist in a medium consisting of uh 13
two different medias 14
the conditions at the field must satisfy 15
are called the boundary conditions 16
so these are the conditions that electrically in the 17
in both mediums is satisfies across that boundary 18
so far electric fields 19
the following boundary conditions are important 20
dialect to to dialectic interface 21
conducted conductor to dialectic interface 22
conductor to free space interface 23
so know that 24
they are helpful in determining the feel 25
on one side of the boundary 26
the feel on other side is none 27
and uh we already uh know that uh 28
we solve these boundary conditions 29
and with the help of these two uh Maxwells 30
uh equations uh Maxwells 31
uh 32
famous first and second law that is uh our Gaza's Law 33
and uh this uh 34
the closed closing trigger of electric fuel is zero 35
that is your er 36
this er considerated property of your electric fee 37
and also in order to find out this 38
electric fields and electric first tentities 39
we decompose these two electric quantities into two 40
autogonell components 41
so the first of all 42
the dialectric to dialect Bondi conditions 43
for this particular case we consider this 44
this uh scenario that uh uh 45
there's uh 46
one medium with dialectic properties of APSRON1 47
the other medium with dialectic permittivity of APSRON2 48
and uh uh 49
we decompose electric fees and we we 50
we apply this uh 51
uh close in 52
tigula of electric wheel across this rectangular box 53
and once we applied all the uh 54
those uh integrations and all those uh 55
simplifications and we came down to a conclusion that 56
um there are two boundary conditions 57
first of all the tangential boundary condition 58
that the potential components of electrical 59
they are equal at the boundary 60
remember at the boundary these two components 61
they are equal at the boundaries 62
between these two different dialectic mediums 63
and uh this means that uh 64
uh traditional component of electric field 65
undergoes no change on the boundary 66
and its continuous across the boundary 67
so that is the intuitiveness from here 68
the second thing is 69
regarding the potential component of your electric 70
prostituency and it undergoes some of the changes 71
because we know the relationship that uh 72
this uh this equal to abstinent e 73
and uh if you plug in that uh constituted relationship 74
then you can see here there's no direct um equal 75
equal Equality between these 76
these uh er potential component of electric flux 10 c 77
so it is said that this electric flux 10 c 78
it is discontinuous across the boundary 79
so once we uh 80
d uh we uh we dealt about these uh normal components 81
then we uh considered this uh gaussian 82
uh as pillar box 83
pillar box box pillar box surface 84
so that we can solve that Gaussian equation 85
and once we solved this equation 86
and we came down to a conclusion that 87
that the difference between the normal company 88
the difference between the normal company 89
is equal to the 90
this surface electric electric as you can 91
as you can see surface area electric 92
electric dance to your electric 93
this surface electric density right 94
surface electric density 95
if it is present on this surface are the interface 96
which is meeting these two mediums 97
like here like here right 98
so this one is present over here 99
then it is equal to basically the difference between B 100
the electricals and CR 101
these two mediums 102
and uh from here you can into that if in case 103
uh no free charge exists at the boundary 104
or the interface between the 105
these two mediums that this uh rose is equal to 0 106
so this means that these two component 107
the normal likely proxancy components that equal 108
and accordingly 109
once we plug in the constitutive relationship 110
then this is the electric field 111
and normal components relationship 112
so from here you can into that normal components 113
normal components of electric presidency 114
are equal at the boundary 115
and the 116
hence that the normal continental likely of electric 117
electricity undergoes no change on the boundary 118
and it is continuous across the boundary 119
now 120
coming towards the normal component of electric field 121
it it does undergo some changes across the boundary 122
and hence it is said that the normal 123
comprehensive electric field 124
it is discontinuous across the bound 125
and accordingly once we 126
once we resolve this electric field 127
and the electric presidentities of these 128
present in these two mediums into two components 129
then uh we found out a relationship 130
very known relationship that 131
that is known as the law reflection 132
so in this case 133
that the ratio of electric permittivity 134
the relative electric permittuities 135
of these two mediums is equal to the 10 liter 136
of the angles that these two electric trees makes 137
uh make with respect to the normal 138
uh to their interface so 139
here we consider that these are two different mediums 140
right 141
this is the interface uh 142
this is the uh Tita one is the angle that uh 143
electrical 144
in medium 1 makes with respect to the normal 145
between these two 146
so this is the normal right between these two media 147
different mediums and accordingly 148
this in the second medium 149
this is the tea tattoo 150
that electrically makes with respect to the normal 151
between these two mediums 152
and here 153
we found out a very important relationship that 154
uh the SO1 of an electric wheel 155
it will uh travel from one medium to another medium 156
so either it will turn towards the medium 157
or it will diverges from the 158
turn towards the normal 159
or it will diverge from the normal 160
and the ratio between the electrical 161
relative permittivity of these two mediums 162
is equal to the ratio of 10 163
10 of these two angles 164
that these two electric freeze make 165
with respect to be normal 166
let's not uh 167
talk about the second condition that is uh 168
conducted to dialectic boundary conditions 169
uh that's how we can find out 170
the only conditions for this scenario that uh 171
if one side of this uh medium 172
it is consisting of this dialectic 173
and the other side is purely conductor 174
we know that uh due to the infinite uh 175
conductivity of the medium and there uh 176
there's no electric field 177
no electric field exist 178
no electric field exists within a conductor 179
so it is the property of an upper uh 180
conductor 181
that due to infinite conductivity of the conductor uh 182
electric field cannot exist inside the 183
or may not exist you can say more 184
more probably may not exist within a conductor 185
so that's that's why we say that the electric field 186
and the electric fuel and the electric frogs 187
dance cheese they are zero inside the conductor medium 188
if we uh took this Assumption and we uh 189
uh took this uh electric field uh 190
this dialectical material of the other medium as ABS 191
1 are and we saw this uh uh 192
this complete uh close interview of edot DL 193
that is your conservated property of that electrical 194
and then we came down to concludion 195
that the financial company 196
since it is a boundary 197
so financial component at the boundary 198
or the interface between these two components 199
and the electric frequency 200
at the interface between these two components 201
it is equal to zero 202
right 203
and the second thing uh that we uh 204
came down to conclusion about this normal components 205
that uh 206
an electric field an electric field must exist 207
this means must exist uh to the uh external 208
external to the conductor and most phenometers surface 209
so the only electric field that exist 210
on the boundary field 211
that is on the boundary field that is here 212
this normal component of electrical and there is uh 213
uh that is the only grid that exist 214
in case of dialectry to conductor 215
uh uh this medium and the boundary condition 216
condition case so in this case you can see that and if 217
if there's a presence of surface identity 218
so that is equal to your um 219
normal component of your electric presidency 220
and accordingly 221
you can find out this normal component of electric fee 222
so remember that uh whenever there's a case that uh 223
on one side there's a dialectric 224
on the other side there's a conductor then uh 225
then uh the initial component of the electric field 226
and electric pregnancy they are zero 227
and the normal components 228
and normal component of this electrical tendency 229
it is equal to the surface identity 230
and the normal component of electric field 231
that is that you can find 232
or using this constitutive relationship 233
and uh it is uh 234
note that in in this scenario 235
only the normal 236
only the only the electric field that exist 237
that is 238
the only normal component of the electric field 239
that is take away from this slight 240
so the third case third case 241
is just the extrapolation of the previous scenario 242
so in this vertical case instead of a dialectic 243
you're only having this free space 244
and for free space 245
we know that the ABS final is equal to one 246
so same scenario inside this uh conductor 247
the electric fuel is going to be zero 248
and outside 249
only the normal component is going to exist 250
and so here you can see that 251
the traditional components of this electric field 252
and the electrical exhausted is equal to 0 253
and the normal component is equal to 0 s 254
and accordingly you can find out 255
the normal component of the electric field 256
so electrophil must approach a conducting surface 257
normally you can see 258
you can see here 259
because only the normal component is existing 260
in this case the financial component is zero 261
and uh we uh 262
you must uh remember that uh uh Runner 263
Runner Day is a conductor around the uh some uh 264
another uh body or some electric surgery 265
it can be utilized uh 266
it can be utilized as a screening 267
as a screening shield right 268
so it can shield the effect of electric 269
and electric beams to entrap 270
to disturb the electric components which are in its 271
in its enclosure 272
so any electric components allow any electric safety 273
which is enclosed by some uh conductive 274
conductive shield uh it will award the 275
it will award the hindrance of external electric field 276
external electric feel to that internal body 277
which is enclosed by this conductor 278
so because why it is like so 279
because the E feel inside 280
this conductor is equal to zero 281
and due to the higher conductivity of this conductor 282
the E field 283
it will reside on this surface of this conductor 284
and this is known as the screen effect as well 285
and uh for this uh pure conductor the 286
this penetration 287
which is known as the skin depth of the electrical uh 288
it is also the minimum 289
so this is the whole uh 290
this is the whole summary of your electric 291
boundary conditions uh different scenarios 292
so at the end 293
this is the take away from these uh three cases 294
the dialectual dry electric boundary conditions 295
conductor to directly boundary conditions 296
and conductor to free space boundary conditions 297
so you can uh uh you can uh 298
remember these boundary conditions to soldier 299
further examples that will discuss uh 300
in our in upcoming slides 301
so first of all the example 5.9 302
so here you you have to first of all uh 303
see that uh what is this uh 304
what is this uh uh this uh coordinate system 305
and uh uh 306
what is the normal to these uh to different mediums 307
and what is this scenario 308
either it is this dialect to 2 309
dialectric or it is dialected to conductor scenario 310
so from this particular case 2 extensive homogeneous 311
right 312
homogeneous means that an isotropic both conditions 313
okay so it is isotropic and homogene and linear as well 314
so Ella yet you can see that in this case in medium 1 315
there is uh it is also exhibiting this uh 316
processing this uh 317
dialect prometivity of apps for No. 1 318
the medium 2 is having this dialect prometivity 319
so this means it is your case number one okay 320
so for uh so what is the interface between these two 321
so the interface between these two is this this one 322
right Z equal to 0 323
so Z equal to 0 plane mean that it is x y plane right 324
you must remember these things 325
that what does the Z is equal to 0 326
from your elementary maths 327
from your calculus 328
that whenever it is Z is equal to zero plane 329
that it is going to be your x 330
y plane so uh in this particular case of whenever 331
whenever there's a boundary there 332
there's a boundary at the Z Z access 333
this means that Z interface between these two 334
or Z plane between these two different mediums 335
then you must include that the normal 336
the normal to these two interfaces right 337
this one uh one 338
uh first medium and the second medium 339
you can hear a see that 340
what is the normal between these two interfaces 341
so this is the plane right 342
this is extra plane so what is normal 343
so in you can see that it is equal to a Z 344
so that is your normal 345
so it is very important that you must uh 346
understand this geometry and this uh 347
uh concept that how to find out this uh 348
normal between these two components otherwise 349
uh if you're not able to uh 350
find out what is the exact normal between these two 351
uh in different mediums uh 352
the boundary that is joining between the 353
these two difference uh 354
these two different mediums that uh uh 355
then you will not be able to solve the uh problems 356
uh correctly so first of all 357
the most important thing is 358
you must understand 359
how to find out the normal between these two 360
and these two different mediums 361
so uh Z is uh or entertainment 362
you can also say is so you can also remember that if 363
if some boundaries given like this 364
Z is equal to 0 or X is equal to 0 365
y is equal to 0 366
so by default the normal will be E x normal to Z 367
it is Z plane which is actually plane 368
so what what is from your maths or geometry 369
it is a Z so here this is your normal component right 370
and rest of 371
and this is your potential component of electric fuel 372
okay so here we have 373
given that a uniform electric field exist 374
for Z is greater than zero right 375
so for the first medium the information of electricity 376
we are given with right 377
and we have to find out the electricity 378
in the second medium we have to find out the 379
these angles 380
that electric and magnetic electricity needs 381
with the interface right 382
with the interface so remember that Tita 1 383
Tita 1 is the angle it makes within normal errors here 384
taken as alpha alpha 385
it is the angle it makes mixed with the interface 386
so it is 90 - 90 minus the teacher okay 387
90 minus the teacher 388
okay and then 389
we will determine this electric duster 390
using the information of electric fields 391
and then we'll find out the energy within a cube 392
which is present in one of these mediums 393
and we will see how good this is like 394
okay so first of all 395
how to find out first of all 396
resolve this electric feel which is given to you 397
that is your electric feel in this medium one 398
so we know that it comprises two components 399
one is the tenantial 400
and the second one is the normal component 401
so how to find out the normal component 402
we remember from our chapter No. 1 403
that if you want to find out that for want 404
want to find out a scaler component 405
along the direction of a 1 er 406
erector quantity 407
then we simply have to take a start product 408
with that unidractor right 409
so in this case the unidractor is the normal 410
normal lecture in this case 411
which is which is the normal between these two 412
to this interface between these two different mediums 413
so this is your skeler comprehend okay 414
e 1 dot a Z which is the a N so it is a 3 415
so what is your multiply by a Z 416
then it will give you the vector comprehend 417
for that 418
normal right for that now for in the normal direction 419
so this is the the this is the 420
you can say it is the normal 421
Raptor component of this electric field 422
in the z direction this particular case 423
so then how to find out the potential component 424
it is simply the use this uh basic definition 425
so it is the vector minus d normal component 426
so it is your definition component in the medium one 427
so this is the first thing 428
no next in while the boundary condition 429
now 430
the boundary condition says that e tunicial components 431
it is it is continuous or is 432
uh equal in these two different mediums 433
so this means that E2T is equal to E1 right 434
so the uh financial component in your 435
uh this medium to it 436
is 437
the same as the financial component of the medium one 438
okay 439
now to find out the normal component 440
so in this case Royal C is equal to 0 right 441
Royal C is equal to 0 no surface chartancy is given 442
so you can involve this 443
continual relationship between the normal component 444
of this electric 445
electricity and uh uh and from this uh 446
Constitution relationship that these equal apps funny 447
you can uh relate this uh 448
connect this relationship into the electrical farm 449
and then from here 450
you will see that astronaut is cancelling out 451
and you're left with the E to N right 452
so what is it because you are given with Ivan right 453
you know what is Ivan from this Stephan okay 454
so plug in the value of this E 1 n 455
use this uh ratio between this uh altitometuity 456
so this is your normal component in B medium 2 457
as you can see it is also in the Z direction okay 458
in the Z direction 459
that is the normal between these two interfaces 460
and uh 461
now simply how to find out the uh complete vector 462
the electrical vector this is second car 463
second uh medium that is uh 464
you have to uh now simply add up these two uh 465
engine and normal component of electrical in medium to 466
so this is your final expression right 467
so this is the day you 468
you uh proceed from one uh medium to another medium 469
uh 470
once you have given the information of galactic field 471
in one medium and you 472
you have the information of galactic 473
uh your uh electric uh 474
permittivities and you have the information of uh 475
boundary conditions for that particular uh scenario 476
in this particular case 477
it is the dialect to to dialect 478
uh boundary condition case 479
now second step uh the uh part B 480
we need to find out the angle that this electric field 481
that this electrically mix 482
with respect to this interface right 483
so we can uh we can uh 484
and also we can find out this uh 485
uh uh 486
this uh animal between this uh 487
normal and the electrical data as well 488
which is your Miss Cheetah 489
so you can proceed in both ways right 490
you can proceed in this ways 491
either you can find out this teacher 1 492
using this relationship that time teacher one 493
it is equated from here 494
you can see that this mammal component right 495
it is a sternicial component right 496
so this is your E1T magnitude right divided by 497
E1N so from here you can also 498
so this is given in your textbook or the other ways 499
so you can find out the 3 tavan just like this 500
that the the the angle between these 4 alpha one 501
you can see here so how to find out to this alpha one 502
from this part of our ceiling 503
you can see that the 10 alpha 504
the 10 alpha this one alpha 505
it is equal to the uh ratio of the uh 506
this magnitude of the normal company 507
and this traditional company 508
and once you have this found of the magnitude 509
you can take a standing rose 510
so standing rose will give you the Alka 1 511
the angle that like to create a medium 512
one makes 513
with respect to the interface between these two mediums 514
and accordingly 515
you can involve this relationship that is 90 minus 516
because the whole this whole angle is 90 517
90 so it is 90 minus alpha one 518
right 519
uh you can find out the Cheetah one from here 520
so it is coming out to be 60.9 on the other ways 521
as I have already told you 522
that it is given in your taxpayers 523
that you can directly found out 524
this angle between the normal and electrical 525
using this relationship that 10 526
Cheetah is equal to be 527
ratio of the financial and normal compound compound 528
remember these two are different ratios 529
remember this thing 530
because they are dealing with different humanities 531
okay and uh 532
how to find out this uh 533
alpha tool are similar in same manner in the elector 534
in this uh 535
uh medium 2 there's a 10 536
alpha 2 to the ratio of the normal components magnitude 537
to the magnitude of this signature component 538
and you can find all this alpha 2 539
and the Peter 2 is equal to 540
you know that it is 90 minus alpha 2 right 541
and so it is equal to I goes 90 minus alpha 2 542
and also you can find out this Chan 543
t dot two using the same relationship 544
that it is the ratio of you can see here like this one 545
so this is the bite I got a diagram for this uh E2 546
so again from here the normal uh B 547
you don't find out that this is the 90 degree okay 548
so to be very short between the 549
financial component of electric fuel in medium 2 to d 550
normal component in medium two right 551
and uh you can also satisfy this uh 552
uh law affection from here 553
that the ratio of the meridic permittivity 554
uh in these two mediums it is equal to the term of uh 555
these uh tea triangles that the electric fuse make 556
with respect to the normal 557
so these two ratio that equal 558
you can also satisfy this condition of law 559
reflection as well 560
okay so the third part 561
we have to find out the energy densities 562
in both dialectrics 563
we know that from our chapter No. 4 564
that the energy density 565
remember it is density jewels by cubic meter 566
and it is half of the 567
this permittivity in medium 568
that impact that particular medium 569
and the magnitude scarabee electric field 570
now we have the information of electric field 571
in these two mediums 572
we have the information of electric permittivities 573
in these two mediums 574
and we know that in this permittivity 575
it is equal to abstel knot in 2 576
Abstel R1 this is your abstel knot 577
this is your Abstel R1 578
this is the magnitude of electrical in medium 1 579
and this is magnitude of electrical in medium 2 580
this is active permittivity in medium uh uh 2 581
and this is your right 582
to permittivity in your medium one 583
and this is extra knot and this is extra knot right 584
so accordingly you can you can label all these 2 585
3 quantities 586
the next step uh solve this four part 587
we will have to find out the energy 588
within a cube of site 2 m right 589
and the Q okay 590
and it is center at 3 4 591
5 point okay 592
so if this is your geometric for example right 593
so let's find out what is this point first of all 594
this is your X y and Z and this is your t 595
4 and 596
this is your discipline maybe okay 597
these are p point okay 598
these are 3 4MINUS5 599
so first of all 600
we need to find out that in which medium applies okay 601
so the 0 z is equal to 0 plane x y plane 602
let's uh see how to solve this 603
uh by D in this particular case 604
we need to find out the energy or within a cube of uh 605
uh side 2 and let's see 606
uh where this uh uh uh this uh point lies 607
so this is your X y and Z plane 608
okay 609
okay so this is your Z 610
Z quarter zero right 611
the interface between these two mediums Z 612
Z quarter zero that is your X 613
y plane and let's see where this point lies 614
so this is your 615
3 is your 4 616
okay 617
like this one okay 618
and then you have to come downwards at 619
minus 5 so this is your 3 620
4 - 5 point right 621
so this means that it lies in Ger medium 2 622
that is Z is less than zero 623
and so 624
this means that the electric field that you involve 625
you need to involve or according 626
the electric density that you need to involve 627
is the electricity in the medium 2 628
that you need to integrate with a specta volume 629
to find out this energy within the cube 630
we know that the energy it is 631
it is the volume in general of this energy density 632
uh and also who 633
you can see here that the energy density juniors 634
they are individuals uh per cubic meter and uh 635
none of uh 636
I see other things uh 637
how to find out these ranges 638
another important thing uh 639
as you can see that uh 640
uh since it is a cube of side 2 641
this means that uh if it is centred at uh 642
x is equal to 3 643
this means that the limit of the x there 644
ranging from 2 to four and the whole range is of 2 645
okay and the center is 3 646
okay and like as you can say 647
you can see here the center is four 648
so notice 3 upper is 5 649
so the whole range is 2 okay 650
and for this uh z xs 651
uh z is minus 5 so there is minus 6 652
upper is minus 4 and whole range is 2 because it is a Q 653
and since he's equal to minus 5 654
so this means it is lying in the region 2 655
now you have to simply integrate this 656
uh uh 657
this given energy density 658
and since it is a constant value 659
so it will come outside 660
it will come outside and you have to simply solve this 661
uh 662
uh this one in trigger 663
right triple in trigger 664
and intuitively 665
you can also see that since it is a cube 666
is one is going to be two ways to bar 667
so two into two into 2 equal and 2 to raise to bar 3 668
that is al q and will 669
you just multiply with the energistensity of that 670
per chlamydium which in this cases medium 2 671
to find out the energy in this Q 672
let's solve uh 673
this another case example 674
so in this case uh 675
region y is less than 0 consist of a perfectly uh 676
conductor so this is your region 677
uh one 678
so again you must 679
must 680
very importantly know what are the coordinate system 681
how to plot this scenario 682
what is the normal between these two interfaces 683
intuitively you can see that y is less than 0 684
this means that a N is going to be a y 685
okay 686
because this is the interface between these two mediums 687
that is your what is his plan 688
what is this plane this plane is your ZX plane okay 689
and the normal between these two planes is 690
normal between these two planes is 691
this is why 692
so this is the thing 693
this is your starting point for this problem 694
so uh 695
region 1 it is like wise less than zero it 696
it consists of a perfect conductor y region 2 697
it is wise greater than zero uh greater than 0 698
it is consisting of a dialect of medium 699
with permittivity as given 700
if there is a surface chart density 701
so here in this case you 702
you are given with this surface chart density 703
0 s of this 704
2 nano coolums per meter scale on the conductor 705
so we need to determine this electric and uh 706
this electric 707
electricity on both sides in both mediums 708
and what are those the those medium points 709
and the first part is this one 710
so here you can see that here in this vertical case 711
your y it is minus 2 712
so this means you are lying in this region one 713
which is your conductor medium 714
for this your y is 1 715
so this means you're lying in this dialectic medium 716
so uh in first case 717
your point uh e is in the conductor 718
so this uh that is your uh region 719
a rise less than 0 and minus 2 is less than 0 720
hence we know and from one week conditions 721
we know that the electric fee inside the conductor is 722
zero hence accordingly 723
Europe electric flux tenses also zero inside your uh 724
this conductor so at this particular point 725
both these electric uh 726
field and electric flux tensities 727
they are going to be zero 728
for the second point it is in the dialectic medium 729
because 1 is greater than 0 730
for divide component 731
and again from the boundary condition 732
from the boundary condition 733
we know that 734
that this normal component of electric constantity 735
it is equal to the 736
this uh surface identity and uh since 737
uh in the uh in the vector form 738
the normal component uh unit rector is a y 739
so it is this rector form and for the electrical 740
uh again the constitute relationship 741
uh that uh this a d is equal to absent E 742
so divide d this please uh divide d by this absent 743
not into absent R by this absent R 744
because it is other than 745
this medium is other than the free space 746
so this is your electric free linger 747
uh at this uh second point 748
this is your assignment for this uh week 749
uh and uh uh 750
you need to solve these uh 751
two problem exercises at your own time however 752
this submission is not required 753
but it is stressed that uh 754
in order to understand this topic more uh 755
uh deeply 756
you must solve these two uh uh practice exercises 757
so today we we 758
we had a review of electric boundary conditions 759
and then we solved some of the case examples 760
to further clarify these electric boundary conditions 761
in the next lecture uh 762
we are going to study the uh 763
what is uh electric 764
electrostatic boundary value problem 765
and how to solve these types of the problem 766
and what is the significance of uh 767
electric boundary problem 768
electric boundary value problems and uh 769
how they are helpful in uh 770
determining the electric uh uh field and the uh these 771
uh electric potentials so that is the significance 772
and uh for that thing 773
we will also drive the warzones and 774
uh laplash equations which are very famous equations 775
for the electric fields 776
and the magnetic fields as well 777
later on 778
and then we will understand what is the uniqueness 779
uh theorem and there after we solve some of the uh 780
case examples as well 781
here I thank you all 782
if you have any questions 783
uh they would be entertained 784
uh through your emails or online 785
synchronous sessions that we have arranged 786
at the department