this is election No. 23 of a V232 1
electromagnetic field theory 2
today we will study matter of images 3
and this will cover your section 6.6 of textbook 4
this is Atomic week No. 14 of your semester 5
the learning objectives of today's lecture would 6
be to understand a method of images 7
to find out the electric potential and electric field 8
so basically in this uh lecture uh 9
we will cover every way to find out the electric 10
potential electric free landing uh 11
forces uh being produced and active upon uh 12
uh to charges and charge distribution on the other uh 13
charge entities 14
uh the matter of images uh 15
was introduced by Lord Calvin in 18 uh 48 uh 16
and is commonly used to determine the potential 17
electric feed potential electrical 18
electric flux density 19
and corresponding lead surface channel density 20
due to the charges in the presence of the conductors 21
so always remember that 22
the condition for the application of this method 23
of images is only in the presence of conductors 24
or frequently it can be the ground as well 25
because at certain frequencies 26
the grounds are also considered as the protect 27
electric conductors 28
but this method 29
so I mean the the earlier 30
earlier matters that we have already covered 31
the presence or left mass equation 32
can be avoided 33
by realizing the fact that conducting surface in a 34
is m equi potential body 35
so we have already discussed that how this uh uh 36
this uh connective service 37
it is considered to be a equi potential body 38
uh this matter 39
uh doesn't apply to all electrostatic problems 40
but it can reduce a formidable problem to a simple one 41
so in in case 42
uh the charges 43
they are in the vicinity of your conductors 44
or ground services 45
then this method can be utilize effectively 46
to find out the resultant potential electric 47
Freeland electrical fractions 48
let's start this method 49
given charge distribution 50
a congregation above an infinite grounded 51
perfect 52
conducting plane 53
may be replaced by the child configuration 54
configuration itself No. 1 55
image No. 2 and an equal potential surface 56
in place of the conducting plane 57
let's see it one by one 58
so the examples for these chart entities are the 59
No. 1 is the is your point source 60
okay so let's consider these point sources 61
they are present in the vicinity of this 62
perfectly conducting plane 63
which is at potential zero 64
that is also known as your grounded plane 65
that is this one and it is of infinite 66
remember infinite uh extent 67
so in case if it is a point charge 68
then it's equivalent 69
image is replaced with the same magnitude 70
but with d opposite polarity 71
if it is a line charge distribution like this one 72
then is image it is mirrored with the 73
you can see here 74
it is mirrored with the same magnitude of the line 75
charge density but with opposite polarity 76
and if it is the surface identity 77
for example in this case 78
it is given as a negative surface 79
volume volume chart density 80
if it is a volume chart density 81
then in this case it will be again better as a 82
you can see see here that it is equally be sphere 83
mirror sphere 84
but with the opposite quality volume chart density 85
so the examples now we have seen that their pointline 86
volume chart densities they can be surfaced as well 87
about what effect conducting grounded pain 88
given in this figure 89
and correspondently image contributions 90
you can see here that images have been replaced 91
at the same distance at the same distance for example 92
it is each about the this perfect conducting pain 93
so it is at the same distance below this 94
perfect land connecting pain 95
and in place of this perfect connecting pain 96
we have 97
we have replaced that with equal potential surface 98
and what is the charge of 99
what is the potential of that surface 100
for this particular case since it is a grounded one 101
so it is taken as zero potential 102
so that is the three steps we are going to do that 103
first one that we are going to uh 104
consider what is the actual child distribution 105
then we are going to place uh 106
a mirror and the opposite bloody charger uh 107
distribution at the same distance below that uh 108
uh perfect connectic plane 109
and the third one is we're going to replace that 110
perfect connectic plane with the equal potential 111
surface of having same potential 112
important considerations 113
in order to apply this image matter is 114
No. 1 115
image charges must be located in the conductive region 116
so this means that uh 117
they must be in the vicinity of the conducting 118
perfect conducting uh playing or reading 119
something like this to satisfy me by John Secretion 120
the second one is the image charges must be located 121
such that on the conducting surfaces 122
the potentially zero constant 123
so this we have already seen 124
seen that in the previous case 125
the image charges 126
they were placed in the vicinity above grounded 127
connecting surface 128
so it can be a constant potential as well 129
to ensure that the boundary conditions are satisfied 130
let's continue the first uh case case examples 131
so the first case example is 132
there's a point charge above our grounded 133
conducting plane right 134
so grounded conducting plane it is at zero potential 135
so this is the uh your conducting plane with 0 printer 136
and this is a point chart Q at a at a height 137
so there's a point chart Q at a height H above this PC 138
and this is considered to be our 139
end of an infinite extent 140
perfect conducting plane 141
in which configuration is given in your figure B 142
where the electric field at point P is given by 143
so at point B we're going to 144
in order to evaluate electric field at this point 145
we're going to follow all these three steps 146
No. 1 the original charge 147
No. 2 the opposite charge with same uh magnitude 148
but opposite polarity at same height but below that 149
paint and the third thing is 150
we're going to replace this with the zero potential 151
this plan okay 152
we could production plan so these are these three steps 153
now what we are going to do that 154
we're going to simply add up 155
the electric field 156
being produced by the original child distribution 157
this one and this mirrored charge right 158
so we're just going to apply the simple condoms law 159
and the other laws by 160
by using that super position principle 161
so here at point P this uh 162
the distance with 163
with respect to this original charges R1 and for the 164
this uh uh in the mirror 165
mirror image it is R2 uh position back to 166
we like to know our basic formulas 167
to find out the electrical intensity 168
that is you know that it is it is a 169
it is charge it is a 170
you remember that it is a force divided by the charge 171
okay so from this particle of point it is a Q right 172
q r one this is your R1 right 173
this is your since you're using the 174
you're replacing the unit tractor for example 175
this is your unit tractor 176
so you have to replace that with this rector 177
divided magnitude of that rector okay 178
so that's why this cube is coming in the 179
denominator okay 180
so this is the same formula that we have studied 181
in chapter No. 3 182
and the contribution of the electric field intensity 183
due to this middle is 184
is simply 185
you have to replace this position factor with this R2 186
and apply the way of charge with minus Q okay 187
the next determine this disposition rector song 188
which is back to this observation point 189
which is placed at point P x 190
y Z and you can see here that 191
and the position right there 192
it is your destination minus your source right 193
destination 194
minus your source right 195
and this is first plus Q which is placed at 0 196
0 h point 197
whereas this mirror it is placed at 0 198
0 minus h point along this ZX is right 199
so you just need to um 200
apply that distance formula like this one 201
and then 202
in order to find out the magnitude of this position 203
rector you know that it is some of the scale of the uh 204
each component and then you are too scared of that 205
that whole quantity so it is the scale of the 206
some of the scales of the each and each quantity okay 207
and help the respective unit rectors have been replaced 208
okay this one and this one 209
so this is your final electric field at this 210
which is being exerted at this point B 211
due to charge Q in the presence of 212
due to charge Q in this 213
in the presence of this perfectly ground 214
and perfectly grounded conducting pin 215
and this is equivalent to the electric field 216
due to this original charge 217
plus the charge due to 218
plus the electric field due to this mirror mirror 219
charge that is minus Q 220
okay so let's see once when Z is 0 221
he has only Z component only 222
let's see if you're Z this one 223
the Z 224
the Z is 0 this means your point is 225
your point is coming on this plane right 226
your part is coming on the screen 227
for example right 228
and so if this is the case 229
if this is the case then you're going to only have this 230
the component so 231
this means that 232
electric field is normal to the conducting surface 233
so you only have the electric field 234
and that is only normal to this conducted surface 235
and that is your E 236
Z right 237
so if you are evaluating the electric field 238
at this conduct on the surface of this conducting plane 239
then it is only the z component 240
and that is where 241
satisfying your earlier boundary conditions as well 242
so 243
confirming that E is normal to the conducting surface 244
and the potential at p is 245
is the obtained 246
from the equation that we have already discussed 247
so let's see how it is Z 248
let's see if you put this equal to zero okay 249
this is zero 250
this is it okay 251
so this is going to cancel out with this one remember 252
okay and then what else 253
only these two things are left in this case that 254
that is only your C component okay 255
that is only your C component and 256
right 257
okay so 258
uh now we need to 259
find out the potential next uh find out this potential 260
the potential it is again 261
be simply the superposition of the potential 262
contribution due to each uh 263
charge entity that is uh 264
your due to at work or perhaps or not into your 265
this distance rector magnitude 266
or the position vector in this case 267
okay and uh 268
so using equation 269
um this uh 44 to find out the surface uh charge 270
then 2 of the induced charge 271
so in this case uh 272
what is the case that you are evaluating 273
at this Z is equal to 0 point 274
so you are only going to have this 275
normal component of the E field 276
that is in the Z direction 277
so you just solve this equation then you will 278
you will find out that this is your resultant E field 279
and that you will uh uh 280
you just need to simplify the things 281
multiply that is not your 282
so this is the boundary condition 283
that we have already studied 284
that for the conducting plane 285
the uh surface identity it is equal to the uh 286
normal component of your electric cross density or e 287
which is equal in V 288
d optional not times b e n 4 free space medium 289
so 290
today in this charge on the connecting pen is now you 291
you want to find out the charge on the pen 292
you just need to integrate this 293
with respect to surface differential surface element 294
and uh since it is uh an infinite uh 295
conducting plane so it's x and y components 296
they are uh 297
many from minus infinite two positive infinity 298
now how to simplify this integration 299
uh you just need to play around with the uh conversion 300
uh into this vertical into the cynical corner system 301
so use that uh simplification 302
just use the change of variables that uh 303
let's replace this uh with uh its respective 304
uh raw component in the 305
in your cynical coordinate system 306
right and once you reduce this 307
the replacement of this DX d y is Rodeo defy 308
now replace this thing 309
the equation will reduce to this form 310
and accordingly the limits of your radius 311
it will roll it is from 0 to infinity and for fire 312
for fire it is from 0 2 to pi and no 313
you need to integrate this right 314
no let's use this substitution rule right 315
and how you can integrate this 316
just substitute this inner quantity 317
that is in your denominator 318
with you so the derivative with your 0 319
your derivative of Jew is 2 row 0 okay 320
and uh once you will do that then uh your 321
this denominator for example 322
it will come in your uh numinator 323
you just apply this uh your uh respective uh power 324
okay so once you will apply the power 325
so it will become minus 3 by two plus 1 right 326
so it will become your minus 1 by 2 327
okay 328
and once it will again go into the denominator then it 329
will be a positive quantity 330
alright and uh let's see what else uh you need to no 331
you need to uh plug in in the previous uh slide 332
you need to plug in that uh 333
uh measure Linux 334
and then this equation will reduce to the desired form 335
that is your 336
once you will er once you will er 337
er put in the Linux 338
anything divided by the infinity United 339
it is zero right 340
and for that case er 341
you'll see that once you will put in 342
then the dessert will be minus 2 343
that is your er induced charge on the surface 344
it is equal to minus Q okay 345
and it is expected because all flex nines 346
uh will terminate on the conductor 347
would have been terminated on the image charge 348
it is no conductor 349
so since there's a conductor on the in this place 350
so 351
you can assume that there's a negative charge of equal 352
equal magnitude level 353
opposite quality on these circuits of charge 354
so that is the takeaway from that by the first light 355
now let's uh 356
consider the another case that is your line charge 357
uh 358
distribution above our grounded plane connecting plane 359
and uh consider our infinite line charge with uh 360
Dansty Royale and uh 361
uh located at a distance H 362
again from the perfect ground connecting plane 363
uh which is uh your at is 0 364
0 plane this this plane okay 365
and uh as you remember last figure 366
we just need to repair this Q with d or L 367
and uh uh 368
you can uh you can uh just recall your previous uh uh 369
previous uh uh concepts from your lecture number 370
from your chapter number 3 okay 371
and that infinite line charge maybe zoom in this case 372
in this case X is equal to zero right 373
so it is 374
you can assume that if this is your plane right 375
you can assume that it is your it is like this one okay 376
like this one it is your role at okay 377
it is at at height age from your this plane 378
your grounded plane and it is 379
it is oriented at this zero point on the x axis however 380
it is changing from minus infinity to pause 381
to infinite d and b by direction okay 382
so that is the case 383
and its image will be minus roll L right 384
minus roll L like this one okay 385
and uh back at the height minus uh 386
minus 8 below that separate 387
so that the two line chart densities are parallel 388
to YXL y parallel dryaxies because x is equal to zero 389
electrophilic point b is again 390
given by the superposition of their individual 391
electric fields 392
and as you can see here the row l and what is row 1 393
so it is 394
the perpendicular distance 395
between the evaluation point 396
and b 397
and that and your this line chart density okay 398
this will be considered 399
as suggested in our previous chapters 400
and what is ERO it is the unit 401
normal rector between the evolution point and the 402
your line chart distribution 403
and your only changes are your unit rector 404
your this protector distance and your 405
this quality is going to change for the second case for 406
for the images 407
but what are the perpendable rectors in this case 408
we are just considering that 409
we're just conceiving that we have partner 410
but we have to graduate here right when we cheer the X 411
y and Z so what you are going to do that you are 412
no now you can find out this uh uh 413
this perpencular distance between the uh 414
evolution point or observation point and the line 415
the uh identity using this again 416
uh 417
the distance from Mola that this is your destination 418
right is your source y x is equal to 0 419
and probably 12 because it is aligned with the y 420
x is only okay and it is at height age 421
and it's y coordinates raining 422
because it is raining from minus infinity to two 423
one stupid infinity so this is the partner 424
you know this is this is you can say that 425
this position director for this distance vector 426
you can see in this case 427
the distance vector between the evolution point 428
and d source 429
and likewise for the image only the differences 430
this minus H 431
okay now plugging back into this equation 432
and accordingly you can 433
you tell you can use this magnitude as well 434
so that is your X square plus Z minus H square 435
and uh so again 436
and you can see over here that if you put Z equal to 0 437
that is your Z equal zero over here 438
so these two are going to cancel out right 439
and uh what else right 440
you can see it and you put this is equal to zero 441
these two are going to cancel out 442
and you're only left with Z compound again 443
and then again it is confirming your statement 444
that in the vicinity of the surface of your effect 445
electric conductor plane your electric release 446
always normal to that conducting surface 447
the next uh find out the potential at point p 448
uh by using the again 449
the the same formulation that we know that it 450
it would be the the resultant uh 451
potential at devotion point p would be again 452
this super position of the potentials 453
of each individual lines 454
line source that is your original line source 455
and your image line source okay 456
and now you have to take take care of this 457
your 458
see these these signs 459
that is your minus coming out because of this thing 460
and your 461
this another simplification form that is your 462
that the difference between the alarm terms 463
you can replace with that division terms 464
let's uh further simplify this thing that uh you know 465
I need to uh put put in back that uh 466
botanical distance between your line source 467
and evaluation point that is this one okay 468
so substituting 469
the value of distance reactors 470
from previous page equations 471
this equation will reduce to this form 472
so the surface chart density 473
induced from the conducting pain 474
is the pain given by using this boundary condition 475
that is surface chart density 476
it is equal to the normal component of the electric 477
first density 478
and again you just need to put in that Z equal to 0 479
into this electrical equation that we have 480
drive in our previous slide 481
so this is your reduced form okay 482
and not in order to find out the induced charge 483
uh in this case on the conducting change 484
you just need to simply this 485
integrate this supercharger density along with this uh 486
axis okay 487
but in this particular case it is your x axis 488
uh where the things are away okay 489
and uh since it is uh along one line 490
you can say that it is uh your uh 491
so once you will uh integrate that uh uh you 492
you you can see over here that uh it's uh dimensions 493
you can you just uh 494
play around with this equation and uh 495
play with this uh integration right 496
uh so these are the 497
these are the construct with respect to the X right 498
and uh what else the only changes okay 499
this one okay 500
here you have to again use this substitution rule 501
and uh uh 502
just just follow that 503
it will be coming out in terms of 504
once you will integrate that 505
whether it's that to your eh 506
this I'll find this case and put back this 507
this limit standard to be 508
it will be reduced to this form 509
that is equal to minus or L 510
again you can see here that it is uh 511
for equal magnitude but with B opposite polarity 512
so it is again uh confirming your previous case 513
the department source case 514
thing that deflect 515
lines originating from the source 516
will terminate on the image 517
if the conductor is absent 518
however this uh particle case and there's a conductor 519
so the uh these uh lines 520
the electric first lines originating from the source 521
they will terminate on the conductor 522
let's solve this example to understand further 523
what is the significance of this matter of images 524
so our point chart Q is located at point a 525
0 and B so this means that your X component is 526
your X component is a 527
okay your y component it is for example X 528
y and Z you can just take it 529
in this direction okay this direction this is your your 530
for example is your y 531
Y1 kids Y0 in this part of the case 532
okay and then Z then Z then Z component is your B 533
okay this one right 534
and uh between two semi infinite conducting pins 535
so why semi infinite 536
because they are fixed at this point zero 537
but they are infinite in z direction 538
and in this y direction 539
they're intersecting at the right angle 540
as you can see here 541
they are prepared to look at each other 542
90 degree orientation 543
and we need to determine the potential at point P 544
and the force and Q 545
so let's try with one by one 546
first of all 547
since they are in the vicinity of the conducting pin 548
which is grounded in this vertical case 549
we need we need to create uh 550
an image configuation for this thing right 551
so in this case the three images are necessary 552
to satisfy the conditions 553
in the the 554
that we have all discussed on previous slide that uh 555
the poisons condition right 556
poison condition only 557
boundary condition that 558
the images must be located in the vicinity of 559
conducting plane the images must be located 560
site that the conducting planes uh 561
potentially zero as the Audi constant 562
so these these are the conditions 563
and then we will talk about something like 564
the boundary conditions and the set it 565
so yeah the first condition is to 566
is necessary to satisfy the poison equation 567
and the second condition that says 568
to satisfy that boundary conditions okay 569
and if you will specify those conditions 570
then you need to put in that 571
see images in this particular case 572
so how to put that 573
you can see here so the image of this one is here okay 574
minus q at at this you can see that at the 575
at the a distance that is your minus B okay 576
right 577
the second image of this one is on the other side okay 578
on this left side that is that minus 8 distant 579
now the third image it is indeed diagonal one 580
right this one 581
so it is that you can see that it just just label it 582
for example this one and it is a distance you can see 583
you can see that is clear plus B square 584
okay because it is a diagonal 585
we will see what is its position right 586
in the next slide so this is the institution from here 587
and then the provincial point P 588
which is the direction point 589
will be the stupid position or the provincial point B 590
due to this individual point charges okay 591
and accordingly the course at this queue 592
due to these images would be 593
these superposition 594
of the forces being exerted by these images 595
on this 596
uh original charge 597
first of all let's find out this potential 598
so just 599
you need to take care of the displities of the image 600
and the source 601
and then you have to take care of the corresponding 602
distance directors okay 603
for this first case the original original case 604
your point is at a zero 605
so this is your point to that a 0 n b 606
so that is this one okay 607
the second case is your image 608
for the image you're on the left hand side 609
it is minus a 0 n 610
b for the image on the 611
on the bottom okay on the on this 612
yeah 613
yes for the image on the diagonal 614
right on the diagonal on the dial it is minus a 615
0 n minus b point 616
so the location of that to image with the 617
with the charge of plus Q 618
so this is your minus Q is your original source plus Q 619
and this is your termimate that is minus Q 620
and it is at point a 621
0 minus B right 622
and not find not 623
find out the force being exerted 624
on the original charge to you 625
you just need to add up 626
the forces being exerted by the new charge 627
on this or by this images on the original charge okay 628
ingestion while this uh chromosome formulation okay 629
and uh for this case the distance 630
the distance once you will uh solve this example okay 631
now you need to find out the distance 632
for example it is your Q 633
it is your minus Q this is another Q and this is your 634
sorry a plus 2 Q and your minus Q 635
okay 636
and uh in this case in this case 637
uh you will see that 638
you need to find the default 639
being exerted by this minus Q 640
image on this Q 641
then you can see it the distance it is to be okay 642
because it is at B and it is at minus B distance 643
this is your access okay 644
and likewise for this one 645
so this is a sorry this is a 646
a and for this one this is B n minus B 647
okay and the distance would be to B 648
K and what else 649
for this Taiwan this one 650
this one is pretty important 651
okay for here 652
you need to involve this because it is at a zero guess 653
and B and this is at minus A 654
0MINUSB you just enroll this distance formation 655
to take this clear look of that 656
and since you are enrolling this rectors over here 657
unit rectors over here 658
so that's where it is coming out to be 3 / 3 659
the part of three because of this involving this 660
this vector right R vector 661
R q okay 662
that is your basic column stop 663
so once you will simplify the things 664
you take the commons and the 665
so this is your final choose form 666
so electrically 667
due to this system can be determined similarly 668
and charge and use on the planes can be form 669
okay and then if you want to find all the electrical 670
so you know that it is 671
e have to identify simply Q 672
second example in general 673
the method matter of images 674
is used for a system consisting of a point 675
charge between two semi infinite conducting plans 676
so in this in this example 677
we are only considering the simple cases 678
that we only have two infinite 679
semi infinite conducting plans 680
and they can be inclined at any angle 5 number five 681
then the number of images 682
they can be determined in this formulation 683
okay 684
3 60 because you can you can consider that is complete 685
sphere or circle or thing like this 686
the angler distribution between these two planes 687
and minus 1 688
let's apply this 689
given formula on all these cases that we have started 690
okay 691
because the charge and its image line on a circle right 692
this extends your circle for example when the uh 693
the angle distribution between the uh 694
once the once the uh this plane 695
connecting plane is at angle of uh 696
this uh one 80 degree right 697
that is your straight line right 698
connecting plane is straight line 699
uh then you're only going to have one image 700
you can see here 701
so this will be 1 80 degrees so 3 60 / 1 80 is 2 702
2MINUS1 is 1 703
so that's why we only had the one image here 704
and if it is they are at the orientation of 90 degree 705
that we studied in our previous example 706
then uh you can see that 3 707
60 divided by this so 5 it is 4 708
4MINUS1 is your 3 709
so that's why we had the three images 710
and in case they are uh 711
inclined at the angle of 60 degree for this case okay 712
then these charges 713
there will be these number of um images 714
there will be you can see here 6 - 1 it is 715
minus 1 it is five okay 716
and you 717
you need to place this five charges in this case right 718
and what would be the angle distribution between them 719
between them that would that 720
that you know that it is your 721
the angle that is your this one 722
okay this one 723
60 degree 724
what is this 30 degree 1960 725
yeah it is 726
it is sorry 727
it is 6 degree 728
okay 729
right and uh 730
likewise they can be room treated like this okay this 731
this okay 732
uh but you need to take care of these priorities 733
priorities okay 734
priorities 735
priorities 736
and then we will solve the case 737
the cases like 738
for the potential 739
and the problems for the presidential 740
and the electric grid and electric forces 741
just like the example that we solved 742
in our previous slide 743
so today we understood uh 744
another matter that is known as the matter of images 745
to find out the electric potential and electric feed 746
due to point charge or the line charge uh 747
or other charge distribution in the vicinity of 748
remember in the vicinity of uh 749
perfect conducting plane so that we can treat that uh 750
equal to the equal potential surface 751
and we replace that we add another image of uh 752
the source with equal magnitude what 753
but with the opposite planet 754
in the next class we're going to have an oval here 755
question number two 756
alright thank you all 757
if you have any questions 758
uh they would be entertained uh 759
through your emails or online 760
synchronous sessions that we have arranged 761
at the department