this is election No. 13 of a V232 2
electromagnetic field theory 3
today 4
we will study a relationship between electric field 5
and electro potential 6
and also the electric typo 7
this will cover section 4.8 and 4.9 of your textbook 8
this is academic week No. 7 of your semester 9
the learning objectives of today's lecture 10
would be to find out 11
a relationship between electric field 12
and electric potential 13
to derive Maxwell's second equation 14
and to understand the working principle of electric 15
typo 16
first of all uh 17
let's uh drive 18
a relationship between electric field 19
and electric potential and then uh 20
we will drive a Maxwell's second equation 21
so if we start up with this potential difference 22
between two points that is between a and B 23
so it is the difference between potential at point B 24
minus the potential at point a 25
if you're taking the reference point of infinity 26
okay and then we studied that 27
if we want to find out this potential 28
difference between these two points 29
then it can be easily found out by using this 30
integral equation so 31
the integration is of this electric field 32
dotted along with it 33
along with this part of travelling 34
along which we want to find out the electric field 35
interception 36
so uh uh Indian uh Indian medical technique 37
you can find out uh 38
like this one that it is minus a uh 39
in integral of a to B e dot B L 40
so 41
you have to take the line integral of electric wheel 42
along the given part 43
so if this part is from a to B 44
so it can be integrated using this line integral 45
so if you want to find out the potential difference 46
between this B to a 47
then so it will be 48
it will be the same but in the minus direction right 49
so the potential difference same 50
but in the minus direction 51
because in that case you're working 52
you're working against the 53
the original electrical difference so there 54
there is a next gain endure uh 55
endure uh endure uh that uh energy or endure potential 56
potential difference so if you sum up these two things 57
then you can find out that it is a closed integral 58
of electric wheel 59
around this closed bar that is starting from your 60
from your this point a and terminating on point B 61
and then starting from point B and then terminate 62
terminating to B 63
originating point 64
so it is making up a close path in this case 65
so for this a close path integral 66
we can find out that this close integral is 67
equal to zero so 68
this means that no network is being done 69
in moving a charge along a closed part 70
in a electrostatic field 71
so the closed part is from a to B 72
N B to A 73
so this means 74
if we apply the famous strokes theorem to find out 75
or to relate disclose a line intrigue 76
well with the open surface integral 77
by taking the help of this curve 78
of the of this rector field quantity electric field 79
then we can find it out that this close interval 80
it is equal to the surface integral of that cross 81
e dot DS and if we equate this equation 82
this equation with this equation right 83
so this means this this is equal to 0 84
so how this can be equal to 0 85
this can only be equal to zero 86
if the core of electric field 87
the inner quantity is equal to zero 88
because your surface element 89
it cannot be equal to zero right 90
so the only thing that is equal to 0 91
that is your attel crossey or otherwise 92
otherwise the angle between this call and the normal 93
to this surface element is equal to 90 degree 94
than in that case 95
also this quantity can be equal to zero 96
but for the time being 97
we are considering a general case 98
so for the general case 99
this dark cross is equal to zero 100
because this quantity is going to be equal to zero 101
so what we what this implies 102
this implies that electrostatic field is conservating 103
or irritational 104
so that is uh your institution from here 105
and then you can uh we also denote this uh 106
differential equation as a second 107
maximum equation for static electric field 108
so picking on uh 109
the research that we concluded on the previous light 110
uh the uh curl of a vector 111
it is equal to zero depend 112
only if this electric quantity is a gradient 113
of some scaler function 114
so if we take the help of this vector identity 115
that curve of gradient of a scale is equal to 0 116
so we can do not like this way 117
that this gradient of v is equal to electric cream 118
in this case so 119
here 120
we assume that this E is equal to minus gradient of v 121
so why we we choose this minus 122
so this was the question 123
that was only asked during the class 124
so that at that point 125
I told the student to reserve this question 126
for future lecture so here's the answer 127
so what is the answer first of all 128
electric fantastic is the gradient of E right 129
so it is coming out from this vector identity 130
and then a negative sign shows that 131
direction of E is opposite to 132
the direction in which we increases 133
right to the potential 134
increases in the direction opposite to the electrical 135
so for this case if you see here 136
uh the electric field 137
it is uh 138
going away from this positively chart sphere right 139
and if you you are moving a chart 140
if you are moving a chart from a positive chart 141
from here towards here 142
then basically you are going to increase the potential 143
right because you are working against the electricity 144
so because of this that against thing 145
because of that against thing 146
this electric field so 147
this negative sign means the sign is coming into play 148
so negative sign 149
shows that the direction of E is opposite to 150
the direction in which V increases 151
so he is in if you're uh 152
removing this Q positive point 153
charge outside towards this positively charged sphere 154
then you're gaining the potential 155
the potential differences being gained 156
so that's why 157
and is you are moving in the direction of 158
direction opposite to the electric grid 159
so that's why this negative sign is coming into play 160
he is a directed from higher to lower levels 161
and you can see here 162
that he is moving in the opposite direction uh 163
to the uh uh to the direction in which uh 164
as your uh 165
it is potential difference going to increase 166
now let's solve this example of your book 4 167
14 to further clarify the concepts of this work 168
being done in an electric field 169
or in or in the presence of that potential difference 170
okay so what we need to find out 171
we are given the potential potential 172
electric potential like this 173
so it is since it is our Tita and phi 174
so it is your spherical coordinate system 175
and we need to find out the electric proxanity 176
at this particular point 177
and then we need to work calculately 178
well done in moving a positive charge 179
and microcoulums from point at this to point this 180
okay so first of all 181
the electric proxanity 182
we know that it is equal to absolute 183
not electric field 184
and electric field is equal to minus ingredient of v 185
v we are given with 186
since it is in spherical coordinate systems 187
also 188
we have to involve this gradient equation of spherical 189
coordinate system and then uh 190
we can change the uh partial derivative of this uh 191
electric potential scaler 192
scaler quantity with respect to R 193
with respect to data and with respect to file 194
and again take care of these cautions 195
along this partial derivatives 196
and then you can simplify these things 197
so first derivation is with respect to R 198
so we know that the derivative 199
so the R 200
r minus r r scale will be going in the numerator 201
so it will be R r minus 2 202
then you involve this P r rule 203
so it will be coming out with minus 1 or r cube 204
and minus minus will be cancelled with this minus 205
so it will be a positive function 206
for the second case 207
the division is with this fact to this theatre 208
and scientific we know that it is a cause of theatre 209
and for the third case it is five 210
so cause of five we know that it's a delevation 211
it is equal to minus sign of uh sign of uh 5 212
so that's why it is coming out to be positive 213
minus minus positive 214
okay then we have to evaluate this 215
we know that it is r which is t 216
r and this is 5 like this thing 217
and then you can evaluate this expression 218
to find out the electric flux tendency 219
at this vertical point 220
we're gonna let solve this uh part 2 221
which is uh 222
very important in this case because it will 223
uh clarify your concept 224
how to find out the work done 225
while you are moving a particular charge 226
in the presence of some electric grid 227
from one point to another point 228
so again the points given are in sperical coordinates 229
system electric field we have found out 230
using the information of electric potential 231
and so the first method first method 232
will involve our standard integral equation 233
and uh from there 234
we know that well done is equal to minus q times uh 235
edot DL right 236
this 237
we have already contributed in our previous structures 238
so uh since the electric field is conservated 239
the part of integration is immaturial 240
so this is very 241
very important Assumption in simplifying this question 242
the integration in this question 243
because you can see here 244
while moving from point a to B 245
all three coordinates are changing 246
right 247
so in order to simplify this into a line integral 248
so what we do we break down this part into three 249
so 250
since we know that electrostatic field is conservated 251
for the part of integration in material 252
so either will go from a to B like this way 253
or we break up this part into uh 254
three parts and that parts are depending upon the uh 255
the variation of one variable at one time 256
so in the first grade 257
we are wearing this r from 1 to 4 258
and we're keeping this Peter and five costume 259
for the second case we are wearing this cheetah 260
and we're keeping this by constant 261
and we are keeping this R constant 262
which was from the previous time 263
and in the test time we are keeping this R question 264
we are keeping this uh 265
teacher question and we are only changing this 5 266
so these are the three uh 267
parts that we are going to integrate 268
so if we see here 269
and the first part only this R is changing 270
so there a line uh element it is d R a R right 271
this we have Covenant chapter number two 272
and in the second case your teacher is wearing 273
you know that if teacher is wearing then uh 274
then 275
then the line in line segment will be to equal to R 276
d Peter B Peter 277
so here you can involve that uh art rule as well 278
okay the third case only 5 vary is 5 vary 279
and we know that line in line segment 280
for this one is our scientifica D85A5 281
let's uh integrate these uh three segments one by one 282
and uh we know that we are given with this electrical 283
we know these uh line segments 284
uh for which we have to integrate 285
so the first uh first part first uh first line segment 286
this one artist Jenny from 1 to uh 4 right 287
and uh we integrate this with respect to Dr so again 288
uh remember if we take this R R R R cube as a variable 289
we take this quantities uh out our side T10 crossfire 290
we evaluate at the constant at t 10 5 291
we integrate this one or r cube 292
so the resultant will be 293
minus 1 hour minus 1 hour are four I guess 294
and then you can 295
you can simplify this integration 296
and then you will see that oh okay 297
this minus sign is coming over here right 298
and then you can plug in these values 299
for the second case 300
only the integration is where it's back to this 301
the Peter so we know that cost Peter's integration is 302
is sign of Peter right 303
so that's why you are going to see what you are 304
yes so you can uh 305
further put in these values of uh qualify and uh 306
R constant you can find out this value 307
and then in the last step 308
you can uh integrate with respect to 5 signify 309
we know that it is equal to minus cause of 5 310
and then you can spot this 311
putting these constant values for that part 312
to simplify this expression 313
so this is your uh your method number one 314
once you are given the electric related formation 315
and in case you are directly given 316
the information of electric potential 317
as in this case 318
we were initially given the information of electric 319
potential so what we need to do 320
we need to simply involve this formula 321
for the work that with er 322
we derived in our previous lectures 323
so that was work is equal to q times we a b right 324
so we a B 325
we know that it is the difference of b b minus we a 326
so the potential at point B and potential at point a 327
so scaler value we are a scaler function 328
we are given electric for the collective potential 329
we just need to simply plug in the values of this point 330
a 331
and this point B into this uh okay 332
this point uh rather you can take it like this 333
uh uh yes you can take it like this 334
so point B will be coming out over here 335
yes four like this point B will be coming out over here 336
point I said a less a less erased 337
less erased so that it is more clear 338
so point B is coming out over here 339
and point a is coming out over here right 340
so B b minus v a just solve this 341
this right this equation and we will point out B 342
same answer that we found out early 343
so these are the two ways 344
so remember once the electric grid is given 345
you have to involve this uh 346
integral and then uh 347
break down the part of your integration 348
to simplify your line integration 349
and if your potential information is given and you 350
you need to find out the world 351
done between these two points 352
then you can simply involve this equation 353
and use this informational 354
collective potential directly 355
now let's understand what is uh an electric die pole 356
so an electric die pole is form 357
so you can see here so it is die 358
die means two pole so two pole 359
there are two poles so 360
it is made up of two point charges of equal magnitude 361
but opposite polarity 362
opposite sign and they are separated by a very 363
very small distance d 364
we will see here what is this distance d 365
so remember electrodetol 366
it is formed by using two point charges 367
of equal magnitude but opposite polarity 368
and with a very small suppression 369
so let's consider this figure 370
so now win a potential at point P 371
at this point 372
due to an electric tripod 373
is just the sum of potentials due to each point charge 374
that is potential at point 375
so this is your 376
as you can as you can treat this point as a 377
this one point at uh at coordinates r t 10 5 378
so we need to find out the potential at this point 379
to do this positive charge and this minus q charge 380
so we know the definition of this 381
the potential 382
so basically the potential at this point P will be up 383
superposition of potential 384
due to these two opposite polarity charges 385
so that's why we are summing up the potential of 386
individual potential of these two charges 387
which is caused at this point p 388
okay so uh so far this uh uh 389
since uh this negative sign is sign is coming to play 390
right so that's why this uh 391
minus sign is coming in because of this thing right 392
so 393
we are assuming this supposed to charge is at distance 394
radio distance R1 395
this minus Q charge is at radio distance R2 396
and this point from the origin 397
it is at radio distance R 398
and uh 399
this is the smallest station between these two charges 400
that is d and uh this uh p 401
it is making an angle of a t job with respect to this 402
uh 403
z uh z 404
uh z uh 405
z axis because it is in this vertical partner system 406
and how does this d cause of teacher comes 407
uh this 408
the d cause of data come from here 409
that uh uh this d you you if 410
if we assume that it is also Tita 1 right 411
and this is Tita 2 412
so we will take help of this 413
uh Assumption that if you are 414
your distance from this origin is very 415
very greater than the expression between these two uh 416
opposite charger point charges 417
then uh what are these options 418
so we can assume that this R1 419
R1 so we can write it R1 is approximately equal to R2 420
right so we can 421
once you take the Altium of this first quantity 422
like this one then you can plug in this thing 423
that R1 is approximately equal to r 424
2 is approximately equal to r 425
so this will be equal to r r 426
so that is r skip right 427
so it is first uh first simplification okay 428
the second simplification is the projection of this r 429
on this uh 430
other axis so how to find out this projection 431
so we can assume that the difference between this R2 432
and R1 if we assume that this is the difference 433
for example we assume that this is the the projection 434
this is the difference 435
video difference between this R2 and R1 436
so how to find out this thing 437
it is d is very small we can consider that this cheetah 438
it is approximately equal to Peter 1 439
is approximately equal to Peter 2 440
so we just take this a theater 441
we we make the projection of this d on this axis 442
we treat it as an axis 443
we can find out that this R go minus or 1 444
it is the projection of the 445
on this separation or 446
of this difference between these two radio distances 447
so since we are projecting on this difference 448
so 449
that's why we are taking the cost component of this d 450
so there is a d cause of data right 451
so plug in here so this is your 452
electric potential for this electric typo 453
let's uh find out 454
the electric feed for this typo 455
is using the information that we drive on this 456
on the previous slide 457
so since it is in um uh cerical cord system 458
we just uh need to uh 459
take the portion derivative of this uh 460
human quantity with respect to R t 10 5 okay 461
and it is having only variable r and t r 462
so this is going to be zero 463
right because this will be acting as a constant 464
with respect to five so for this thing 465
uh it is uh 466
uh Oscar 467
so Oscar is uh 468
one or Oscar derivative is minus 1 or R2 469
so minus minus well 470
then we cancel out over here 471
and it could be a plus over here 472
right 473
and the question for the second thing 474
it is causal tita we know that it's derivative 475
it is minus sign of tita so again 476
b it is coming out to be positive 477
so that is the uh 478
it's electric feed of the electric typo 479
and then we can simplify by taking the common uh 480
this QD divider for perhaps or not our cube 481
and that is the simplified form of this electric feed 482
due to an electric typo 483
now let's further define some of the other definitions 484
related to this electric typo 485
first of all let's define what is this a d cost data 486
but if you assume that this d 487
d is a line vector line segment 488
right uh 489
since it is oriented along 490
this is the XS so we can uh 491
we can in the vector uh 492
notation 493
we can denote it like a d a Z first do not replace a Z 494
an a Z will be directed from minus q two q 495
so remember this thing for the convention of this book 496
so it is directed from for the tripod 497
it is directed from minus Q 2 q k 498
then uh we define another terminology 499
that is the dipal movement 500
so what is dipal movement 501
so dipal movement 502
is the product of the magnitude of charge time b 503
separation between these two charges 504
so in this case it would consider it a 505
d as a erector quantity right 506
like this way okay 507
so this is you can treat it like d 508
a Z right 509
and then we can we can involve what 510
we can involve this this thing right 511
this thing so we can involve this thing 512
and this definition of the Z 513
if you see here if you see here the magnitude of a r 514
it is equal to one so basically 515
we are taking the dark part out of this d 516
along this radial direction 517
along this radial direction 518
because we are taking this projection 519
we are taking this projection of d 520
on this difference between R2 521
minus R1 522
right and which is approximately equal to 523
the cause of data 524
okay so uh 525
let's further play around with this 526
this thing so we define the uh vector 527
uh diaper moment that is equal to q 528
uh this uh d 529
uh vector d right 530
and then we can uh write down this uh 531
this thing right this thing that 532
q 533
q d q d right 534
this thing 535
so we multiply both sides with q right 536
we multiply both sides with q 537
so it is equal to q d cost data 538
and must we simplify in the form of this 539
drop product notation 540
we can simplify like this thing right 541
so you can play around that it is equal to 542
so multiply q d right 543
q d vector so multiplying both sides with this q 544
Q and Q right so it would be Q d dot a r right 545
and now we can uh involve this definition of your 546
type of moment like this 547
and if you if you open up this dot product 548
so it is in other words 549
it is equal to q d cause of Twp magnitude of this a 550
a R unit factor is 1 and the magnitude of this p is 551
p magnitude is equal to curing 552
like this okay so 553
this is another form of this electric vector potential 554
because of this electric iPhone 555
since we know that unit vector is the vector 556
divided by gratitude of the vector 557
so we can further donate 558
denote this equation 559
in this dot product of the vector form 560
so if your diple centre is at our bar 561
this means if your diple is not at this origin 562
rather it is at some another position back to our bar 563
so you just be extend your 564
simply extend your case to the standard case 565
that you can take the distance 566
you can treat this position 567
vector as a distance vector between this R 568
and R but it is just like the same case 569
that we considered in our coolants law 570
for once 571
your point charge is not located at the origin 572
k and then uh what else so again 573
we can uh write this uh 574
electrical in the vector form 575
like this thing that are in the diaper moment form 576
like this thing so this QD will define 577
it is equal to this magnitude of diaper movement 578
like this thing and rest things are staying the same 579
okay 580
okay so only way define this thing additionally 581
so here so you can 582
you can summarize this line like this way 583
that it is the 584
electric potential form in the dipole 585
right in the dipole moment 586
using the dipole movement information 587
and it is the uh this second equation 588
is the electric feeder of the dipole 589
using the dipole moment in formation 590
okay so there are some notes about the electric 591
electric Dibol so our point charge is a monopole 592
so it is a single 593
single charge and its electric grid varies inversely 594
with respect to our skill 595
so it is in mostly proportionate to our skill 596
that is electric wheel 597
electric fee for the single point charge 598
it is inversely proportional to R scale 599
and we found out that it's potential 600
it is inversely proportional to R 601
so it is one part less than the R scale right 602
so it is one part less than the electric fee right 603
so likewise 604
if we see over here from the last page equations 605
that the electrode 606
page due to a dye ball varies inversely 607
uh as RQ right 608
RQ this we saw 609
and then we saw that the potential it varies 610
uh inversely with respect to RQ 611
so again it is one less than the uh 612
this inverse relationship is 613
one less than the electrician relationship 614
and then we can extend this case to the 615
to the higher order multi multiples 616
so electrically due to successive high order multiples 617
such as a quadruple quadruple 618
which says quadruple as well 619
consisting of two die poles or an octable 620
uh consisting of two uh 621
quadruples and a dear anniversary proportion to ask uh 622
R R4 623
R5 and R R6 onwards 624
and likewise the potentials they will be 1 625
they will be inversely proportional to all our part 626
which is one less than the 627
in worst relationship for the electric field 628
so that is the extension of these two 629
uh first two cases 630
let's solve this example for the 631
for the clarification of this electric dipole 632
so two dipoles with dipole movement 633
so here two dipoles are given 634
and dipole movement is given right 635
p is given in this case 636
and uh for the first case it is having minus 5 637
easy for the second dipole it is 9 638
80 and the location of these two dipole is 0 639
0MINUS2 and 0 0 3 so we see here 640
so the first iPhone it is at if it is X 641
y Z corner so it is X and y both are 0 only Z having 642
so this is your first iPhone 643
right 644
this is your second iPod okay 645
and we simply involve this summation of 646
we need to find a differential act origin 647
so this is our point of interest right 648
so we simply add up the potential because of these two 649
uh electric die poles 650
uh uh at this point at this origin 651
so we find out the potential due to this electric die 652
the first diepole at this origin 653
we find out the potential due to this second diepole 654
at this region and then we sum some them up 655
these two diepoles diepoles potential okay 656
so for P1P1 is given as minus 5 and Nano Coolance a Z 657
and what is R1 so R1 is your discuss factor 658
uh for this first iPhone with respect to this origin 659
so this is your destination right 660
this is your destination 661
this is your source 1 this is your source 2 662
and you just need to subtract the destination 663
coordinate from the source coordinates 664
and likewise for the second one 665
and you can find out that it is 2 a Z and minus a Z 666
okay then simply plug in these uh values 667
and since it is a dot product right 668
so both are uh easy so easy dotted with easy 669
easy go to one and then minus 9 is coming into place 670
so it is minus 10 nanoquodoms and then what else 671
okay R1 I one okay I one I 1 672
2 so I one magnitude is 2 so it can 673
it will come out to be 8 and R2 magnitude is 3 674
so it will come out to be 27 675
and then you can simplify to find out the potential 676
that is minus 20.25 voltz 677
so I'm here today's lecture 678
so first of all 679
we found out our relationship between electric and 680
electric field and electric potential 681
that how 682
if we are given the information of electric potential 683
then how we can find out be this electrical 684
and if we are given the information of this electrical 685
then how we can find out the 686
this potential electric potential 687
so these are the two very useful relationships 688
then based on this property of 689
of this consegrated property of electrostatic fee 690
we drive this Maxwell's second equation 691
that current of electrical is equal to zero 692
then uh we discussed this uh electric uh typo 693
and then we found out it's uh 694
potential and the electric free equations 695
uh and uh and this standard equation of these uh typos 696
in the next class 697
we are going to study this electric flux lines and uh 698
what are the equiperential surfaces 699
and what are the equiperential lines 700
and then we will also determine the energy density 701
in an electrostatic fee 702
here I thank you all if you have any questions 703
that will be entertained to your emails or online 704
synchron 705
a session that we have arranged at the department