1 00:00:01,600 --> 00:00:04,466 this is election No. 20 of a B 2 2 00:00:04,466 --> 00:00:05,266 3 2 3 00:00:05,666 --> 00:00:06,466 Today 4 00:00:07,000 --> 00:00:10,966 we will understand what is electrostatic boundary value 5 00:00:10,966 --> 00:00:13,200 problems and their significance 6 00:00:13,366 --> 00:00:17,533 this will color your sections of a chapter No. 6 7 00:00:17,533 --> 00:00:20,366 that is a 6.12 6.4 8 00:00:22,066 --> 00:00:23,966 so the learning objectives of uh 9 00:00:23,966 --> 00:00:26,100 today's election would be uh 10 00:00:26,100 --> 00:00:29,733 to understand electrostatic boundary problems 11 00:00:29,733 --> 00:00:34,000 to derive by zones and that last equations 12 00:00:34,000 --> 00:00:36,666 to understand what is uniqueness here 13 00:00:36,800 --> 00:00:40,066 and thereafter to solve the case examples 14 00:00:41,966 --> 00:00:43,700 uh in previous chapters uh 15 00:00:43,700 --> 00:00:47,300 we have uh determined this uh electric uh field 16 00:00:47,300 --> 00:00:50,833 yeah either using this uh columns law or causes law 17 00:00:50,900 --> 00:00:51,300 uh 18 00:00:51,300 --> 00:00:56,300 when the charge distribution is known are this uh uh 19 00:00:56,300 --> 00:00:58,466 this relationship uh 20 00:00:58,466 --> 00:00:58,933 that is 21 00:00:58,933 --> 00:01:01,533 the relationship between electric field and the uh 22 00:01:01,533 --> 00:01:02,600 electric potential 23 00:01:02,600 --> 00:01:06,266 once this electric potential is not in more common 24 00:01:06,266 --> 00:01:08,700 most practical applications 25 00:01:08,900 --> 00:01:13,666 neither of these charge distribution nor the potential 26 00:01:13,666 --> 00:01:16,066 electorponentialist distribution is known 27 00:01:16,066 --> 00:01:17,566 so what is the realt 28 00:01:18,166 --> 00:01:21,566 so that is the the realt is given in this chapter 29 00:01:21,866 --> 00:01:22,700 uh uh 30 00:01:22,700 --> 00:01:23,766 that's a practical 31 00:01:23,766 --> 00:01:26,500 electrostatic problems are considered 32 00:01:26,500 --> 00:01:30,000 when only electrostatic conditions 33 00:01:30,733 --> 00:01:34,033 lactostatic conditions at some boundaries 34 00:01:34,300 --> 00:01:36,866 electrostatic conditions at some boundaries 35 00:01:36,866 --> 00:01:40,100 so these conditions can be charges or the potentials 36 00:01:40,100 --> 00:01:43,600 or the electric fields at some boundaries are given 37 00:01:43,700 --> 00:01:46,300 and from there when we can find out the electric field 38 00:01:46,300 --> 00:01:48,133 and the potentials um 39 00:01:48,133 --> 00:01:52,233 in other mediums or in other uh in other locations 40 00:01:52,966 --> 00:01:54,600 and such problems uh 41 00:01:54,600 --> 00:01:56,733 as well is using this uh 42 00:01:56,733 --> 00:02:00,066 poisons or lap class equations are the metro 43 00:02:00,066 --> 00:02:02,066 images that we will discuss 44 00:02:02,066 --> 00:02:04,400 in our next coming lectures 45 00:02:06,333 --> 00:02:06,800 very important 46 00:02:06,800 --> 00:02:09,933 indications are from your electrostatic point of view 47 00:02:09,933 --> 00:02:12,866 that is prysons and laplassecutions 48 00:02:12,866 --> 00:02:14,733 and let's see how to drive them 49 00:02:14,733 --> 00:02:17,266 so they can be easily drive from the causes law 50 00:02:17,266 --> 00:02:18,666 so what was the causes law 51 00:02:18,666 --> 00:02:22,966 it was a delta de v quatu robe the volume chart density 52 00:02:23,800 --> 00:02:24,000 and 53 00:02:24,000 --> 00:02:27,300 if we use the constitutive relationship between this d 54 00:02:27,300 --> 00:02:29,266 and the electric field and also 55 00:02:29,266 --> 00:02:31,766 the relationship between this electric field 56 00:02:31,766 --> 00:02:35,133 and elective potential and plug in this relationship 57 00:02:35,133 --> 00:02:38,533 this relationship between this discretion 58 00:02:38,533 --> 00:02:40,200 then this is your uh 59 00:02:40,200 --> 00:02:43,133 relationship for in homogeneous medium 60 00:02:43,133 --> 00:02:44,800 so why in homogeneous 61 00:02:44,800 --> 00:02:48,500 uh because it is not constant over anti medium 62 00:02:48,566 --> 00:02:51,866 that's why we cannot take this outside 63 00:02:52,100 --> 00:02:54,400 we cannot take this outside 64 00:02:54,400 --> 00:02:56,966 and treat this as a constant 65 00:02:56,966 --> 00:02:59,566 we cannot do this thing however 66 00:02:59,866 --> 00:03:03,000 if uh the medium is homogeneous then we we 67 00:03:03,000 --> 00:03:05,500 we can drive a poison equation 68 00:03:05,500 --> 00:03:08,266 this means that the ABS non electric permitivity 69 00:03:08,266 --> 00:03:09,600 it is the cost trying to cost 70 00:03:09,600 --> 00:03:12,900 the medium in which we need to find out the potential 71 00:03:13,100 --> 00:03:14,366 and the electric fee 72 00:03:14,566 --> 00:03:19,800 so and alternate this in particular case 73 00:03:19,800 --> 00:03:22,300 you can in this particular case but you can do that 74 00:03:22,300 --> 00:03:24,500 no you can treat this as a question it 75 00:03:24,500 --> 00:03:27,200 and it can come outside of the integration 76 00:03:27,200 --> 00:03:28,866 and once you solve this 77 00:03:29,200 --> 00:03:31,833 take this on the right hand side and 78 00:03:33,800 --> 00:03:35,966 you treat this Dal dot d 79 00:03:35,966 --> 00:03:40,900 you know that it is equal to your Dal square okay 80 00:03:41,200 --> 00:03:42,166 Dal square 81 00:03:43,566 --> 00:03:46,600 that is your lapation operator as well 82 00:03:48,300 --> 00:03:51,133 okay so in case there's 83 00:03:51,133 --> 00:03:53,533 there's no charges to present in the medium 84 00:03:53,533 --> 00:03:57,200 this means that the it is free of the charge region 85 00:03:57,200 --> 00:04:00,800 then it can simply convert to a homogeneous yes 86 00:04:00,800 --> 00:04:01,133 see that 87 00:04:01,133 --> 00:04:04,166 it is homogeneous second order differential equation 88 00:04:04,166 --> 00:04:06,833 it is non homogeneous okay 89 00:04:07,000 --> 00:04:09,366 because this right hand side is not equal to zero 90 00:04:09,533 --> 00:04:12,500 this is homogeneous secondary differential equations 91 00:04:13,900 --> 00:04:16,300 and this aggression is known as the leplassequision 92 00:04:16,300 --> 00:04:19,166 for a charge free region 93 00:04:20,166 --> 00:04:21,000 so the class 94 00:04:21,000 --> 00:04:23,866 and thousands equations are useful in solving 95 00:04:23,866 --> 00:04:26,000 electrostatic field problems 96 00:04:26,000 --> 00:04:27,866 and various other problems as well 97 00:04:27,866 --> 00:04:30,400 and we what are those problems as well 98 00:04:30,466 --> 00:04:33,500 and where these equations can be utilized 99 00:04:33,800 --> 00:04:37,966 this is only by simply replacing the variables 100 00:04:37,966 --> 00:04:39,900 in these two standard equations 101 00:04:39,900 --> 00:04:43,233 so if we treat this v as a magnet 102 00:04:43,466 --> 00:04:46,466 this electric potential as a magnetic potential 103 00:04:46,466 --> 00:04:47,633 then we can 104 00:04:47,933 --> 00:04:51,300 we can utilize the same equation in v magnetic statics 105 00:04:51,300 --> 00:04:53,366 we'll see in that coming textures as well 106 00:04:53,366 --> 00:04:54,733 and also this so we 107 00:04:54,733 --> 00:04:55,066 if it 108 00:04:55,066 --> 00:04:58,466 if it is treated as treated as a temperature variable 109 00:04:58,466 --> 00:05:01,733 then it can be utilizing the heat conduction problems 110 00:05:01,733 --> 00:05:04,100 and if it is treated as a stress function 111 00:05:04,100 --> 00:05:07,833 then it can be utilizing the fluid dynamics problem 112 00:05:08,600 --> 00:05:12,400 this food food uh food flow dynamics problem 113 00:05:14,266 --> 00:05:16,666 and uh you must remember from your chapter No. 2 114 00:05:16,666 --> 00:05:18,500 what are these uh uh 115 00:05:18,500 --> 00:05:19,700 second order uh 116 00:05:19,700 --> 00:05:22,366 differentiate equations or the the liplash 117 00:05:22,366 --> 00:05:25,266 uh liplashing operators in the second order 118 00:05:25,266 --> 00:05:29,100 uh form and uh you must uh 119 00:05:29,366 --> 00:05:31,433 use them and you must remember them 120 00:05:31,533 --> 00:05:33,566 according to their respective card 121 00:05:33,566 --> 00:05:36,300 uh cardination uh this uh coordinate system 122 00:05:38,166 --> 00:05:40,700 and similarly for deposits equation 123 00:05:40,700 --> 00:05:42,566 you can see that the right anxiety 124 00:05:42,566 --> 00:05:45,200 it is not equal to 0 so 125 00:05:45,200 --> 00:05:49,266 this means it is in homogeneous second order equation 126 00:05:49,266 --> 00:05:51,566 in homogeneous 127 00:05:53,133 --> 00:05:55,666 second order 128 00:05:57,066 --> 00:05:58,166 differential 129 00:05:59,800 --> 00:06:04,566 equation right RPD partial differential equation 130 00:06:04,600 --> 00:06:07,900 second order but it is homogeneous 131 00:06:10,366 --> 00:06:11,400 second order 132 00:06:13,966 --> 00:06:17,700 Piti right so you must remember that 133 00:06:17,700 --> 00:06:19,366 what equation you need to involve 134 00:06:19,366 --> 00:06:22,066 according to its cardigan or cardinal 135 00:06:22,066 --> 00:06:24,100 is the cardinal system either Cartigian 136 00:06:24,100 --> 00:06:27,300 cynticol or strical system and heavy 137 00:06:27,300 --> 00:06:30,133 it is known as the potential according to its 138 00:06:30,133 --> 00:06:32,966 uh the coordinate system 139 00:06:33,900 --> 00:06:35,800 remember these multipliers 140 00:06:35,800 --> 00:06:38,800 always remember these uh cautions 141 00:06:38,800 --> 00:06:42,300 these multipliers these cautions okay 142 00:06:42,300 --> 00:06:44,633 these scope multipliers the cautions 143 00:06:44,733 --> 00:06:46,500 either they are outside of the uh 144 00:06:46,500 --> 00:06:49,366 this differential or they are inside the differential 145 00:06:51,600 --> 00:06:51,866 no no 146 00:06:51,866 --> 00:06:54,466 no be threatening what is unique testing them 147 00:06:54,533 --> 00:06:59,000 there are several matters of solving a given problem 148 00:06:59,000 --> 00:07:02,566 so we know that either it is an analytical problem 149 00:07:02,566 --> 00:07:05,400 or it can be solved using some graphical approach 150 00:07:05,400 --> 00:07:07,700 or it can be solved using some grammarical method 151 00:07:07,700 --> 00:07:11,633 or it can be solved uh to your experimental uh studies 152 00:07:12,533 --> 00:07:16,100 so solving Le class equation in different ways 153 00:07:16,533 --> 00:07:18,200 will give you different answers 154 00:07:18,200 --> 00:07:23,233 but the question is if a solution to Le class equation 155 00:07:23,300 --> 00:07:24,833 two different ways 156 00:07:24,933 --> 00:07:28,366 satisfy a given set of boundary condition 157 00:07:28,366 --> 00:07:30,400 a given set of boundary condition 158 00:07:30,766 --> 00:07:32,666 is this the only possible solution 159 00:07:33,100 --> 00:07:34,466 either we solve this equation 160 00:07:34,466 --> 00:07:35,866 for this given boundary condition 161 00:07:35,866 --> 00:07:38,333 to this 3 form matters so 162 00:07:38,333 --> 00:07:41,366 the solution that we are going to find out using this 163 00:07:41,400 --> 00:07:42,200 for matter 164 00:07:42,266 --> 00:07:46,166 is it going to be the same for all this matters or not 165 00:07:46,500 --> 00:07:47,700 yes it is seen 166 00:07:47,700 --> 00:07:51,333 because it is going to satisfy a given set of 167 00:07:51,333 --> 00:07:52,700 boundary conditions so 168 00:07:52,700 --> 00:07:56,066 either resolve that Lepause equation using analytical 169 00:07:56,066 --> 00:07:58,866 medical graphical or experimental way 170 00:07:59,400 --> 00:08:00,466 but if if if 171 00:08:00,466 --> 00:08:02,733 the solution is going to satisfy 172 00:08:02,733 --> 00:08:05,166 with given set of boundary condition 173 00:08:05,166 --> 00:08:06,800 then the solution will be the same 174 00:08:06,800 --> 00:08:09,466 for all these four different matters 175 00:08:09,466 --> 00:08:12,600 and that's why we say that there's only one solution 176 00:08:12,600 --> 00:08:14,766 and that solution is unique 177 00:08:14,766 --> 00:08:16,966 if it is going to satisfy a given set of 178 00:08:16,966 --> 00:08:18,066 boundary conditions 179 00:08:19,366 --> 00:08:22,966 so unique mesteinum uh states that any solution of uh 180 00:08:22,966 --> 00:08:25,700 leclase equation which satisfies uh 181 00:08:25,700 --> 00:08:30,066 steam boundary conditions must be the only solution 182 00:08:30,166 --> 00:08:32,866 regardless of the method used 183 00:08:33,300 --> 00:08:35,666 premium applies to any solution of poisons 184 00:08:35,666 --> 00:08:38,000 or lip gloss equation in a given region 185 00:08:38,000 --> 00:08:40,100 or closed surface 186 00:08:40,933 --> 00:08:44,366 so it is valid for both lip gloss and poison equation 187 00:08:45,733 --> 00:08:46,666 so pre acquisition 188 00:08:46,666 --> 00:08:49,566 for solving any of the body valley problems 189 00:08:49,800 --> 00:08:53,466 you must remember all these three uh pre acts 190 00:08:54,100 --> 00:08:55,600 these three steps as well 191 00:08:55,600 --> 00:08:58,166 before proceeding or solving these uh uh 192 00:08:58,166 --> 00:09:02,833 these problems so a rare solution is uniquely becomes 193 00:09:03,200 --> 00:09:05,800 so rare solution is uniquely 194 00:09:05,800 --> 00:09:07,466 describes a problem 195 00:09:07,466 --> 00:09:09,866 that we have only a single solution 196 00:09:09,866 --> 00:09:11,633 to these types of the problems 197 00:09:11,900 --> 00:09:14,766 so first of all appropriate differentiate equation 198 00:09:14,766 --> 00:09:16,566 so either it will be a class equation 199 00:09:16,566 --> 00:09:19,300 or it will be a presence equation 200 00:09:19,400 --> 00:09:21,966 a solution vision where you need to solve the question 201 00:09:22,266 --> 00:09:24,366 prescribe boundary conditions 202 00:09:24,366 --> 00:09:27,400 require bound boundary conditions or boundary problems 203 00:09:27,500 --> 00:09:30,666 a problem does not have a unique solution 204 00:09:30,766 --> 00:09:32,966 and cannot be solved completely 205 00:09:33,266 --> 00:09:37,233 if any of these three items is missing 206 00:09:37,466 --> 00:09:38,500 that either is nor 207 00:09:38,500 --> 00:09:41,200 it cannot be described in a differential equation form 208 00:09:41,200 --> 00:09:41,866 or either 209 00:09:41,866 --> 00:09:45,133 you don't have some particular boundary conditions 210 00:09:45,133 --> 00:09:48,966 then you will have uh many possible equation 211 00:09:48,966 --> 00:09:52,133 that in finitely many possible equation 212 00:09:52,133 --> 00:09:55,200 solutions that we see in our Mathematica terminology 213 00:09:55,300 --> 00:09:55,900 however 214 00:09:55,900 --> 00:09:59,266 if it is going to satisfy all these three period acts 215 00:09:59,266 --> 00:10:01,000 that that the problem 216 00:10:01,000 --> 00:10:02,666 it is going to have a particular 217 00:10:02,666 --> 00:10:03,866 differentiate equation 218 00:10:03,933 --> 00:10:06,166 it is being solved in a particular region 219 00:10:06,300 --> 00:10:10,166 given region and it is going to satisfy the particular 220 00:10:10,333 --> 00:10:13,866 bond conditions then you will have a unique solution 221 00:10:15,766 --> 00:10:18,366 so general procedure for swarming poisons of 222 00:10:18,366 --> 00:10:22,300 for our laplassecrations and these are that for laplas 223 00:10:22,300 --> 00:10:24,800 you know that the wall of chart density 224 00:10:24,800 --> 00:10:25,866 it is equal to 0 225 00:10:25,866 --> 00:10:31,200 for the poisons it is not equal to 0 and if uh 226 00:10:31,200 --> 00:10:33,733 they say there's only one variable involved 227 00:10:33,733 --> 00:10:34,066 then 228 00:10:34,066 --> 00:10:37,800 you can solve this using a direct integration method 229 00:10:37,800 --> 00:10:39,566 so direct integration when we 230 00:10:39,566 --> 00:10:41,566 is a functional only single variable 231 00:10:42,066 --> 00:10:44,200 and um if one of them are able 232 00:10:44,200 --> 00:10:46,266 and they can be seperable from each other then 233 00:10:46,266 --> 00:10:48,800 then then you can solve for this uh 234 00:10:48,800 --> 00:10:52,166 variable b using this separation of a even matter 235 00:10:52,166 --> 00:10:54,200 we will see these matters in the upcoming uh 236 00:10:54,466 --> 00:10:56,300 uh examples as well 237 00:10:56,466 --> 00:10:58,966 solution at this point is not unique 238 00:10:59,000 --> 00:11:03,733 but expressed in terms of unknown integration costumes 239 00:11:03,733 --> 00:11:03,933 so 240 00:11:03,933 --> 00:11:07,133 this means that as far as there are multiple costumes 241 00:11:07,133 --> 00:11:10,000 and solutions are going to be multiple in finite 242 00:11:10,000 --> 00:11:13,566 many solutions so how to find out a unique solution 243 00:11:13,900 --> 00:11:16,633 you need to involve the boundary conditions 244 00:11:16,700 --> 00:11:17,966 apply the boundary conditions 245 00:11:17,966 --> 00:11:19,733 to determine a unique solution 246 00:11:19,733 --> 00:11:22,000 for V that is your electric connection 247 00:11:22,000 --> 00:11:24,533 in this electrostatic case 248 00:11:24,533 --> 00:11:27,800 boundary problems imposingly given boundary conditions 249 00:11:27,800 --> 00:11:29,866 makes the solution unique 250 00:11:30,700 --> 00:11:32,700 and having obtained the potential 251 00:11:32,700 --> 00:11:34,200 you can find out the electricity 252 00:11:34,200 --> 00:11:35,633 by using this relationship 253 00:11:35,866 --> 00:11:37,800 having found out this electrical 254 00:11:37,800 --> 00:11:39,833 you can find out the electrical business 255 00:11:40,000 --> 00:11:41,666 using this relationship 256 00:11:42,466 --> 00:11:46,366 and if desired you can find out this charge as well 257 00:11:46,366 --> 00:11:49,500 inducing on a conductor using this relationship 258 00:11:49,500 --> 00:11:50,700 that charges the 259 00:11:50,900 --> 00:11:55,266 uh is the surface integral of the uh circuit identity 260 00:11:55,266 --> 00:11:57,466 where uh for this conductors 261 00:11:57,466 --> 00:11:58,600 we know that uh 262 00:11:58,800 --> 00:12:01,200 this rule as is equal to normal electrotroxancy 263 00:12:01,200 --> 00:12:02,233 from this uh 264 00:12:02,500 --> 00:12:04,533 boundary conditions that we discussed in check 265 00:12:04,533 --> 00:12:06,033 in our chapter number five 266 00:12:06,600 --> 00:12:09,933 which is the normal component of the electrical stansty 267 00:12:09,933 --> 00:12:12,066 it is equal to this sufficial dansty 268 00:12:12,266 --> 00:12:15,000 and there on if we integrate this ruas 269 00:12:15,000 --> 00:12:16,166 ruas is equal to bn 270 00:12:16,166 --> 00:12:18,733 d and can be found out using this relationship 271 00:12:18,733 --> 00:12:19,600 and if we were 272 00:12:19,600 --> 00:12:22,233 E can be found out this using this relationship 273 00:12:22,500 --> 00:12:24,866 and the V can be found out using this 274 00:12:25,566 --> 00:12:28,433 uh based on this boundary problem 275 00:12:28,900 --> 00:12:30,000 and again uh 276 00:12:30,000 --> 00:12:32,633 you can find out this capacity as well between those 277 00:12:32,866 --> 00:12:35,000 between these two conductors 278 00:12:35,066 --> 00:12:35,600 even if 279 00:12:35,600 --> 00:12:39,600 if you know that this charge coming out from here right 280 00:12:39,600 --> 00:12:42,300 and this potential coming out from this 281 00:12:42,300 --> 00:12:43,800 over boundary problem 282 00:12:44,066 --> 00:12:46,066 so you can also proceed in this way as well 283 00:12:49,400 --> 00:12:51,766 the next show were some of the examples 284 00:12:52,066 --> 00:12:55,966 to further clarify the concept of body valley problems 285 00:12:55,966 --> 00:12:59,800 that how we can enroll these poisons and labour 286 00:12:59,800 --> 00:13:03,533 secretions to find out the electric potentials 287 00:13:03,533 --> 00:13:04,800 and electric fields 288 00:13:05,000 --> 00:13:07,566 so here very important example in your 289 00:13:07,566 --> 00:13:09,100 this is your practical example 290 00:13:09,300 --> 00:13:09,733 the current 291 00:13:09,733 --> 00:13:13,500 carrying components in high voltage power equipment 292 00:13:13,500 --> 00:13:14,766 must be cooled 293 00:13:15,300 --> 00:13:17,666 so cooling phenomena is to be exist 294 00:13:18,266 --> 00:13:21,700 uh to clean because the heat is uh you know uh 295 00:13:21,700 --> 00:13:23,700 it is eminating from these uh 296 00:13:23,700 --> 00:13:24,900 high voltage components 297 00:13:24,900 --> 00:13:26,866 and this company must be cooled down 298 00:13:26,933 --> 00:13:28,400 so that they are not uh 299 00:13:28,400 --> 00:13:30,933 uh hampered or there's no uh 300 00:13:30,933 --> 00:13:35,533 you know fire hazards or there's no damages to the uh 301 00:13:35,533 --> 00:13:36,866 overheated components 302 00:13:36,866 --> 00:13:39,900 so you must uh uh have the mechanism 303 00:13:39,900 --> 00:13:42,633 mechanism to cool down these electric components 304 00:13:43,166 --> 00:13:45,266 uh because uh due to the again 305 00:13:45,266 --> 00:13:46,700 uh because they are crank and compromise 306 00:13:46,700 --> 00:13:48,400 so due to the only clouds 307 00:13:48,400 --> 00:13:51,800 uh they are going to eminate or dissipate the energy 308 00:13:51,866 --> 00:13:54,266 uh or dissipate the heat in their surroundings 309 00:13:54,266 --> 00:13:56,066 and the surrounding has to be cooled down 310 00:13:56,266 --> 00:13:58,566 so there must be some uh phenomena 311 00:13:58,933 --> 00:14:02,466 and to carry out of any heat caused by the army classes 312 00:14:02,466 --> 00:14:03,266 okay 313 00:14:05,066 --> 00:14:07,200 physical to ice 314 00:14:11,166 --> 00:14:11,466 so 315 00:14:11,466 --> 00:14:15,366 a means of pumping is based on the force transmitted 316 00:14:15,366 --> 00:14:18,966 to the cooling fluids to the cooling fluids 317 00:14:19,000 --> 00:14:23,600 so a cooling fluids so inside this tube a fluid 318 00:14:24,166 --> 00:14:25,733 a cooling a coolant 319 00:14:25,733 --> 00:14:28,133 you can see as well a coolant or cooling fluid 320 00:14:28,133 --> 00:14:32,100 it is uh uh residing inside this um 321 00:14:33,100 --> 00:14:35,500 and inside the you know and 322 00:14:35,500 --> 00:14:38,533 and you can you can see that it is residing 323 00:14:38,533 --> 00:14:42,466 residing inside this are circulating as 324 00:14:42,466 --> 00:14:45,366 as well inside this coolant two 325 00:14:46,466 --> 00:14:48,866 okay so uh 326 00:14:49,266 --> 00:14:50,566 transmit to calling uh 327 00:14:50,566 --> 00:14:53,666 so a means of pumping you must must have a some 328 00:14:53,666 --> 00:14:56,833 some procedure of pumping of this uh 329 00:14:57,533 --> 00:14:58,633 this electric 330 00:14:59,000 --> 00:15:03,133 this coolant in inside this tube so that the 331 00:15:03,133 --> 00:15:06,500 it can cool down the sounding okay 332 00:15:06,500 --> 00:15:10,366 so electro so this phenomenon is known as the electro 333 00:15:10,500 --> 00:15:13,466 hydrodynamic pumping and it is more 334 00:15:13,466 --> 00:15:14,666 it is more than this uh 335 00:15:14,666 --> 00:15:18,266 figure and that the region between electro electrodes 336 00:15:18,266 --> 00:15:19,566 that is electrodes 337 00:15:19,566 --> 00:15:22,266 contains a uniform chart and steel or not 338 00:15:22,266 --> 00:15:25,100 which is generated by the left electrode and collector 339 00:15:25,100 --> 00:15:26,933 Gatby right electrode 340 00:15:26,933 --> 00:15:29,133 and it is going to pumping 341 00:15:29,133 --> 00:15:31,533 pumping this uh or you can uh 342 00:15:31,533 --> 00:15:33,466 you can see that it is going to 343 00:15:34,733 --> 00:15:41,466 circulate this uh uh this coolant in its uh in its uh 344 00:15:41,466 --> 00:15:44,266 ultra parameter or inside this so that uh 345 00:15:45,100 --> 00:15:49,633 the sounding is cooldown so a region between electrodes 346 00:15:50,100 --> 00:15:50,533 the region 347 00:15:50,533 --> 00:15:54,200 between these electrodes contains a uniform Cha chaunsy 348 00:15:54,200 --> 00:15:55,566 so here you have this unit 349 00:15:56,333 --> 00:15:58,166 this means it is going to work 350 00:15:58,166 --> 00:16:00,566 going to be a Byzance problem 351 00:16:00,566 --> 00:16:03,666 boundary problems and the left uh left 352 00:16:03,666 --> 00:16:05,600 left toad and then eclectic on the right 353 00:16:05,600 --> 00:16:06,966 left toad and accordingly 354 00:16:06,966 --> 00:16:09,800 the things are going to be sinus idles 355 00:16:09,800 --> 00:16:13,300 so that the coolant it can be uh it can be circulate 356 00:16:13,300 --> 00:16:18,533 uh in this uh complete uh tube and what else 357 00:16:18,533 --> 00:16:19,466 so the left side 358 00:16:19,466 --> 00:16:21,766 we can assume that the potential at the left side 359 00:16:21,766 --> 00:16:22,833 it is be not 360 00:16:23,000 --> 00:16:25,200 and on the right side it is going to collect it down 361 00:16:25,200 --> 00:16:26,533 so the potential on the right side 362 00:16:26,533 --> 00:16:29,266 it is assumed as the ground or zero vote 363 00:16:29,666 --> 00:16:31,466 so on both boundaries we know that 364 00:16:31,466 --> 00:16:33,766 we know the potentials inside 365 00:16:33,766 --> 00:16:37,866 we know that the there's a chart density inside this 366 00:16:38,533 --> 00:16:40,166 and it is constant in this case 367 00:16:42,200 --> 00:16:43,766 so unlike the traditional 368 00:16:43,766 --> 00:16:45,566 Thermal cooling technologies 369 00:16:45,566 --> 00:16:47,766 that rely on mechanical pumps 370 00:16:47,933 --> 00:16:48,466 ESD cooling 371 00:16:48,466 --> 00:16:52,600 uses electric fields to pump the coolant to the tube 372 00:16:52,600 --> 00:16:53,766 so that is the electric field 373 00:16:53,766 --> 00:16:55,566 which is generated by this 374 00:16:56,200 --> 00:16:58,433 applying of this external voltages 375 00:16:59,500 --> 00:17:00,366 electro supply 376 00:17:00,366 --> 00:17:02,900 the voltage that pushes the coolant to the tube 377 00:17:03,100 --> 00:17:05,166 ESTB systems what is the advantage 378 00:17:05,166 --> 00:17:08,266 they are lightweight they consume a little bar 379 00:17:08,500 --> 00:17:10,200 because they are involved with this 380 00:17:10,400 --> 00:17:11,766 electrostatic phenomena 381 00:17:13,866 --> 00:17:14,600 now 382 00:17:14,600 --> 00:17:17,633 let's solve this example to find out this differential 383 00:17:19,100 --> 00:17:21,466 uh since uh Robbie is not equal to 0 384 00:17:21,466 --> 00:17:24,466 so we can treat this problem as a present equation 385 00:17:25,366 --> 00:17:27,800 and this is a positive question okay 386 00:17:27,933 --> 00:17:31,200 and this is a Slender Girl farm right this is a slender 387 00:17:31,566 --> 00:17:34,933 so we can involve this slendrical coordinates 388 00:17:34,933 --> 00:17:38,266 systems form of this presents equation okay 389 00:17:38,266 --> 00:17:41,166 to body conditions we are given at z equal to zero 390 00:17:41,366 --> 00:17:43,166 at the start of this tube 391 00:17:43,166 --> 00:17:45,133 at these equal to 0 the potential is v 392 00:17:45,133 --> 00:17:47,466 not at the end of this tube 393 00:17:47,466 --> 00:17:49,866 at these equal t the potential is 0 where the 394 00:17:50,300 --> 00:17:52,466 the coolant is going to collect it 395 00:17:52,466 --> 00:17:54,400 and then the cycle will reverse 396 00:17:54,400 --> 00:17:57,066 and the coolant will circulate 397 00:17:57,066 --> 00:17:58,100 in the opposite direction 398 00:17:58,100 --> 00:17:59,566 and the phenomenon will repeat 399 00:18:00,733 --> 00:18:02,600 and uh show that uh 400 00:18:02,600 --> 00:18:05,333 the V depends only on C okay 401 00:18:05,333 --> 00:18:06,000 there's a Z 402 00:18:06,000 --> 00:18:09,266 only depended upon the distance from the electrodes 403 00:18:10,566 --> 00:18:13,533 so you plug in uh these two values 404 00:18:13,533 --> 00:18:15,666 the only variable the only variable 405 00:18:15,666 --> 00:18:18,566 on which these changes in the potential 406 00:18:18,566 --> 00:18:22,600 they are depending is only in the z direction 407 00:18:22,600 --> 00:18:24,333 right in the z direction 408 00:18:24,333 --> 00:18:27,666 because these changes are only with respect to this Z 409 00:18:27,933 --> 00:18:29,500 so this is going to be zero 410 00:18:30,666 --> 00:18:32,366 the changes are for potential 411 00:18:32,366 --> 00:18:35,466 with respect to this fi company is going to be zero 412 00:18:35,566 --> 00:18:37,600 the changes with respect to this raw company 413 00:18:37,600 --> 00:18:38,800 is going to be zero 414 00:18:38,866 --> 00:18:41,466 and the only changes that we have in this uh 415 00:18:41,500 --> 00:18:44,766 vertical phenomena that is with respect to just Z 416 00:18:44,933 --> 00:18:47,600 because these changes are with respect to Z coordinate 417 00:18:48,900 --> 00:18:51,166 so this is the second order differential equation 418 00:18:51,166 --> 00:18:52,433 that you need to solve 419 00:18:54,100 --> 00:18:54,900 okay 420 00:18:55,533 --> 00:18:56,700 so uh 421 00:18:56,700 --> 00:18:59,600 it only involves one variable that is your potential 422 00:18:59,700 --> 00:19:02,166 so we integrated once right 423 00:19:02,333 --> 00:19:04,666 uh this is the integration constant 424 00:19:04,900 --> 00:19:07,433 the right hand side it is only having one constant 425 00:19:07,866 --> 00:19:08,466 right 426 00:19:08,466 --> 00:19:13,266 so integration of constant with respect to Z is Z right 427 00:19:13,333 --> 00:19:16,366 so integrated twice to find out this v 428 00:19:16,700 --> 00:19:17,600 you can see that 429 00:19:17,600 --> 00:19:19,700 first integration eliminates the first 430 00:19:19,700 --> 00:19:23,200 uh the second elevator of this differential 431 00:19:23,366 --> 00:19:25,133 the second integration will eliminate 432 00:19:25,133 --> 00:19:26,366 the first elevator of uh 433 00:19:26,366 --> 00:19:31,133 this differential so this way if you twice 434 00:19:31,133 --> 00:19:33,666 do twice integrate this equation 435 00:19:33,900 --> 00:19:36,133 you are going to eliminate this 436 00:19:36,133 --> 00:19:37,700 second order differential 437 00:19:37,966 --> 00:19:41,566 so that it is reverse phenomena right 438 00:19:41,600 --> 00:19:46,400 so integrate it twice to eliminate this 439 00:19:48,366 --> 00:19:51,466 second order differential equation 440 00:19:51,466 --> 00:19:54,000 so this will cancel out okay 441 00:19:54,166 --> 00:19:55,366 and right hand size you need 442 00:19:55,366 --> 00:19:57,466 you need to solve the equations accordingly 443 00:19:57,533 --> 00:19:58,933 since it is uh us 444 00:19:58,933 --> 00:20:00,900 uh second uh tricing integration 445 00:20:00,900 --> 00:20:03,033 so you are going to have two 446 00:20:03,266 --> 00:20:06,766 two constants here in the second um 447 00:20:07,000 --> 00:20:09,833 integration since there's no one 448 00:20:10,100 --> 00:20:11,133 one along with this 449 00:20:11,133 --> 00:20:13,666 so it's integration is going to be Z 450 00:20:14,866 --> 00:20:16,466 for this first integration 451 00:20:16,466 --> 00:20:18,900 we have this Z here so 452 00:20:18,900 --> 00:20:23,066 each integration is going to be Z squared divided by 2 453 00:20:23,266 --> 00:20:25,900 okay and this is Rona divided by absmont 454 00:20:25,900 --> 00:20:29,300 it is going to be constant okay 455 00:20:29,300 --> 00:20:31,766 so we need to find out this a and B integration 456 00:20:31,766 --> 00:20:34,966 questions so let's apply the boundary conditions 457 00:20:35,000 --> 00:20:36,866 at these equal to zero we have the potential 458 00:20:36,866 --> 00:20:40,700 these will be not so put in this equation okay 459 00:20:41,533 --> 00:20:43,400 so first time is going to be 0 460 00:20:43,400 --> 00:20:44,866 second time is going to be 0 461 00:20:44,866 --> 00:20:47,933 and we are left on living B for your B constant 462 00:20:47,933 --> 00:20:51,700 it is equal to B not now plug in this um 463 00:20:53,400 --> 00:20:55,366 be not into this okay 464 00:20:55,866 --> 00:20:57,666 and I'll put in this second differential 465 00:20:57,666 --> 00:21:00,700 second boundary condition that your is equal to 0 466 00:21:00,700 --> 00:21:02,366 at this d is equal to 467 00:21:03,000 --> 00:21:05,500 as is equal to 3 that is your this end 468 00:21:06,133 --> 00:21:08,833 this end and the first case does this end 469 00:21:10,066 --> 00:21:13,266 now her plug in this says Z is equal to d 470 00:21:13,533 --> 00:21:15,966 Z is equal to d and v is equal to 0 471 00:21:16,333 --> 00:21:18,200 and solve the equation to find out this 472 00:21:18,200 --> 00:21:21,900 second question that is your is equal to this one 473 00:21:23,066 --> 00:21:26,066 and now you plug in this a and B into this 474 00:21:26,066 --> 00:21:29,033 uh into this uh 475 00:21:30,100 --> 00:21:32,166 standard equation of this uh 476 00:21:33,000 --> 00:21:35,266 what age in terms of the question 477 00:21:35,266 --> 00:21:40,233 then you can find out this that plug in this e 478 00:21:41,766 --> 00:21:42,366 I'm being 479 00:21:42,366 --> 00:21:45,666 today's to find out the final equation of this 480 00:21:46,300 --> 00:21:49,666 uh voltage for this particular boundary problem 481 00:21:50,300 --> 00:21:51,966 so what is that uh 482 00:21:51,966 --> 00:21:53,766 final equation 483 00:21:54,533 --> 00:21:57,266 that you can simply replace this thing with this 484 00:21:57,266 --> 00:21:59,266 and this thing with this okay 485 00:22:00,500 --> 00:22:02,666 and the next thing you need to find out is 486 00:22:02,966 --> 00:22:03,866 the electrical 487 00:22:03,866 --> 00:22:07,833 so that is the minus gradient of differential 488 00:22:08,333 --> 00:22:10,566 and since it is clinical partner system 489 00:22:10,866 --> 00:22:13,600 and the dependencies are with respect to Z only 490 00:22:13,600 --> 00:22:16,200 so these two are going to be the zero and you 491 00:22:16,200 --> 00:22:17,666 need to find out this thing 492 00:22:17,866 --> 00:22:18,500 and this thing 493 00:22:18,500 --> 00:22:20,666 you have already found out in your previous slide 494 00:22:20,666 --> 00:22:23,666 as well or you can again integrate 495 00:22:23,966 --> 00:22:25,300 you can again difference 496 00:22:25,300 --> 00:22:28,466 differentiate this equation with respect to Z 497 00:22:28,600 --> 00:22:29,800 to find out this thing 498 00:22:29,800 --> 00:22:31,366 and if you differentiate this with this 499 00:22:31,366 --> 00:22:34,666 with respect to Z then this is going to be zero 500 00:22:34,866 --> 00:22:38,900 this is going to be one and this is going to be 2 z 501 00:22:38,900 --> 00:22:39,700 right 502 00:22:40,466 --> 00:22:41,300 2 2 503 00:22:41,366 --> 00:22:42,900 2 2 is going to be cancel out 504 00:22:42,900 --> 00:22:45,666 and you are going to laptop with Ronald Z 505 00:22:45,800 --> 00:22:47,100 and you have a option on 506 00:22:47,100 --> 00:22:48,466 and since there is a negative over here 507 00:22:48,466 --> 00:22:49,766 so it is going to be positive 508 00:22:49,766 --> 00:22:51,200 and it is going to be negative 509 00:22:52,000 --> 00:22:54,666 and I plug in this constant E over here 510 00:22:54,666 --> 00:22:56,966 so this is your electrical equation 511 00:22:58,366 --> 00:23:01,200 okay now we need to find out in that force 512 00:23:01,800 --> 00:23:05,700 F force we know that it is equal to q e and what is q 513 00:23:05,700 --> 00:23:08,866 it is d volume and trigger larger row V 514 00:23:09,166 --> 00:23:12,500 and what is row V row V is constant in this case right 515 00:23:13,500 --> 00:23:16,800 okay so it would come outside right 516 00:23:16,966 --> 00:23:18,466 and the volume and trigger 517 00:23:18,466 --> 00:23:22,066 we can divide this into the surface integral 518 00:23:22,066 --> 00:23:25,000 and into the third varying dimension 519 00:23:25,133 --> 00:23:26,833 that is your d dimension 520 00:23:27,533 --> 00:23:30,000 and this uh surface integral 521 00:23:30,000 --> 00:23:31,933 we can simply treat this at the 522 00:23:31,933 --> 00:23:38,000 treated as the surface area of this conductor right 523 00:23:39,200 --> 00:23:42,566 okay and we are now left with only one integration 524 00:23:42,566 --> 00:23:45,666 that is with the spectrum z divering 525 00:23:45,800 --> 00:23:47,633 divering waver 526 00:23:48,500 --> 00:23:50,866 and e e 527 00:23:50,866 --> 00:23:51,466 e e 528 00:23:51,466 --> 00:23:53,566 you can involve from here right 529 00:23:54,400 --> 00:23:57,600 so the first term integration with the start of Z is 530 00:23:57,600 --> 00:23:58,966 since it is on the constant 531 00:23:58,966 --> 00:24:00,400 so it is going to be Z 532 00:24:00,766 --> 00:24:03,700 second time it is going to be Z square divided by two 533 00:24:04,133 --> 00:24:05,166 return time time 534 00:24:05,166 --> 00:24:06,400 time it is the constant 535 00:24:06,400 --> 00:24:09,900 so it is going to be Z so that is your force 536 00:24:10,300 --> 00:24:13,933 you further simplify this for this integration 537 00:24:13,933 --> 00:24:15,000 from 0 to D 538 00:24:15,533 --> 00:24:18,766 and you will see that for the zero 539 00:24:18,966 --> 00:24:23,600 all terms are going to be 0 for the Z equal to d 540 00:24:23,900 --> 00:24:25,866 so this is going to be d square 541 00:24:25,900 --> 00:24:27,700 this is going to be d square 542 00:24:27,800 --> 00:24:30,066 these two are going to be cancelled out 543 00:24:30,100 --> 00:24:32,633 and you are left out with this first term 544 00:24:32,800 --> 00:24:33,600 okay 545 00:24:34,300 --> 00:24:34,700 and then 546 00:24:34,700 --> 00:24:36,966 this day is going to be cancelled out over here 547 00:24:37,000 --> 00:24:39,533 and this is the only thing that you are left out 548 00:24:39,533 --> 00:24:41,866 after the intergation no 549 00:24:41,866 --> 00:24:44,266 we need to find a deforsed pressure 550 00:24:44,266 --> 00:24:46,066 that is your force by Unitarian 551 00:24:46,100 --> 00:24:49,000 so activate the s as is going to be cancelled out 552 00:24:49,166 --> 00:24:51,466 and now you plug in your zonaut 553 00:24:51,466 --> 00:24:53,366 the given or not and v not 554 00:24:53,400 --> 00:24:56,633 so this is your final pressure 555 00:24:57,766 --> 00:25:00,800 which is applied by the external potential 556 00:25:00,800 --> 00:25:02,800 or accordingly the electric feel 557 00:25:02,800 --> 00:25:05,833 the applied electric feel within this tube 558 00:25:05,900 --> 00:25:07,200 so that is the pressure 559 00:25:07,200 --> 00:25:11,633 which is applied on the coolant for pumping in 560 00:25:12,000 --> 00:25:13,366 into this tube 561 00:25:16,533 --> 00:25:17,966 let's follow this 562 00:25:17,966 --> 00:25:22,566 another example to find out this what is an electrical 563 00:25:22,566 --> 00:25:27,666 so what is this semi infinite conducting pain at 564 00:25:28,966 --> 00:25:31,766 5 equal to 0 and 5 0 5 by 2 565 00:25:31,933 --> 00:25:35,966 so I start driving geometric semi infinite plane 566 00:25:35,966 --> 00:25:38,200 connecting plane at 5 0 0 567 00:25:38,300 --> 00:25:41,366 so this is your x y z X is uh plane 568 00:25:41,366 --> 00:25:43,000 so add X is equal to of 5 569 00:25:43,000 --> 00:25:46,700 we're not at 5 equal to 0 at this point 570 00:25:47,000 --> 00:25:50,200 so it is semi infinite plane like this one semi y 571 00:25:50,200 --> 00:25:52,200 it is a half plane 572 00:25:52,766 --> 00:25:55,766 but it is infinite in this extra action 573 00:25:56,266 --> 00:25:58,700 second manager of 5 is equal to 574 00:25:59,300 --> 00:26:02,266 second plane is at 5 is equal to 5 by 6 575 00:26:02,266 --> 00:26:03,266 that is your 576 00:26:03,300 --> 00:26:07,866 I guess 30 degree angle by by 6 wanted to divide yeah 577 00:26:07,866 --> 00:26:13,900 it is 30 degree angle 30 degree angle okay by by six 578 00:26:15,766 --> 00:26:18,400 okay so that is another plane 579 00:26:18,966 --> 00:26:21,433 second infinite plane in this direction 580 00:26:21,900 --> 00:26:23,633 an infinite in this direction 581 00:26:24,000 --> 00:26:28,166 the boundary conditions that we are given is that at 5 582 00:26:28,166 --> 00:26:32,566 5 0 we have the voltage zero right 583 00:26:32,566 --> 00:26:35,233 so this means it is a ground this thing 584 00:26:37,800 --> 00:26:38,566 okay the 585 00:26:38,566 --> 00:26:41,933 and the second bomb definition that at 586 00:26:41,933 --> 00:26:45,233 at second infinite plane it is visical to v not 587 00:26:46,200 --> 00:26:49,166 and in between these two in between these two plane 588 00:26:49,166 --> 00:26:50,000 it is 589 00:26:51,900 --> 00:26:54,300 an insulating gap here 590 00:26:54,300 --> 00:26:58,066 you can can see that the gap is over here right 591 00:26:58,200 --> 00:27:00,400 so this is the gap between these two planes 592 00:27:01,900 --> 00:27:03,400 so that they are not shorted 593 00:27:04,100 --> 00:27:05,466 and we need to find out the 594 00:27:06,533 --> 00:27:08,466 potential and electrical 595 00:27:08,966 --> 00:27:13,466 in the region between these players that is your 596 00:27:15,266 --> 00:27:19,033 this region you can see okay between this region 597 00:27:20,700 --> 00:27:21,600 so 598 00:27:24,466 --> 00:27:25,366 let's go 599 00:27:26,100 --> 00:27:28,100 what is this particular case 600 00:27:29,366 --> 00:27:31,066 uh in this case 601 00:27:31,066 --> 00:27:35,200 uh you can see that you're not provided with the 602 00:27:35,933 --> 00:27:38,500 any of the information of uh density 603 00:27:38,533 --> 00:27:42,800 so this means that row v is equal to 0 right 604 00:27:43,100 --> 00:27:45,966 so it will be equal to zero that it is going to be a 605 00:27:46,300 --> 00:27:48,233 the class equation problem 606 00:27:48,666 --> 00:27:51,000 and since the things are given in a row 5 z 607 00:27:51,000 --> 00:27:54,000 so it is a splendidal partner system problem 608 00:27:54,466 --> 00:27:57,366 so variations are being specified 609 00:27:57,700 --> 00:28:00,766 so rest of the variations are zero 610 00:28:01,500 --> 00:28:02,566 because we have seen that 611 00:28:02,566 --> 00:28:04,666 the boundary conditions are given with respect to 612 00:28:04,766 --> 00:28:06,166 by corner system 613 00:28:07,066 --> 00:28:10,000 so we'll only need to solve this 614 00:28:11,000 --> 00:28:12,833 homogeneous differential equation 615 00:28:12,966 --> 00:28:16,566 and you can sense at sense rose equal 616 00:28:16,566 --> 00:28:19,800 zeros excluded due to the insulating gap right 617 00:28:19,800 --> 00:28:20,400 there's a gap 618 00:28:20,400 --> 00:28:23,900 so at rose equal to 0 and we don't have any potential 619 00:28:23,900 --> 00:28:27,500 or you can simply eliminate this factor as well 620 00:28:27,500 --> 00:28:30,433 because using this elementary mat 621 00:28:30,466 --> 00:28:32,700 that you can take it on the right hand side 622 00:28:32,700 --> 00:28:34,233 so it would be eliminated 623 00:28:34,500 --> 00:28:37,466 and you are only left out with this equation to solve 624 00:28:38,300 --> 00:28:38,933 so again 625 00:28:38,933 --> 00:28:41,600 you need to intigate it twice to get rid of this 626 00:28:41,600 --> 00:28:43,400 second order differentiation 627 00:28:43,533 --> 00:28:45,300 to solve this potential 628 00:28:46,000 --> 00:28:48,633 first integration is going to provide you this 629 00:28:50,300 --> 00:28:51,266 a only 630 00:28:51,533 --> 00:28:55,800 the second generation is going to provide this y and B 631 00:28:55,900 --> 00:28:57,866 okay so it's a second dimension 632 00:28:59,566 --> 00:29:00,700 applying boundary conditional 633 00:29:00,700 --> 00:29:03,566 to find out these integration constants 634 00:29:03,566 --> 00:29:06,300 so at 5 is equal to 0 b is equal to 0 635 00:29:06,300 --> 00:29:08,600 so this will give you b is equal to zero 636 00:29:09,400 --> 00:29:11,233 and plug in this over here 637 00:29:11,400 --> 00:29:13,266 and not integrate this second 638 00:29:13,266 --> 00:29:14,400 uh not 639 00:29:14,400 --> 00:29:16,000 plug in this second boundary condition 640 00:29:16,000 --> 00:29:17,866 that at 5 is equal to 5 not 641 00:29:17,866 --> 00:29:21,366 which is equal to v not so v not is equal to this one 642 00:29:21,366 --> 00:29:25,066 and this is your is equal to v not to verify not 643 00:29:26,166 --> 00:29:28,866 now plug in these two constants into the this one 644 00:29:28,866 --> 00:29:31,266 b is equal 0 is equal to this one 645 00:29:31,600 --> 00:29:36,900 and this is your final form of uh potential in the uh 646 00:29:36,900 --> 00:29:39,300 in terms of this varying uh variable 647 00:29:39,300 --> 00:29:40,566 that is your fine 648 00:29:41,700 --> 00:29:44,066 okay now when we need to find out the electrical field 649 00:29:44,066 --> 00:29:46,800 second part that is minus gradient of b 650 00:29:46,966 --> 00:29:50,600 so there are two ways there are two ways either uh 651 00:29:50,600 --> 00:29:51,533 you in uh 652 00:29:51,533 --> 00:29:54,500 you differentiate this equation with the spectify 653 00:29:54,733 --> 00:29:58,966 or you can also uh bring in 654 00:29:58,966 --> 00:30:02,900 bring in this d read by right 655 00:30:03,300 --> 00:30:06,600 defire from this first integration start 656 00:30:08,166 --> 00:30:09,566 and then uh 657 00:30:09,566 --> 00:30:13,033 uh know what would you need to find out substituting 658 00:30:13,366 --> 00:30:14,133 you need to evaluate 659 00:30:14,133 --> 00:30:17,666 basically you need to evaluate this uh uh 660 00:30:17,666 --> 00:30:22,500 electrical at this uh point and where uh and this uh 661 00:30:23,866 --> 00:30:25,133 vertical boundary conditions 662 00:30:25,133 --> 00:30:28,966 so what are these boundary conditions that uh 663 00:30:28,966 --> 00:30:31,400 at this uh 5 equal final 664 00:30:31,400 --> 00:30:34,600 that is your 5 by 60 second boundary condition 665 00:30:34,600 --> 00:30:36,833 your B is equal to V not okay 666 00:30:37,133 --> 00:30:40,500 so this is your final form of potentials 667 00:30:40,500 --> 00:30:46,300 and electric field reading with respect to this 5 okay 668 00:30:47,466 --> 00:30:50,366 and what else 669 00:30:51,400 --> 00:30:53,000 and here you can see that 670 00:30:53,333 --> 00:30:55,566 the variations are with respect to your 671 00:30:55,566 --> 00:30:56,466 for electrically 672 00:30:56,466 --> 00:30:59,666 the variations are with respect to your look 673 00:31:01,166 --> 00:31:03,766 k and for potential degraations 674 00:31:03,766 --> 00:31:05,233 that will structure your fine 675 00:31:05,933 --> 00:31:08,733 so let's um find out whether these two equations 676 00:31:08,733 --> 00:31:10,700 they satisfy our boundary values 677 00:31:10,766 --> 00:31:12,866 uh boundary conditions that that we have 678 00:31:12,866 --> 00:31:14,366 we have been provided earlier 679 00:31:14,366 --> 00:31:16,400 you can see that at 5 equal to 0 680 00:31:16,400 --> 00:31:20,666 this is evaluating to be equal to 0 681 00:31:20,800 --> 00:31:22,266 and at 5 equal 5 by 6 682 00:31:22,266 --> 00:31:26,700 this is evaluating it equal to hundred 683 00:31:26,933 --> 00:31:30,466 so plug in this 5 equal 5 by 6 right 684 00:31:30,466 --> 00:31:33,666 5 by six so this will cancel out 685 00:31:33,666 --> 00:31:34,700 this will cancel out 686 00:31:34,700 --> 00:31:36,500 and the voltage is going to be hundred 687 00:31:36,566 --> 00:31:38,766 so this means that for this poison equation 688 00:31:38,766 --> 00:31:41,400 boundary weight problem we have satisfied the 689 00:31:41,566 --> 00:31:44,433 these two conditions boundary boundary conditions 690 00:31:45,800 --> 00:31:48,000 let's all disturb the example 691 00:31:48,166 --> 00:31:51,966 this is given in your uh central partner system 692 00:31:52,100 --> 00:31:54,300 uh cervical partner system okay 693 00:31:54,900 --> 00:31:57,100 uh two connecting cones 694 00:31:57,200 --> 00:32:01,700 so first connecting cone it is your inner cone right 695 00:32:01,900 --> 00:32:03,866 it is at uh 5 by 10 there 696 00:32:03,866 --> 00:32:04,666 this one 697 00:32:05,100 --> 00:32:07,966 and the second connecting cone it is your outer cone 698 00:32:08,366 --> 00:32:11,666 uh both are uh infinite in extent yeah 699 00:32:11,666 --> 00:32:14,600 infinite in this uh in this in this 700 00:32:14,600 --> 00:32:17,466 in this extent and separated by uh 701 00:32:17,566 --> 00:32:21,100 infinite infine infinite infinite decimal 702 00:32:21,100 --> 00:32:22,733 infinite decimal disabilities 703 00:32:22,733 --> 00:32:25,466 infinite decimal gap that is this gap 704 00:32:26,866 --> 00:32:30,233 with the insulation right so that they are not shorted 705 00:32:31,900 --> 00:32:35,100 and uh we have given the boundary condition that 706 00:32:35,100 --> 00:32:38,133 for the first corn it is at zero potential 707 00:32:38,133 --> 00:32:41,533 for the second corn it is at 50 potential 708 00:32:41,533 --> 00:32:43,900 that is where we not okay 709 00:32:44,700 --> 00:32:46,900 and again we need to find out the electrical 710 00:32:46,900 --> 00:32:50,466 and the electrical potential in this area 711 00:32:50,466 --> 00:32:53,366 in between these two planes here 712 00:32:57,066 --> 00:33:01,366 let's solve this uh by treating it uh as a uh 713 00:33:01,366 --> 00:33:03,066 boundary belly problem so again 714 00:33:03,066 --> 00:33:05,766 in this case we don't have the information of uh 715 00:33:06,066 --> 00:33:08,266 1 inch identity we don't have it right 716 00:33:09,100 --> 00:33:11,600 so it is a left last equation case 717 00:33:11,733 --> 00:33:15,133 and since it is in uh your cervical corner system 718 00:33:15,133 --> 00:33:16,066 so this is the 719 00:33:16,066 --> 00:33:18,300 and the changes that we respect to treat only 720 00:33:18,300 --> 00:33:23,266 so we are left with only this equation of this uh 721 00:33:23,266 --> 00:33:26,400 second order uh differential equation 722 00:33:26,900 --> 00:33:30,766 and again the changes at your data is equal to 0 723 00:33:30,766 --> 00:33:33,666 and as since I since r is equal to 0 724 00:33:33,666 --> 00:33:37,166 and at the data is equal to 0 and pi are excluded 725 00:33:37,166 --> 00:33:38,300 we can um 726 00:33:39,600 --> 00:33:42,066 multiply these two terms on both sides 727 00:33:42,066 --> 00:33:44,000 and also mathematically it can be excluded 728 00:33:44,000 --> 00:33:46,933 because it is treating as a constant 729 00:33:46,933 --> 00:33:49,566 because the variations are with respect to this 730 00:33:50,100 --> 00:33:50,900 the teacher 731 00:33:52,200 --> 00:33:58,300 okay so since R is equal to 0 and r is equal to 0 732 00:33:58,700 --> 00:34:02,933 you're uh it is both conditions of cheetah 733 00:34:02,933 --> 00:34:05,433 Cheetah is vating from uh zero to pi 734 00:34:05,466 --> 00:34:07,466 they can be excluded uh 735 00:34:07,466 --> 00:34:09,766 because uh the mortgages are going to be zero 736 00:34:09,766 --> 00:34:12,200 at these two particle points 737 00:34:12,200 --> 00:34:14,866 so magnetically can be eliminated as well 738 00:34:14,866 --> 00:34:17,200 by bringing them on the right hand side 739 00:34:17,200 --> 00:34:19,233 so this is the thing that we need to fall 740 00:34:20,100 --> 00:34:24,466 integrated ones we are left out with this thing right 741 00:34:24,800 --> 00:34:26,366 on the right hand side we have the constant 742 00:34:26,366 --> 00:34:29,400 we know that integration of the constant uh 0 743 00:34:29,400 --> 00:34:32,300 it is a constant and the differentiation of the 744 00:34:33,700 --> 00:34:36,300 with the special tea diet is very important concept 745 00:34:36,300 --> 00:34:38,433 it is constant 746 00:34:38,600 --> 00:34:41,200 and the differentiation of a constant is 747 00:34:43,733 --> 00:34:47,600 zero because it is constant 748 00:34:48,533 --> 00:34:49,366 so you must remember 749 00:34:49,366 --> 00:34:52,400 this differentiation and integration of the constant 750 00:34:52,866 --> 00:34:54,966 that the integration of 0 751 00:34:54,966 --> 00:34:59,400 it is some constant and the integration of some 752 00:34:59,400 --> 00:35:02,000 differentiation of constant it is zero 753 00:35:03,700 --> 00:35:06,800 okay so bring this on the right hand side 754 00:35:06,800 --> 00:35:08,400 or we are left out with this thing 755 00:35:08,400 --> 00:35:10,666 but we have to take these second integration 756 00:35:10,666 --> 00:35:12,700 so that we can eliminate this thing 757 00:35:12,700 --> 00:35:14,766 and we are left out with the prevention only 758 00:35:15,866 --> 00:35:17,100 okay so in this case 759 00:35:17,100 --> 00:35:20,600 we need to enroll this uh geometric identity 760 00:35:20,600 --> 00:35:23,333 your uh that uh sign of teacher 761 00:35:23,333 --> 00:35:26,066 it is equal to sign of teacher 762 00:35:26,066 --> 00:35:29,566 it is equal to 2 times cause of teacher by 2 763 00:35:29,566 --> 00:35:31,100 and sign of teacher by two 764 00:35:31,933 --> 00:35:36,900 uh what is this property that is your sign of 765 00:35:39,466 --> 00:35:40,866 A plus b 766 00:35:42,133 --> 00:35:43,700 it is equal to 2 767 00:35:45,100 --> 00:35:47,866 times cause affair n 768 00:35:50,600 --> 00:35:51,766 sign off 769 00:35:54,966 --> 00:35:59,100 B essence a and B both are Peter by two 770 00:36:00,733 --> 00:36:01,533 right 771 00:36:02,200 --> 00:36:06,200 and here you can also treat it like A's t W 2 772 00:36:06,933 --> 00:36:12,066 b is t W2 and you can write this in terms of this 773 00:36:12,733 --> 00:36:17,966 okay so we already have this sign of Tita 774 00:36:18,300 --> 00:36:20,566 so we have converted it like this okay 775 00:36:20,566 --> 00:36:23,733 so multiply and divide this by this Cobra Peter by two 776 00:36:23,733 --> 00:36:27,200 so that we can have the uh 777 00:36:27,200 --> 00:36:31,033 we can have what we can convert this thing into a 778 00:36:31,933 --> 00:36:34,033 into a this thing 779 00:36:35,066 --> 00:36:36,966 this thing so divide it over here 780 00:36:36,966 --> 00:36:39,633 so this is your 10 Peter by two 781 00:36:39,933 --> 00:36:43,633 and this thing is your core square Peter right 782 00:36:43,700 --> 00:36:45,333 and you take it in the numerator 783 00:36:45,333 --> 00:36:47,633 so it is your second square Peter by two 784 00:36:48,133 --> 00:36:48,566 and now 785 00:36:48,566 --> 00:36:51,600 involve this very important property in English 786 00:36:51,600 --> 00:36:52,100 and property 787 00:36:52,100 --> 00:36:55,700 if you have the databasive of the denominator 788 00:36:55,700 --> 00:36:56,900 in the numerator 789 00:36:57,500 --> 00:37:00,066 we know that deteriorated of the stand theater 790 00:37:00,066 --> 00:37:01,866 it is equal to Sequence Square Theater 791 00:37:01,866 --> 00:37:03,766 so we can treat it like that 792 00:37:03,766 --> 00:37:08,066 this is this integration is equal to Alan of the 793 00:37:09,966 --> 00:37:12,833 this denominator 794 00:37:13,700 --> 00:37:16,733 so here the denominator is tantita by 2 795 00:37:16,733 --> 00:37:18,000 so integration of this is 796 00:37:18,000 --> 00:37:22,000 it is Allen of tantita by 2+ the constrat 797 00:37:22,133 --> 00:37:24,633 this e constrat of integration 798 00:37:25,666 --> 00:37:27,966 okay and also you know that one of our X 799 00:37:27,966 --> 00:37:30,400 it is equal to alanoff 800 00:37:30,400 --> 00:37:33,566 integration is equal to alanoff x plus 801 00:37:34,900 --> 00:37:36,500 construct okay 802 00:37:36,566 --> 00:37:38,966 so you can also use this chain rule as well 803 00:37:38,966 --> 00:37:41,400 this this substitution rule as well 804 00:37:42,366 --> 00:37:44,666 so these are the techniques of integration 805 00:37:44,666 --> 00:37:46,933 that's why I constantly say that 806 00:37:46,933 --> 00:37:48,066 you must know 807 00:37:48,066 --> 00:37:50,733 what are the differentiation and integration rules 808 00:37:50,733 --> 00:37:52,933 and how to find out the differentiation 809 00:37:52,933 --> 00:37:54,100 and integrations 810 00:37:54,100 --> 00:37:57,000 to have success in this particular course 811 00:37:58,200 --> 00:37:59,166 but the things are simple 812 00:37:59,166 --> 00:38:02,033 over here we have we have found out this voltage 813 00:38:02,066 --> 00:38:04,200 no we need to find out these costumes 814 00:38:04,466 --> 00:38:06,366 and let's solve this equation 815 00:38:06,366 --> 00:38:08,566 at this given boundary conditions 816 00:38:09,166 --> 00:38:10,800 so for the first boundary condition 817 00:38:10,800 --> 00:38:12,300 at Peter's equal to Peter 1 818 00:38:12,300 --> 00:38:14,366 we have this water is equal to 0 819 00:38:14,366 --> 00:38:17,333 so from we plug in this to this boundary condition 820 00:38:17,333 --> 00:38:19,800 so this is your constant b 821 00:38:20,933 --> 00:38:25,266 okay now plugging this b over here 822 00:38:26,500 --> 00:38:28,800 what does uh 823 00:38:28,800 --> 00:38:32,833 Peter is equal to Peter 2 right 824 00:38:33,533 --> 00:38:36,300 Peter is equal to Peter 2 and reads equal to 825 00:38:36,300 --> 00:38:38,366 we not and uh 826 00:38:38,366 --> 00:38:40,200 what we are now going to find out 827 00:38:40,400 --> 00:38:43,800 we are going to find out simply this think that 828 00:38:46,066 --> 00:38:48,900 that in the first step in the previous step 829 00:38:49,200 --> 00:38:50,566 you know that Eleanor 830 00:38:53,500 --> 00:38:54,533 a divided by B 831 00:38:54,533 --> 00:39:01,600 it is equal to Ellenoff a minus Ellenoff B 832 00:39:02,333 --> 00:39:05,033 so we simplified in this previous stat that 833 00:39:05,100 --> 00:39:06,200 I don't know if uh 834 00:39:06,266 --> 00:39:09,500 tta by 2 minus E I don't know tta uh 835 00:39:09,500 --> 00:39:14,500 10 tta 1 / 2 we have simplified after putting in this 836 00:39:14,500 --> 00:39:15,300 uh 837 00:39:17,066 --> 00:39:18,433 uh this found out uh 838 00:39:18,700 --> 00:39:21,066 question of integration that was your b 839 00:39:21,333 --> 00:39:24,533 uh you can simplify this subtraction of Alan 840 00:39:24,533 --> 00:39:26,866 Alan terms in the form of this 841 00:39:28,166 --> 00:39:30,266 division of these two in a terms 842 00:39:30,266 --> 00:39:31,900 and then take this Eleanor 843 00:39:32,066 --> 00:39:34,800 Eleanor outside of this division term 844 00:39:35,666 --> 00:39:37,100 so remember this identity 845 00:39:37,100 --> 00:39:38,466 the logic media identity 846 00:39:38,466 --> 00:39:41,266 that the difference of Ellenoff two terms 847 00:39:41,266 --> 00:39:45,200 it is equal to the Ellenoff ratio of these two terms 848 00:39:45,366 --> 00:39:48,566 so the division of the two terms 849 00:39:48,566 --> 00:39:49,933 uh Allen of two terms 850 00:39:49,933 --> 00:39:52,333 it can be written as the uh 851 00:39:52,333 --> 00:39:54,700 difference between the Allen of these two terms 852 00:39:54,700 --> 00:39:55,800 in the other way 853 00:39:56,866 --> 00:39:59,066 okay so once we have done this thing 854 00:39:59,066 --> 00:40:02,166 you can plug in this second boundary condition 855 00:40:02,400 --> 00:40:05,266 and you can simplify this thing 856 00:40:05,266 --> 00:40:07,800 to find out the second question right 857 00:40:08,133 --> 00:40:11,266 uh E and not plug in this into this 858 00:40:11,300 --> 00:40:12,500 okay into this 859 00:40:12,500 --> 00:40:15,866 to find out this standard potential 860 00:40:16,000 --> 00:40:18,700 in terms of your varying constant 861 00:40:18,700 --> 00:40:21,900 that is your variable variable in your 862 00:40:21,900 --> 00:40:25,100 in your variable or variable coordinate 863 00:40:25,100 --> 00:40:28,500 that is your teacher in this case so this is your final 864 00:40:29,966 --> 00:40:32,266 equation in terms of your data 865 00:40:32,266 --> 00:40:35,033 only for this vertical vulnerable problem 866 00:40:36,500 --> 00:40:37,766 we found out in London London 867 00:40:37,766 --> 00:40:38,866 we need to find out this 868 00:40:38,866 --> 00:40:41,200 uh electrify and this electrify 869 00:40:41,200 --> 00:40:43,433 it is minus period of fee 870 00:40:43,800 --> 00:40:45,900 so this is your diss track right 871 00:40:46,766 --> 00:40:48,800 uh since changes are over the special teacher 872 00:40:48,800 --> 00:40:49,366 so that's 873 00:40:49,366 --> 00:40:52,133 that's part of this gradient terms are going to be 0 874 00:40:52,133 --> 00:40:54,600 that is your changes with spectify 875 00:40:54,600 --> 00:40:59,100 and changes with respect to your R there 0 876 00:40:59,766 --> 00:41:01,100 and your you need to solve 877 00:41:01,100 --> 00:41:02,166 well solve this 878 00:41:02,166 --> 00:41:06,566 only tell the differentiate part of this gradient term 879 00:41:07,766 --> 00:41:09,100 so either you can 880 00:41:10,466 --> 00:41:12,666 take its differentiation okay 881 00:41:12,700 --> 00:41:17,200 or you can basically uh juicy 882 00:41:17,733 --> 00:41:20,133 uh first integration 883 00:41:20,133 --> 00:41:24,966 uh that we found out uh in our previous life 884 00:41:25,300 --> 00:41:26,466 the first integration 885 00:41:26,466 --> 00:41:28,900 once we done the first integration step 886 00:41:28,900 --> 00:41:32,300 you can take this uh DVR data from data as well 887 00:41:32,300 --> 00:41:36,366 but remember to multiply that with this minus 1 or R 888 00:41:37,600 --> 00:41:42,366 so this is your electric field and it is in terms of 889 00:41:42,366 --> 00:41:45,200 you can see that it is in terms of your boat 890 00:41:45,366 --> 00:41:46,566 teatime car 891 00:41:48,466 --> 00:41:50,266 let's uh verify 892 00:41:50,566 --> 00:41:53,566 let's find out these evaluate these to uh 893 00:41:53,566 --> 00:41:55,533 uh potential electric Protection 894 00:41:55,533 --> 00:41:59,300 electric field at the given uh boundary conditions 895 00:41:59,300 --> 00:42:02,200 so at the time teacher 1 896 00:42:02,200 --> 00:42:04,133 we are given that it is power 2 897 00:42:04,133 --> 00:42:05,400 power 10 teacher 2 898 00:42:05,400 --> 00:42:08,366 it is power 6 and we not it is equal to 50 899 00:42:08,366 --> 00:42:10,566 so this is your final expression 900 00:42:10,566 --> 00:42:12,166 for the electric potential 901 00:42:12,166 --> 00:42:15,233 and the electric fee and you can find out this 902 00:42:15,700 --> 00:42:17,433 you can verify this 903 00:42:19,400 --> 00:42:21,266 electric potential equation 904 00:42:21,266 --> 00:42:23,700 for this boundary value conditions as well 905 00:42:23,766 --> 00:42:26,166 that at t ties equal 5 or 10 906 00:42:26,166 --> 00:42:30,000 it it should evaluate equal to 0 because uh 907 00:42:30,000 --> 00:42:31,200 you will see that you 908 00:42:31,200 --> 00:42:33,433 you can value it at your own times 909 00:42:33,466 --> 00:42:37,400 and then the t tie is equal to uh 910 00:42:37,400 --> 00:42:43,400 by the 6 it will value equal to be not that is your 50 911 00:42:43,733 --> 00:42:46,866 okay and then it will also satisfy your 912 00:42:47,533 --> 00:42:50,266 what is on the question this lapation equation as well 913 00:42:50,366 --> 00:42:53,100 so you double differentiate this thing 914 00:42:53,500 --> 00:42:56,200 with respect to this data 915 00:42:56,300 --> 00:42:57,466 then you will will 916 00:42:57,466 --> 00:43:00,700 you will find out that it is going to be equal to zero 917 00:43:02,133 --> 00:43:03,866 message from your lecture 918 00:43:04,200 --> 00:43:06,266 for this electrostatic boundary problems 919 00:43:06,266 --> 00:43:07,700 we understood what are the 920 00:43:08,166 --> 00:43:10,366 what is the significance of these 921 00:43:10,766 --> 00:43:12,600 electrostatic boundary problems 922 00:43:12,600 --> 00:43:15,800 and how to find out this poisons 923 00:43:15,800 --> 00:43:17,400 and the class questions 924 00:43:17,400 --> 00:43:20,400 and what is the geneticness theorem 925 00:43:20,466 --> 00:43:22,533 and then uh how 926 00:43:22,533 --> 00:43:24,166 how to involve this uh 927 00:43:24,266 --> 00:43:27,266 electrostatic boundary value problems and uh 928 00:43:27,266 --> 00:43:29,466 these poisons and laprasecrations 929 00:43:29,466 --> 00:43:31,100 to find out the electric uh 930 00:43:31,100 --> 00:43:33,033 field and electric potentials 931 00:43:33,100 --> 00:43:35,833 and by solving these boundary value problems 932 00:43:36,466 --> 00:43:40,100 based upon these by zones and the class equations 933 00:43:40,966 --> 00:43:44,166 next time we will determine this 934 00:43:44,166 --> 00:43:46,866 the resistance and the capacity 935 00:43:47,300 --> 00:43:50,166 and by solving these Bondi value problems 936 00:43:53,733 --> 00:43:54,966 here I thank you all 937 00:43:54,966 --> 00:43:56,533 if you have any questions 938 00:43:56,533 --> 00:43:58,066 uh they will be entertained 939 00:43:58,066 --> 00:44:00,666 uh with your emails or online 940 00:44:00,666 --> 00:44:03,133 synchronous sessions that we have arranged 941 00:44:03,133 --> 00:44:04,366 at the department