1 00:00:01,533 --> 00:00:05,166 this is election No. 19 of a B 2 3 2 2 00:00:05,266 --> 00:00:09,400 Today we will have an overview of electric field 3 00:00:09,400 --> 00:00:10,733 boundary conditions 4 00:00:10,733 --> 00:00:13,200 and thereafter we will solve some of 5 00:00:13,266 --> 00:00:15,000 some of case examples 6 00:00:15,500 --> 00:00:19,000 this will cover your rest of uh section 5.9 7 00:00:19,000 --> 00:00:22,566 this is academic week No. 11 of your semester 8 00:00:24,400 --> 00:00:26,900 so today we will uh review uh 9 00:00:26,900 --> 00:00:28,333 electronic conditions 10 00:00:28,333 --> 00:00:31,266 and thereafter we will solve multiple examples 11 00:00:32,933 --> 00:00:34,600 first of all the boundary conditions 12 00:00:34,600 --> 00:00:39,066 uh run the field exist in a medium consisting of uh 13 00:00:39,100 --> 00:00:40,700 two different medias 14 00:00:40,733 --> 00:00:43,366 the conditions at the field must satisfy 15 00:00:43,766 --> 00:00:45,400 are called the boundary conditions 16 00:00:45,400 --> 00:00:48,866 so these are the conditions that electrically in the 17 00:00:49,400 --> 00:00:54,233 in both mediums is satisfies across that boundary 18 00:00:54,733 --> 00:00:56,900 so far electric fields 19 00:00:56,900 --> 00:00:59,133 the following boundary conditions are important 20 00:00:59,133 --> 00:01:01,300 dialect to to dialectic interface 21 00:01:01,400 --> 00:01:04,133 conducted conductor to dialectic interface 22 00:01:04,133 --> 00:01:06,366 conductor to free space interface 23 00:01:06,600 --> 00:01:07,600 so know that 24 00:01:07,600 --> 00:01:10,533 they are helpful in determining the feel 25 00:01:10,533 --> 00:01:12,433 on one side of the boundary 26 00:01:12,600 --> 00:01:15,300 the feel on other side is none 27 00:01:17,133 --> 00:01:19,400 and uh we already uh know that uh 28 00:01:19,400 --> 00:01:21,866 we solve these boundary conditions 29 00:01:21,866 --> 00:01:25,266 and with the help of these two uh Maxwells 30 00:01:25,266 --> 00:01:26,600 uh equations uh Maxwells 31 00:01:26,600 --> 00:01:26,733 uh 32 00:01:26,733 --> 00:01:30,100 famous first and second law that is uh our Gaza's Law 33 00:01:30,100 --> 00:01:32,033 and uh this uh 34 00:01:32,300 --> 00:01:35,533 the closed closing trigger of electric fuel is zero 35 00:01:35,533 --> 00:01:37,166 that is your er 36 00:01:37,166 --> 00:01:40,866 this er considerated property of your electric fee 37 00:01:41,166 --> 00:01:43,500 and also in order to find out this 38 00:01:43,500 --> 00:01:46,233 electric fields and electric first tentities 39 00:01:46,266 --> 00:01:50,466 we decompose these two electric quantities into two 40 00:01:50,466 --> 00:01:52,500 autogonell components 41 00:01:55,166 --> 00:01:56,166 so the first of all 42 00:01:56,166 --> 00:01:58,866 the dialectric to dialect Bondi conditions 43 00:01:58,866 --> 00:02:01,800 for this particular case we consider this 44 00:02:02,533 --> 00:02:06,533 this uh scenario that uh uh 45 00:02:06,533 --> 00:02:07,300 there's uh 46 00:02:07,300 --> 00:02:10,600 one medium with dialectic properties of APSRON1 47 00:02:10,600 --> 00:02:14,066 the other medium with dialectic permittivity of APSRON2 48 00:02:14,133 --> 00:02:15,833 and uh uh 49 00:02:15,900 --> 00:02:19,200 we decompose electric fees and we we 50 00:02:19,200 --> 00:02:20,600 we apply this uh 51 00:02:21,266 --> 00:02:21,900 uh close in 52 00:02:21,900 --> 00:02:25,100 tigula of electric wheel across this rectangular box 53 00:02:25,166 --> 00:02:27,000 and once we applied all the uh 54 00:02:27,000 --> 00:02:30,500 those uh integrations and all those uh 55 00:02:30,500 --> 00:02:33,766 simplifications and we came down to a conclusion that 56 00:02:33,966 --> 00:02:35,900 um there are two boundary conditions 57 00:02:35,900 --> 00:02:38,166 first of all the tangential boundary condition 58 00:02:38,166 --> 00:02:41,166 that the potential components of electrical 59 00:02:41,200 --> 00:02:43,800 they are equal at the boundary 60 00:02:43,866 --> 00:02:46,533 remember at the boundary these two components 61 00:02:46,533 --> 00:02:48,800 they are equal at the boundaries 62 00:02:48,800 --> 00:02:51,566 between these two different dialectic mediums 63 00:02:52,300 --> 00:02:55,466 and uh this means that uh 64 00:02:55,466 --> 00:02:58,033 uh traditional component of electric field 65 00:02:58,066 --> 00:03:01,966 undergoes no change on the boundary 66 00:03:02,066 --> 00:03:04,766 and its continuous across the boundary 67 00:03:04,766 --> 00:03:07,300 so that is the intuitiveness from here 68 00:03:07,666 --> 00:03:08,533 the second thing is 69 00:03:08,533 --> 00:03:11,133 regarding the potential component of your electric 70 00:03:11,133 --> 00:03:14,600 prostituency and it undergoes some of the changes 71 00:03:14,600 --> 00:03:17,266 because we know the relationship that uh 72 00:03:17,266 --> 00:03:20,633 this uh this equal to abstinent e 73 00:03:20,800 --> 00:03:24,266 and uh if you plug in that uh constituted relationship 74 00:03:24,266 --> 00:03:28,266 then you can see here there's no direct um equal 75 00:03:28,266 --> 00:03:29,933 equal Equality between these 76 00:03:29,933 --> 00:03:33,900 these uh er potential component of electric flux 10 c 77 00:03:33,900 --> 00:03:38,000 so it is said that this electric flux 10 c 78 00:03:38,000 --> 00:03:40,666 it is discontinuous across the boundary 79 00:03:41,766 --> 00:03:43,133 so once we uh 80 00:03:43,133 --> 00:03:47,166 d uh we uh we dealt about these uh normal components 81 00:03:47,166 --> 00:03:49,900 then we uh considered this uh gaussian 82 00:03:49,900 --> 00:03:52,500 uh as pillar box 83 00:03:53,300 --> 00:03:56,466 pillar box box pillar box surface 84 00:03:56,466 --> 00:03:59,466 so that we can solve that Gaussian equation 85 00:03:59,466 --> 00:04:01,466 and once we solved this equation 86 00:04:01,466 --> 00:04:03,600 and we came down to a conclusion that 87 00:04:03,666 --> 00:04:06,000 that the difference between the normal company 88 00:04:06,000 --> 00:04:07,766 the difference between the normal company 89 00:04:07,766 --> 00:04:08,833 is equal to the 90 00:04:09,466 --> 00:04:12,966 this surface electric electric as you can 91 00:04:12,966 --> 00:04:14,866 as you can see surface area electric 92 00:04:15,400 --> 00:04:16,900 electric dance to your electric 93 00:04:17,366 --> 00:04:21,466 this surface electric density right 94 00:04:21,466 --> 00:04:23,000 surface electric density 95 00:04:23,300 --> 00:04:27,700 if it is present on this surface are the interface 96 00:04:27,700 --> 00:04:30,933 which is meeting these two mediums 97 00:04:30,933 --> 00:04:33,166 like here like here right 98 00:04:33,166 --> 00:04:34,800 so this one is present over here 99 00:04:34,800 --> 00:04:37,866 then it is equal to basically the difference between B 100 00:04:38,966 --> 00:04:40,266 the electricals and CR 101 00:04:41,266 --> 00:04:42,566 these two mediums 102 00:04:43,400 --> 00:04:48,166 and uh from here you can into that if in case 103 00:04:48,200 --> 00:04:51,366 uh no free charge exists at the boundary 104 00:04:51,366 --> 00:04:52,700 or the interface between the 105 00:04:52,700 --> 00:04:55,900 these two mediums that this uh rose is equal to 0 106 00:04:55,900 --> 00:04:58,300 so this means that these two component 107 00:04:58,300 --> 00:05:01,566 the normal likely proxancy components that equal 108 00:05:01,800 --> 00:05:02,366 and accordingly 109 00:05:02,366 --> 00:05:06,000 once we plug in the constitutive relationship 110 00:05:06,366 --> 00:05:08,133 then this is the electric field 111 00:05:08,133 --> 00:05:10,000 and normal components relationship 112 00:05:10,000 --> 00:05:13,366 so from here you can into that normal components 113 00:05:13,366 --> 00:05:15,800 normal components of electric presidency 114 00:05:15,866 --> 00:05:17,866 are equal at the boundary 115 00:05:18,100 --> 00:05:18,500 and the 116 00:05:18,500 --> 00:05:21,400 hence that the normal continental likely of electric 117 00:05:21,400 --> 00:05:25,200 electricity undergoes no change on the boundary 118 00:05:25,200 --> 00:05:27,866 and it is continuous across the boundary 119 00:05:28,466 --> 00:05:28,566 now 120 00:05:28,566 --> 00:05:30,966 coming towards the normal component of electric field 121 00:05:31,000 --> 00:05:35,966 it it does undergo some changes across the boundary 122 00:05:35,966 --> 00:05:38,500 and hence it is said that the normal 123 00:05:38,500 --> 00:05:40,700 comprehensive electric field 124 00:05:40,700 --> 00:05:43,800 it is discontinuous across the bound 125 00:05:46,266 --> 00:05:48,033 and accordingly once we 126 00:05:48,100 --> 00:05:53,266 once we resolve this electric field 127 00:05:53,266 --> 00:05:56,366 and the electric presidentities of these 128 00:05:56,566 --> 00:05:59,800 present in these two mediums into two components 129 00:05:59,800 --> 00:06:02,266 then uh we found out a relationship 130 00:06:02,266 --> 00:06:05,066 very known relationship that 131 00:06:05,100 --> 00:06:06,800 that is known as the law reflection 132 00:06:06,800 --> 00:06:08,000 so in this case 133 00:06:08,466 --> 00:06:11,400 that the ratio of electric permittivity 134 00:06:11,400 --> 00:06:13,266 the relative electric permittuities 135 00:06:13,266 --> 00:06:16,766 of these two mediums is equal to the 10 liter 136 00:06:16,766 --> 00:06:21,100 of the angles that these two electric trees makes 137 00:06:21,100 --> 00:06:23,666 uh make with respect to the normal 138 00:06:23,733 --> 00:06:25,966 uh to their interface so 139 00:06:25,966 --> 00:06:28,700 here we consider that these are two different mediums 140 00:06:28,700 --> 00:06:29,500 right 141 00:06:30,200 --> 00:06:32,166 this is the interface uh 142 00:06:32,166 --> 00:06:35,200 this is the uh Tita one is the angle that uh 143 00:06:35,200 --> 00:06:35,966 electrical 144 00:06:35,966 --> 00:06:38,833 in medium 1 makes with respect to the normal 145 00:06:38,866 --> 00:06:40,100 between these two 146 00:06:40,700 --> 00:06:44,933 so this is the normal right between these two media 147 00:06:44,933 --> 00:06:47,066 different mediums and accordingly 148 00:06:47,066 --> 00:06:49,100 this in the second medium 149 00:06:49,100 --> 00:06:50,133 this is the tea tattoo 150 00:06:50,133 --> 00:06:53,600 that electrically makes with respect to the normal 151 00:06:53,600 --> 00:06:55,100 between these two mediums 152 00:06:55,100 --> 00:06:55,966 and here 153 00:06:55,966 --> 00:06:59,566 we found out a very important relationship that 154 00:07:00,100 --> 00:07:03,466 uh the SO1 of an electric wheel 155 00:07:03,466 --> 00:07:06,466 it will uh travel from one medium to another medium 156 00:07:06,466 --> 00:07:09,333 so either it will turn towards the medium 157 00:07:09,333 --> 00:07:11,766 or it will diverges from the 158 00:07:12,333 --> 00:07:13,566 turn towards the normal 159 00:07:13,566 --> 00:07:16,233 or it will diverge from the normal 160 00:07:16,366 --> 00:07:18,800 and the ratio between the electrical 161 00:07:18,800 --> 00:07:21,400 relative permittivity of these two mediums 162 00:07:21,666 --> 00:07:25,300 is equal to the ratio of 10 163 00:07:25,366 --> 00:07:27,333 10 of these two angles 164 00:07:27,333 --> 00:07:29,366 that these two electric freeze make 165 00:07:29,366 --> 00:07:31,200 with respect to be normal 166 00:07:33,566 --> 00:07:34,300 let's not uh 167 00:07:34,300 --> 00:07:37,566 talk about the second condition that is uh 168 00:07:37,700 --> 00:07:40,066 conducted to dialectic boundary conditions 169 00:07:40,066 --> 00:07:41,700 uh that's how we can find out 170 00:07:41,700 --> 00:07:44,600 the only conditions for this scenario that uh 171 00:07:44,600 --> 00:07:46,766 if one side of this uh medium 172 00:07:46,766 --> 00:07:48,900 it is consisting of this dialectic 173 00:07:48,900 --> 00:07:51,266 and the other side is purely conductor 174 00:07:51,266 --> 00:07:54,000 we know that uh due to the infinite uh 175 00:07:54,000 --> 00:07:56,700 conductivity of the medium and there uh 176 00:07:56,700 --> 00:07:58,400 there's no electric field 177 00:07:58,400 --> 00:08:00,600 no electric field exist 178 00:08:00,866 --> 00:08:03,500 no electric field exists within a conductor 179 00:08:03,500 --> 00:08:05,733 so it is the property of an upper uh 180 00:08:05,733 --> 00:08:06,266 conductor 181 00:08:06,266 --> 00:08:10,200 that due to infinite conductivity of the conductor uh 182 00:08:10,200 --> 00:08:12,466 electric field cannot exist inside the 183 00:08:12,466 --> 00:08:14,566 or may not exist you can say more 184 00:08:14,566 --> 00:08:17,200 more probably may not exist within a conductor 185 00:08:17,200 --> 00:08:20,400 so that's that's why we say that the electric field 186 00:08:20,566 --> 00:08:23,566 and the electric fuel and the electric frogs 187 00:08:23,566 --> 00:08:26,900 dance cheese they are zero inside the conductor medium 188 00:08:27,600 --> 00:08:30,966 if we uh took this Assumption and we uh 189 00:08:30,966 --> 00:08:34,466 uh took this uh electric field uh 190 00:08:34,466 --> 00:08:37,200 this dialectical material of the other medium as ABS 191 00:08:37,200 --> 00:08:40,666 1 are and we saw this uh uh 192 00:08:40,666 --> 00:08:44,400 this complete uh close interview of edot DL 193 00:08:44,400 --> 00:08:49,233 that is your conservated property of that electrical 194 00:08:49,333 --> 00:08:50,966 and then we came down to concludion 195 00:08:50,966 --> 00:08:53,166 that the financial company 196 00:08:54,100 --> 00:08:55,333 since it is a boundary 197 00:08:55,333 --> 00:08:57,666 so financial component at the boundary 198 00:08:57,666 --> 00:08:59,933 or the interface between these two components 199 00:08:59,933 --> 00:09:02,066 and the electric frequency 200 00:09:02,200 --> 00:09:04,666 at the interface between these two components 201 00:09:04,666 --> 00:09:06,400 it is equal to zero 202 00:09:07,400 --> 00:09:08,200 right 203 00:09:09,900 --> 00:09:12,400 and the second thing uh that we uh 204 00:09:12,466 --> 00:09:15,500 came down to conclusion about this normal components 205 00:09:15,500 --> 00:09:16,966 that uh 206 00:09:16,966 --> 00:09:20,666 an electric field an electric field must exist 207 00:09:20,866 --> 00:09:25,333 this means must exist uh to the uh external 208 00:09:25,333 --> 00:09:29,566 external to the conductor and most phenometers surface 209 00:09:29,600 --> 00:09:32,466 so the only electric field that exist 210 00:09:32,800 --> 00:09:33,733 on the boundary field 211 00:09:33,733 --> 00:09:36,300 that is on the boundary field that is here 212 00:09:36,700 --> 00:09:40,100 this normal component of electrical and there is uh 213 00:09:40,300 --> 00:09:42,600 uh that is the only grid that exist 214 00:09:42,600 --> 00:09:44,833 in case of dialectry to conductor 215 00:09:44,866 --> 00:09:48,466 uh uh this medium and the boundary condition 216 00:09:48,466 --> 00:09:52,266 condition case so in this case you can see that and if 217 00:09:52,266 --> 00:09:55,466 if there's a presence of surface identity 218 00:09:55,466 --> 00:09:57,766 so that is equal to your um 219 00:09:58,566 --> 00:10:01,633 normal component of your electric presidency 220 00:10:01,800 --> 00:10:02,566 and accordingly 221 00:10:02,566 --> 00:10:05,266 you can find out this normal component of electric fee 222 00:10:05,266 --> 00:10:08,966 so remember that uh whenever there's a case that uh 223 00:10:09,100 --> 00:10:10,700 on one side there's a dialectric 224 00:10:10,700 --> 00:10:13,700 on the other side there's a conductor then uh 225 00:10:14,000 --> 00:10:17,600 then uh the initial component of the electric field 226 00:10:17,600 --> 00:10:19,700 and electric pregnancy they are zero 227 00:10:19,700 --> 00:10:21,200 and the normal components 228 00:10:21,200 --> 00:10:23,500 and normal component of this electrical tendency 229 00:10:23,500 --> 00:10:25,633 it is equal to the surface identity 230 00:10:25,733 --> 00:10:28,166 and the normal component of electric field 231 00:10:28,166 --> 00:10:29,333 that is that you can find 232 00:10:29,333 --> 00:10:31,700 or using this constitutive relationship 233 00:10:31,800 --> 00:10:33,733 and uh it is uh 234 00:10:33,733 --> 00:10:36,166 note that in in this scenario 235 00:10:36,166 --> 00:10:37,333 only the normal 236 00:10:37,333 --> 00:10:40,166 only the only the electric field that exist 237 00:10:40,166 --> 00:10:40,600 that is 238 00:10:40,600 --> 00:10:43,366 the only normal component of the electric field 239 00:10:43,366 --> 00:10:46,033 that is take away from this slight 240 00:10:47,000 --> 00:10:48,366 so the third case third case 241 00:10:48,366 --> 00:10:53,333 is just the extrapolation of the previous scenario 242 00:10:53,333 --> 00:10:56,066 so in this vertical case instead of a dialectic 243 00:10:56,066 --> 00:10:57,900 you're only having this free space 244 00:10:57,900 --> 00:10:58,600 and for free space 245 00:10:58,600 --> 00:11:01,266 we know that the ABS final is equal to one 246 00:11:01,333 --> 00:11:03,900 so same scenario inside this uh conductor 247 00:11:03,900 --> 00:11:05,700 the electric fuel is going to be zero 248 00:11:05,733 --> 00:11:06,500 and outside 249 00:11:06,500 --> 00:11:09,100 only the normal component is going to exist 250 00:11:09,366 --> 00:11:11,366 and so here you can see that 251 00:11:11,366 --> 00:11:14,100 the traditional components of this electric field 252 00:11:14,100 --> 00:11:16,666 and the electrical exhausted is equal to 0 253 00:11:16,666 --> 00:11:19,166 and the normal component is equal to 0 s 254 00:11:19,300 --> 00:11:20,700 and accordingly you can find out 255 00:11:20,700 --> 00:11:22,933 the normal component of the electric field 256 00:11:22,933 --> 00:11:26,866 so electrophil must approach a conducting surface 257 00:11:27,100 --> 00:11:28,333 normally you can see 258 00:11:28,333 --> 00:11:28,866 you can see here 259 00:11:28,866 --> 00:11:31,300 because only the normal component is existing 260 00:11:31,300 --> 00:11:34,500 in this case the financial component is zero 261 00:11:35,333 --> 00:11:36,200 and uh we uh 262 00:11:36,200 --> 00:11:39,500 you must uh remember that uh uh Runner 263 00:11:39,500 --> 00:11:42,933 Runner Day is a conductor around the uh some uh 264 00:11:42,933 --> 00:11:46,200 another uh body or some electric surgery 265 00:11:46,200 --> 00:11:48,100 it can be utilized uh 266 00:11:48,100 --> 00:11:50,200 it can be utilized as a screening 267 00:11:50,200 --> 00:11:52,233 as a screening shield right 268 00:11:52,300 --> 00:11:54,366 so it can shield the effect of electric 269 00:11:54,366 --> 00:11:58,566 and electric beams to entrap 270 00:11:59,100 --> 00:12:03,700 to disturb the electric components which are in its 271 00:12:03,700 --> 00:12:04,766 in its enclosure 272 00:12:04,766 --> 00:12:07,966 so any electric components allow any electric safety 273 00:12:07,966 --> 00:12:11,900 which is enclosed by some uh conductive 274 00:12:11,900 --> 00:12:14,633 conductive shield uh it will award the 275 00:12:14,733 --> 00:12:18,233 it will award the hindrance of external electric field 276 00:12:19,000 --> 00:12:22,600 external electric feel to that internal body 277 00:12:22,600 --> 00:12:24,900 which is enclosed by this conductor 278 00:12:25,200 --> 00:12:27,166 so because why it is like so 279 00:12:27,166 --> 00:12:28,666 because the E feel inside 280 00:12:28,666 --> 00:12:30,566 this conductor is equal to zero 281 00:12:31,133 --> 00:12:33,900 and due to the higher conductivity of this conductor 282 00:12:34,066 --> 00:12:35,566 the E field 283 00:12:35,566 --> 00:12:39,366 it will reside on this surface of this conductor 284 00:12:40,333 --> 00:12:42,466 and this is known as the screen effect as well 285 00:12:42,466 --> 00:12:46,200 and uh for this uh pure conductor the 286 00:12:46,366 --> 00:12:47,566 this penetration 287 00:12:47,566 --> 00:12:50,900 which is known as the skin depth of the electrical uh 288 00:12:50,900 --> 00:12:52,866 it is also the minimum 289 00:12:55,166 --> 00:12:57,066 so this is the whole uh 290 00:12:57,066 --> 00:12:59,366 this is the whole summary of your electric 291 00:12:59,366 --> 00:13:01,900 boundary conditions uh different scenarios 292 00:13:01,900 --> 00:13:03,100 so at the end 293 00:13:03,100 --> 00:13:06,200 this is the take away from these uh three cases 294 00:13:06,200 --> 00:13:08,600 the dialectual dry electric boundary conditions 295 00:13:08,733 --> 00:13:10,800 conductor to directly boundary conditions 296 00:13:10,800 --> 00:13:13,366 and conductor to free space boundary conditions 297 00:13:13,366 --> 00:13:16,733 so you can uh uh you can uh 298 00:13:16,733 --> 00:13:19,166 remember these boundary conditions to soldier 299 00:13:19,166 --> 00:13:21,766 further examples that will discuss uh 300 00:13:21,766 --> 00:13:23,833 in our in upcoming slides 301 00:13:25,133 --> 00:13:27,600 so first of all the example 5.9 302 00:13:27,666 --> 00:13:30,633 so here you you have to first of all uh 303 00:13:30,966 --> 00:13:33,466 see that uh what is this uh 304 00:13:33,466 --> 00:13:37,233 what is this uh uh this uh coordinate system 305 00:13:37,333 --> 00:13:39,000 and uh uh 306 00:13:39,000 --> 00:13:42,766 what is the normal to these uh to different mediums 307 00:13:42,766 --> 00:13:44,266 and what is this scenario 308 00:13:44,266 --> 00:13:48,133 either it is this dialect to 2 309 00:13:48,133 --> 00:13:52,200 dialectric or it is dialected to conductor scenario 310 00:13:52,300 --> 00:13:56,100 so from this particular case 2 extensive homogeneous 311 00:13:56,100 --> 00:13:56,366 right 312 00:13:56,366 --> 00:13:59,600 homogeneous means that an isotropic both conditions 313 00:13:59,600 --> 00:14:04,700 okay so it is isotropic and homogene and linear as well 314 00:14:04,700 --> 00:14:09,666 so Ella yet you can see that in this case in medium 1 315 00:14:09,666 --> 00:14:13,266 there is uh it is also exhibiting this uh 316 00:14:13,266 --> 00:14:14,400 processing this uh 317 00:14:14,566 --> 00:14:16,733 dialect prometivity of apps for No. 1 318 00:14:16,733 --> 00:14:19,566 the medium 2 is having this dialect prometivity 319 00:14:19,566 --> 00:14:22,533 so this means it is your case number one okay 320 00:14:22,533 --> 00:14:25,933 so for uh so what is the interface between these two 321 00:14:25,933 --> 00:14:30,333 so the interface between these two is this this one 322 00:14:30,333 --> 00:14:32,400 right Z equal to 0 323 00:14:33,266 --> 00:14:37,600 so Z equal to 0 plane mean that it is x y plane right 324 00:14:37,733 --> 00:14:39,100 you must remember these things 325 00:14:39,100 --> 00:14:41,166 that what does the Z is equal to 0 326 00:14:41,166 --> 00:14:43,266 from your elementary maths 327 00:14:43,266 --> 00:14:44,366 from your calculus 328 00:14:44,366 --> 00:14:46,500 that whenever it is Z is equal to zero plane 329 00:14:46,500 --> 00:14:48,133 that it is going to be your x 330 00:14:48,133 --> 00:14:52,200 y plane so uh in this particular case of whenever 331 00:14:52,400 --> 00:14:54,300 whenever there's a boundary there 332 00:14:54,300 --> 00:14:57,133 there's a boundary at the Z Z access 333 00:14:57,133 --> 00:15:00,333 this means that Z interface between these two 334 00:15:00,333 --> 00:15:02,933 or Z plane between these two different mediums 335 00:15:02,933 --> 00:15:07,100 then you must include that the normal 336 00:15:07,200 --> 00:15:10,000 the normal to these two interfaces right 337 00:15:10,533 --> 00:15:12,266 this one uh one 338 00:15:12,366 --> 00:15:14,233 uh first medium and the second medium 339 00:15:14,466 --> 00:15:15,666 you can hear a see that 340 00:15:15,666 --> 00:15:17,600 what is the normal between these two interfaces 341 00:15:17,600 --> 00:15:19,666 so this is the plane right 342 00:15:19,666 --> 00:15:22,066 this is extra plane so what is normal 343 00:15:22,066 --> 00:15:25,166 so in you can see that it is equal to a Z 344 00:15:25,666 --> 00:15:26,866 so that is your normal 345 00:15:26,866 --> 00:15:29,900 so it is very important that you must uh 346 00:15:29,900 --> 00:15:32,666 understand this geometry and this uh 347 00:15:33,000 --> 00:15:35,766 uh concept that how to find out this uh 348 00:15:35,766 --> 00:15:38,733 normal between these two components otherwise 349 00:15:38,733 --> 00:15:40,500 uh if you're not able to uh 350 00:15:40,500 --> 00:15:43,466 find out what is the exact normal between these two 351 00:15:43,466 --> 00:15:45,700 uh in different mediums uh 352 00:15:45,700 --> 00:15:47,466 the boundary that is joining between the 353 00:15:47,466 --> 00:15:49,000 these two difference uh 354 00:15:49,000 --> 00:15:51,566 these two different mediums that uh uh 355 00:15:51,566 --> 00:15:54,400 then you will not be able to solve the uh problems 356 00:15:54,400 --> 00:15:56,466 uh correctly so first of all 357 00:15:56,466 --> 00:15:58,100 the most important thing is 358 00:15:58,100 --> 00:16:00,133 you must understand 359 00:16:00,133 --> 00:16:02,466 how to find out the normal between these two 360 00:16:02,466 --> 00:16:04,566 and these two different mediums 361 00:16:05,600 --> 00:16:08,500 so uh Z is uh or entertainment 362 00:16:08,500 --> 00:16:11,433 you can also say is so you can also remember that if 363 00:16:11,766 --> 00:16:14,466 if some boundaries given like this 364 00:16:14,466 --> 00:16:16,700 Z is equal to 0 or X is equal to 0 365 00:16:16,700 --> 00:16:17,700 y is equal to 0 366 00:16:17,700 --> 00:16:21,166 so by default the normal will be E x normal to Z 367 00:16:21,166 --> 00:16:23,400 it is Z plane which is actually plane 368 00:16:23,400 --> 00:16:26,300 so what what is from your maths or geometry 369 00:16:26,300 --> 00:16:31,400 it is a Z so here this is your normal component right 370 00:16:31,400 --> 00:16:32,133 and rest of 371 00:16:32,133 --> 00:16:34,866 and this is your potential component of electric fuel 372 00:16:36,100 --> 00:16:37,933 okay so here we have 373 00:16:37,933 --> 00:16:41,800 given that a uniform electric field exist 374 00:16:41,800 --> 00:16:44,600 for Z is greater than zero right 375 00:16:44,600 --> 00:16:48,566 so for the first medium the information of electricity 376 00:16:48,766 --> 00:16:50,266 we are given with right 377 00:16:50,400 --> 00:16:51,933 and we have to find out the electricity 378 00:16:51,933 --> 00:16:54,600 in the second medium we have to find out the 379 00:16:54,933 --> 00:16:55,933 these angles 380 00:16:55,933 --> 00:16:58,833 that electric and magnetic electricity needs 381 00:16:58,866 --> 00:17:00,333 with the interface right 382 00:17:00,333 --> 00:17:03,166 with the interface so remember that Tita 1 383 00:17:03,166 --> 00:17:07,000 Tita 1 is the angle it makes within normal errors here 384 00:17:07,000 --> 00:17:08,766 taken as alpha alpha 385 00:17:08,766 --> 00:17:11,766 it is the angle it makes mixed with the interface 386 00:17:11,766 --> 00:17:15,700 so it is 90 - 90 minus the teacher okay 387 00:17:15,700 --> 00:17:18,266 90 minus the teacher 388 00:17:19,500 --> 00:17:20,933 okay and then 389 00:17:20,933 --> 00:17:23,666 we will determine this electric duster 390 00:17:23,666 --> 00:17:26,000 using the information of electric fields 391 00:17:26,100 --> 00:17:29,866 and then we'll find out the energy within a cube 392 00:17:30,733 --> 00:17:32,666 which is present in one of these mediums 393 00:17:32,666 --> 00:17:36,500 and we will see how good this is like 394 00:17:38,400 --> 00:17:40,266 okay so first of all 395 00:17:40,266 --> 00:17:41,400 how to find out first of all 396 00:17:41,400 --> 00:17:45,000 resolve this electric feel which is given to you 397 00:17:45,000 --> 00:17:48,166 that is your electric feel in this medium one 398 00:17:48,333 --> 00:17:51,166 so we know that it comprises two components 399 00:17:51,166 --> 00:17:52,200 one is the tenantial 400 00:17:52,200 --> 00:17:54,166 and the second one is the normal component 401 00:17:54,166 --> 00:17:56,466 so how to find out the normal component 402 00:17:56,466 --> 00:17:59,133 we remember from our chapter No. 1 403 00:17:59,133 --> 00:18:02,466 that if you want to find out that for want 404 00:18:02,466 --> 00:18:04,333 want to find out a scaler component 405 00:18:04,333 --> 00:18:07,033 along the direction of a 1 er 406 00:18:07,166 --> 00:18:07,933 erector quantity 407 00:18:07,933 --> 00:18:10,400 then we simply have to take a start product 408 00:18:10,400 --> 00:18:12,366 with that unidractor right 409 00:18:12,366 --> 00:18:16,000 so in this case the unidractor is the normal 410 00:18:16,066 --> 00:18:17,700 normal lecture in this case 411 00:18:17,733 --> 00:18:20,033 which is which is the normal between these two 412 00:18:20,600 --> 00:18:24,566 to this interface between these two different mediums 413 00:18:24,600 --> 00:18:26,966 so this is your skeler comprehend okay 414 00:18:26,966 --> 00:18:32,400 e 1 dot a Z which is the a N so it is a 3 415 00:18:32,400 --> 00:18:34,366 so what is your multiply by a Z 416 00:18:34,366 --> 00:18:37,466 then it will give you the vector comprehend 417 00:18:39,133 --> 00:18:39,933 for that 418 00:18:41,700 --> 00:18:45,733 normal right for that now for in the normal direction 419 00:18:45,733 --> 00:18:48,700 so this is the the this is the 420 00:18:48,700 --> 00:18:51,966 you can say it is the normal 421 00:18:52,000 --> 00:18:54,600 Raptor component of this electric field 422 00:18:54,600 --> 00:18:57,000 in the z direction this particular case 423 00:18:57,000 --> 00:18:59,066 so then how to find out the potential component 424 00:18:59,066 --> 00:19:02,800 it is simply the use this uh basic definition 425 00:19:02,800 --> 00:19:05,066 so it is the vector minus d normal component 426 00:19:05,066 --> 00:19:08,966 so it is your definition component in the medium one 427 00:19:08,966 --> 00:19:10,300 so this is the first thing 428 00:19:10,300 --> 00:19:12,200 no next in while the boundary condition 429 00:19:12,566 --> 00:19:12,700 now 430 00:19:12,700 --> 00:19:15,466 the boundary condition says that e tunicial components 431 00:19:15,466 --> 00:19:17,600 it is it is continuous or is 432 00:19:17,900 --> 00:19:20,066 uh equal in these two different mediums 433 00:19:20,066 --> 00:19:23,566 so this means that E2T is equal to E1 right 434 00:19:23,566 --> 00:19:26,566 so the uh financial component in your 435 00:19:26,566 --> 00:19:28,166 uh this medium to it 436 00:19:28,166 --> 00:19:28,300 is 437 00:19:28,300 --> 00:19:31,466 the same as the financial component of the medium one 438 00:19:31,533 --> 00:19:32,333 okay 439 00:19:33,300 --> 00:19:35,866 now to find out the normal component 440 00:19:35,866 --> 00:19:38,800 so in this case Royal C is equal to 0 right 441 00:19:38,966 --> 00:19:42,466 Royal C is equal to 0 no surface chartancy is given 442 00:19:42,566 --> 00:19:43,800 so you can involve this 443 00:19:43,800 --> 00:19:45,966 continual relationship between the normal component 444 00:19:45,966 --> 00:19:47,133 of this electric 445 00:19:47,133 --> 00:19:51,200 electricity and uh uh and from this uh 446 00:19:51,200 --> 00:19:55,500 Constitution relationship that these equal apps funny 447 00:19:56,066 --> 00:19:58,233 you can uh relate this uh 448 00:19:58,800 --> 00:20:02,000 connect this relationship into the electrical farm 449 00:20:02,000 --> 00:20:03,266 and then from here 450 00:20:03,266 --> 00:20:05,800 you will see that astronaut is cancelling out 451 00:20:05,800 --> 00:20:08,433 and you're left with the E to N right 452 00:20:08,666 --> 00:20:14,400 so what is it because you are given with Ivan right 453 00:20:14,600 --> 00:20:18,366 you know what is Ivan from this Stephan okay 454 00:20:18,700 --> 00:20:20,800 so plug in the value of this E 1 n 455 00:20:21,066 --> 00:20:24,133 use this uh ratio between this uh altitometuity 456 00:20:24,133 --> 00:20:27,766 so this is your normal component in B medium 2 457 00:20:27,766 --> 00:20:30,366 as you can see it is also in the Z direction okay 458 00:20:30,366 --> 00:20:31,633 in the Z direction 459 00:20:31,966 --> 00:20:34,300 that is the normal between these two interfaces 460 00:20:35,566 --> 00:20:36,066 and uh 461 00:20:36,066 --> 00:20:40,133 now simply how to find out the uh complete vector 462 00:20:40,133 --> 00:20:42,066 the electrical vector this is second car 463 00:20:42,066 --> 00:20:44,566 second uh medium that is uh 464 00:20:44,566 --> 00:20:47,900 you have to uh now simply add up these two uh 465 00:20:48,200 --> 00:20:51,566 engine and normal component of electrical in medium to 466 00:20:51,566 --> 00:20:54,066 so this is your final expression right 467 00:20:54,566 --> 00:20:56,166 so this is the day you 468 00:20:56,166 --> 00:20:59,766 you uh proceed from one uh medium to another medium 469 00:20:59,900 --> 00:21:00,333 uh 470 00:21:00,333 --> 00:21:02,533 once you have given the information of galactic field 471 00:21:02,533 --> 00:21:04,166 in one medium and you 472 00:21:04,166 --> 00:21:05,700 you have the information of galactic 473 00:21:05,700 --> 00:21:07,366 uh your uh electric uh 474 00:21:07,366 --> 00:21:10,100 permittivities and you have the information of uh 475 00:21:10,333 --> 00:21:13,533 boundary conditions for that particular uh scenario 476 00:21:13,533 --> 00:21:14,666 in this particular case 477 00:21:14,666 --> 00:21:16,433 it is the dialect to to dialect 478 00:21:16,666 --> 00:21:18,400 uh boundary condition case 479 00:21:19,300 --> 00:21:22,500 now second step uh the uh part B 480 00:21:22,500 --> 00:21:26,566 we need to find out the angle that this electric field 481 00:21:27,300 --> 00:21:29,300 that this electrically mix 482 00:21:29,300 --> 00:21:32,100 with respect to this interface right 483 00:21:32,500 --> 00:21:34,066 so we can uh we can uh 484 00:21:34,066 --> 00:21:36,100 and also we can find out this uh 485 00:21:36,466 --> 00:21:37,666 uh uh 486 00:21:37,900 --> 00:21:39,900 this uh animal between this uh 487 00:21:40,166 --> 00:21:42,500 normal and the electrical data as well 488 00:21:42,500 --> 00:21:44,200 which is your Miss Cheetah 489 00:21:44,400 --> 00:21:47,000 so you can proceed in both ways right 490 00:21:47,133 --> 00:21:48,700 you can proceed in this ways 491 00:21:48,700 --> 00:21:51,966 either you can find out this teacher 1 492 00:21:51,966 --> 00:21:55,900 using this relationship that time teacher one 493 00:21:56,500 --> 00:21:57,933 it is equated from here 494 00:21:57,933 --> 00:22:00,900 you can see that this mammal component right 495 00:22:00,900 --> 00:22:03,100 it is a sternicial component right 496 00:22:03,100 --> 00:22:10,700 so this is your E1T magnitude right divided by 497 00:22:12,700 --> 00:22:15,533 E1N so from here you can also 498 00:22:15,533 --> 00:22:18,566 so this is given in your textbook or the other ways 499 00:22:18,566 --> 00:22:21,133 so you can find out the 3 tavan just like this 500 00:22:21,133 --> 00:22:24,800 that the the the angle between these 4 alpha one 501 00:22:24,800 --> 00:22:27,966 you can see here so how to find out to this alpha one 502 00:22:27,966 --> 00:22:29,700 from this part of our ceiling 503 00:22:29,933 --> 00:22:31,900 you can see that the 10 alpha 504 00:22:31,900 --> 00:22:33,866 the 10 alpha this one alpha 505 00:22:33,866 --> 00:22:37,466 it is equal to the uh ratio of the uh 506 00:22:37,466 --> 00:22:39,300 this magnitude of the normal company 507 00:22:39,300 --> 00:22:41,000 and this traditional company 508 00:22:42,066 --> 00:22:44,800 and once you have this found of the magnitude 509 00:22:44,800 --> 00:22:46,200 you can take a standing rose 510 00:22:46,200 --> 00:22:48,166 so standing rose will give you the Alka 1 511 00:22:48,166 --> 00:22:51,500 the angle that like to create a medium 512 00:22:51,500 --> 00:22:52,000 one makes 513 00:22:52,000 --> 00:22:54,666 with respect to the interface between these two mediums 514 00:22:54,866 --> 00:22:55,533 and accordingly 515 00:22:55,533 --> 00:22:59,166 you can involve this relationship that is 90 minus 516 00:22:59,166 --> 00:23:01,566 because the whole this whole angle is 90 517 00:23:01,566 --> 00:23:04,800 90 so it is 90 minus alpha one 518 00:23:06,400 --> 00:23:07,200 right 519 00:23:07,766 --> 00:23:10,366 uh you can find out the Cheetah one from here 520 00:23:10,366 --> 00:23:13,966 so it is coming out to be 60.9 on the other ways 521 00:23:13,966 --> 00:23:15,533 as I have already told you 522 00:23:15,533 --> 00:23:17,200 that it is given in your taxpayers 523 00:23:17,200 --> 00:23:18,500 that you can directly found out 524 00:23:18,500 --> 00:23:21,600 this angle between the normal and electrical 525 00:23:21,866 --> 00:23:23,666 using this relationship that 10 526 00:23:23,666 --> 00:23:24,766 Cheetah is equal to be 527 00:23:25,100 --> 00:23:27,866 ratio of the financial and normal compound compound 528 00:23:27,866 --> 00:23:31,300 remember these two are different ratios 529 00:23:31,300 --> 00:23:32,433 remember this thing 530 00:23:32,466 --> 00:23:35,500 because they are dealing with different humanities 531 00:23:37,400 --> 00:23:39,400 okay and uh 532 00:23:39,566 --> 00:23:41,033 how to find out this uh 533 00:23:41,066 --> 00:23:44,366 alpha tool are similar in same manner in the elector 534 00:23:44,366 --> 00:23:45,300 in this uh 535 00:23:45,566 --> 00:23:47,800 uh medium 2 there's a 10 536 00:23:47,800 --> 00:23:50,800 alpha 2 to the ratio of the normal components magnitude 537 00:23:50,800 --> 00:23:53,266 to the magnitude of this signature component 538 00:23:53,266 --> 00:23:54,933 and you can find all this alpha 2 539 00:23:54,933 --> 00:23:56,400 and the Peter 2 is equal to 540 00:23:56,400 --> 00:24:00,066 you know that it is 90 minus alpha 2 right 541 00:24:00,066 --> 00:24:05,633 and so it is equal to I goes 90 minus alpha 2 542 00:24:06,166 --> 00:24:07,800 and also you can find out this Chan 543 00:24:07,800 --> 00:24:11,666 t dot two using the same relationship 544 00:24:11,666 --> 00:24:16,533 that it is the ratio of you can see here like this one 545 00:24:16,533 --> 00:24:21,333 so this is the bite I got a diagram for this uh E2 546 00:24:21,333 --> 00:24:23,600 so again from here the normal uh B 547 00:24:24,466 --> 00:24:27,700 you don't find out that this is the 90 degree okay 548 00:24:27,766 --> 00:24:29,466 so to be very short between the 549 00:24:29,733 --> 00:24:33,000 financial component of electric fuel in medium 2 to d 550 00:24:33,333 --> 00:24:37,866 normal component in medium two right 551 00:24:38,066 --> 00:24:40,833 and uh you can also satisfy this uh 552 00:24:41,166 --> 00:24:43,133 uh law affection from here 553 00:24:43,133 --> 00:24:46,700 that the ratio of the meridic permittivity 554 00:24:46,900 --> 00:24:51,166 uh in these two mediums it is equal to the term of uh 555 00:24:51,166 --> 00:24:53,700 these uh tea triangles that the electric fuse make 556 00:24:53,700 --> 00:24:56,100 with respect to the normal 557 00:24:56,400 --> 00:24:58,166 so these two ratio that equal 558 00:24:58,166 --> 00:25:00,500 you can also satisfy this condition of law 559 00:25:00,500 --> 00:25:01,466 reflection as well 560 00:25:04,266 --> 00:25:05,700 okay so the third part 561 00:25:05,800 --> 00:25:08,133 we have to find out the energy densities 562 00:25:08,133 --> 00:25:10,500 in both dialectrics 563 00:25:10,533 --> 00:25:13,300 we know that from our chapter No. 4 564 00:25:13,300 --> 00:25:15,200 that the energy density 565 00:25:15,200 --> 00:25:19,600 remember it is density jewels by cubic meter 566 00:25:19,800 --> 00:25:22,366 and it is half of the 567 00:25:23,200 --> 00:25:24,666 this permittivity in medium 568 00:25:24,666 --> 00:25:26,566 that impact that particular medium 569 00:25:26,566 --> 00:25:28,966 and the magnitude scarabee electric field 570 00:25:29,066 --> 00:25:30,733 now we have the information of electric field 571 00:25:30,733 --> 00:25:31,900 in these two mediums 572 00:25:31,900 --> 00:25:34,700 we have the information of electric permittivities 573 00:25:34,700 --> 00:25:35,800 in these two mediums 574 00:25:35,800 --> 00:25:38,200 and we know that in this permittivity 575 00:25:38,200 --> 00:25:40,266 it is equal to abstel knot in 2 576 00:25:40,500 --> 00:25:43,100 Abstel R1 this is your abstel knot 577 00:25:43,100 --> 00:25:44,600 this is your Abstel R1 578 00:25:44,600 --> 00:25:47,466 this is the magnitude of electrical in medium 1 579 00:25:47,466 --> 00:25:50,100 and this is magnitude of electrical in medium 2 580 00:25:50,100 --> 00:25:54,200 this is active permittivity in medium uh uh 2 581 00:25:54,200 --> 00:25:55,700 and this is your right 582 00:25:55,700 --> 00:25:58,166 to permittivity in your medium one 583 00:25:58,166 --> 00:26:02,100 and this is extra knot and this is extra knot right 584 00:26:02,200 --> 00:26:06,266 so accordingly you can you can label all these 2 585 00:26:06,566 --> 00:26:09,266 3 quantities 586 00:26:10,866 --> 00:26:13,066 the next step uh solve this four part 587 00:26:13,200 --> 00:26:14,966 we will have to find out the energy 588 00:26:14,966 --> 00:26:18,066 within a cube of site 2 m right 589 00:26:18,700 --> 00:26:20,300 and the Q okay 590 00:26:20,300 --> 00:26:23,266 and it is center at 3 4 591 00:26:23,266 --> 00:26:24,966 5 point okay 592 00:26:24,966 --> 00:26:27,866 so if this is your geometric for example right 593 00:26:28,066 --> 00:26:30,466 so let's find out what is this point first of all 594 00:26:30,466 --> 00:26:36,100 this is your X y and Z and this is your t 595 00:26:37,400 --> 00:26:40,233 4 and 596 00:26:42,166 --> 00:26:45,200 this is your discipline maybe okay 597 00:26:45,200 --> 00:26:46,933 these are p point okay 598 00:26:46,933 --> 00:26:51,166 these are 3 4MINUS5 599 00:26:51,200 --> 00:26:52,000 so first of all 600 00:26:52,000 --> 00:26:57,633 we need to find out that in which medium applies okay 601 00:26:58,066 --> 00:27:02,633 so the 0 z is equal to 0 plane x y plane 602 00:27:04,466 --> 00:27:06,533 let's uh see how to solve this 603 00:27:06,533 --> 00:27:08,866 uh by D in this particular case 604 00:27:08,866 --> 00:27:13,800 we need to find out the energy or within a cube of uh 605 00:27:13,800 --> 00:27:15,600 uh side 2 and let's see 606 00:27:15,600 --> 00:27:20,700 uh where this uh uh uh this uh point lies 607 00:27:20,700 --> 00:27:25,400 so this is your X y and Z plane 608 00:27:26,100 --> 00:27:26,900 okay 609 00:27:29,466 --> 00:27:30,533 okay so this is your Z 610 00:27:30,533 --> 00:27:32,066 Z quarter zero right 611 00:27:32,200 --> 00:27:33,800 the interface between these two mediums Z 612 00:27:33,800 --> 00:27:35,300 Z quarter zero that is your X 613 00:27:35,300 --> 00:27:39,066 y plane and let's see where this point lies 614 00:27:39,066 --> 00:27:40,100 so this is your 615 00:27:41,466 --> 00:27:44,466 3 is your 4 616 00:27:45,866 --> 00:27:46,666 okay 617 00:27:50,500 --> 00:27:51,800 like this one okay 618 00:27:51,800 --> 00:27:54,433 and then you have to come downwards at 619 00:27:55,766 --> 00:27:58,500 minus 5 so this is your 3 620 00:27:58,533 --> 00:28:02,966 4 - 5 point right 621 00:28:03,533 --> 00:28:07,500 so this means that it lies in Ger medium 2 622 00:28:08,366 --> 00:28:11,300 that is Z is less than zero 623 00:28:11,466 --> 00:28:11,900 and so 624 00:28:11,900 --> 00:28:15,200 this means that the electric field that you involve 625 00:28:15,200 --> 00:28:17,100 you need to involve or according 626 00:28:17,100 --> 00:28:19,600 the electric density that you need to involve 627 00:28:19,600 --> 00:28:22,900 is the electricity in the medium 2 628 00:28:22,900 --> 00:28:25,800 that you need to integrate with a specta volume 629 00:28:25,800 --> 00:28:29,266 to find out this energy within the cube 630 00:28:29,266 --> 00:28:31,100 we know that the energy it is 631 00:28:31,100 --> 00:28:35,433 it is the volume in general of this energy density 632 00:28:35,600 --> 00:28:36,800 uh and also who 633 00:28:36,800 --> 00:28:39,300 you can see here that the energy density juniors 634 00:28:39,300 --> 00:28:43,000 they are individuals uh per cubic meter and uh 635 00:28:43,000 --> 00:28:43,900 none of uh 636 00:28:44,000 --> 00:28:45,500 I see other things uh 637 00:28:45,500 --> 00:28:46,866 how to find out these ranges 638 00:28:46,866 --> 00:28:48,766 another important thing uh 639 00:28:48,766 --> 00:28:50,500 as you can see that uh 640 00:28:50,500 --> 00:28:53,800 uh since it is a cube of side 2 641 00:28:53,800 --> 00:28:57,500 this means that uh if it is centred at uh 642 00:28:57,700 --> 00:28:58,800 x is equal to 3 643 00:28:58,800 --> 00:29:01,800 this means that the limit of the x there 644 00:29:01,866 --> 00:29:07,600 ranging from 2 to four and the whole range is of 2 645 00:29:07,600 --> 00:29:10,400 okay and the center is 3 646 00:29:10,933 --> 00:29:12,600 okay and like as you can say 647 00:29:12,600 --> 00:29:15,600 you can see here the center is four 648 00:29:16,166 --> 00:29:19,300 so notice 3 upper is 5 649 00:29:19,300 --> 00:29:22,600 so the whole range is 2 okay 650 00:29:23,000 --> 00:29:25,566 and for this uh z xs 651 00:29:25,666 --> 00:29:30,500 uh z is minus 5 so there is minus 6 652 00:29:30,700 --> 00:29:36,200 upper is minus 4 and whole range is 2 because it is a Q 653 00:29:37,366 --> 00:29:39,266 and since he's equal to minus 5 654 00:29:39,266 --> 00:29:42,033 so this means it is lying in the region 2 655 00:29:42,100 --> 00:29:43,800 now you have to simply integrate this 656 00:29:43,800 --> 00:29:45,766 uh uh 657 00:29:45,766 --> 00:29:48,466 this given energy density 658 00:29:48,466 --> 00:29:50,200 and since it is a constant value 659 00:29:50,200 --> 00:29:52,166 so it will come outside 660 00:29:52,166 --> 00:29:55,533 it will come outside and you have to simply solve this 661 00:29:55,533 --> 00:29:56,333 uh 662 00:29:56,533 --> 00:29:58,500 uh this one in trigger 663 00:29:58,500 --> 00:29:59,866 right triple in trigger 664 00:29:59,900 --> 00:30:01,333 and intuitively 665 00:30:01,333 --> 00:30:04,733 you can also see that since it is a cube 666 00:30:04,733 --> 00:30:08,400 is one is going to be two ways to bar 667 00:30:08,466 --> 00:30:15,900 so two into two into 2 equal and 2 to raise to bar 3 668 00:30:16,400 --> 00:30:18,000 that is al q and will 669 00:30:18,000 --> 00:30:20,500 you just multiply with the energistensity of that 670 00:30:20,500 --> 00:30:23,133 per chlamydium which in this cases medium 2 671 00:30:23,133 --> 00:30:25,800 to find out the energy in this Q 672 00:30:29,166 --> 00:30:30,200 let's solve uh 673 00:30:30,200 --> 00:30:32,000 this another case example 674 00:30:32,000 --> 00:30:33,533 so in this case uh 675 00:30:33,533 --> 00:30:39,066 region y is less than 0 consist of a perfectly uh 676 00:30:39,066 --> 00:30:40,800 conductor so this is your region 677 00:30:40,800 --> 00:30:41,766 uh one 678 00:30:41,900 --> 00:30:43,500 so again you must 679 00:30:43,500 --> 00:30:43,966 must 680 00:30:43,966 --> 00:30:46,666 very importantly know what are the coordinate system 681 00:30:46,800 --> 00:30:48,966 how to plot this scenario 682 00:30:48,966 --> 00:30:51,666 what is the normal between these two interfaces 683 00:30:53,066 --> 00:30:56,366 intuitively you can see that y is less than 0 684 00:30:56,366 --> 00:31:00,300 this means that a N is going to be a y 685 00:31:00,700 --> 00:31:01,500 okay 686 00:31:01,966 --> 00:31:05,066 because this is the interface between these two mediums 687 00:31:05,400 --> 00:31:07,766 that is your what is his plan 688 00:31:08,366 --> 00:31:12,766 what is this plane this plane is your ZX plane okay 689 00:31:12,766 --> 00:31:15,166 and the normal between these two planes is 690 00:31:15,700 --> 00:31:18,000 normal between these two planes is 691 00:31:19,766 --> 00:31:21,800 this is why 692 00:31:24,333 --> 00:31:25,733 so this is the thing 693 00:31:25,733 --> 00:31:28,466 this is your starting point for this problem 694 00:31:29,366 --> 00:31:30,300 so uh 695 00:31:30,300 --> 00:31:33,066 region 1 it is like wise less than zero it 696 00:31:33,066 --> 00:31:36,166 it consists of a perfect conductor y region 2 697 00:31:36,166 --> 00:31:38,600 it is wise greater than zero uh greater than 0 698 00:31:38,600 --> 00:31:41,166 it is consisting of a dialect of medium 699 00:31:41,166 --> 00:31:42,966 with permittivity as given 700 00:31:42,966 --> 00:31:45,000 if there is a surface chart density 701 00:31:45,000 --> 00:31:46,400 so here in this case you 702 00:31:46,400 --> 00:31:48,466 you are given with this surface chart density 703 00:31:48,700 --> 00:31:50,166 0 s of this 704 00:31:50,166 --> 00:31:53,800 2 nano coolums per meter scale on the conductor 705 00:31:53,800 --> 00:31:56,800 so we need to determine this electric and uh 706 00:31:56,800 --> 00:31:57,566 this electric 707 00:31:57,566 --> 00:32:00,933 electricity on both sides in both mediums 708 00:32:00,933 --> 00:32:04,233 and what are those the those medium points 709 00:32:04,333 --> 00:32:06,533 and the first part is this one 710 00:32:06,533 --> 00:32:09,766 so here you can see that here in this vertical case 711 00:32:09,766 --> 00:32:11,566 your y it is minus 2 712 00:32:11,566 --> 00:32:15,300 so this means you are lying in this region one 713 00:32:15,300 --> 00:32:16,833 which is your conductor medium 714 00:32:16,933 --> 00:32:18,866 for this your y is 1 715 00:32:18,866 --> 00:32:22,266 so this means you're lying in this dialectic medium 716 00:32:23,466 --> 00:32:25,133 so uh in first case 717 00:32:25,133 --> 00:32:27,966 your point uh e is in the conductor 718 00:32:27,966 --> 00:32:30,533 so this uh that is your uh region 719 00:32:30,533 --> 00:32:34,100 a rise less than 0 and minus 2 is less than 0 720 00:32:34,100 --> 00:32:36,533 hence we know and from one week conditions 721 00:32:36,533 --> 00:32:39,600 we know that the electric fee inside the conductor is 722 00:32:39,600 --> 00:32:41,133 zero hence accordingly 723 00:32:41,133 --> 00:32:45,300 Europe electric flux tenses also zero inside your uh 724 00:32:45,300 --> 00:32:47,766 this conductor so at this particular point 725 00:32:47,766 --> 00:32:49,066 both these electric uh 726 00:32:49,066 --> 00:32:50,933 field and electric flux tensities 727 00:32:50,933 --> 00:32:52,400 they are going to be zero 728 00:32:53,966 --> 00:32:56,366 for the second point it is in the dialectic medium 729 00:32:56,366 --> 00:32:58,200 because 1 is greater than 0 730 00:32:58,200 --> 00:33:00,866 for divide component 731 00:33:00,966 --> 00:33:03,966 and again from the boundary condition 732 00:33:04,066 --> 00:33:05,100 from the boundary condition 733 00:33:05,100 --> 00:33:05,966 we know that 734 00:33:06,966 --> 00:33:10,733 that this normal component of electric constantity 735 00:33:10,733 --> 00:33:11,866 it is equal to the 736 00:33:13,100 --> 00:33:16,466 this uh surface identity and uh since 737 00:33:16,466 --> 00:33:18,366 uh in the uh in the vector form 738 00:33:18,366 --> 00:33:20,800 the normal component uh unit rector is a y 739 00:33:20,800 --> 00:33:25,700 so it is this rector form and for the electrical 740 00:33:25,733 --> 00:33:28,000 uh again the constitute relationship 741 00:33:28,066 --> 00:33:31,000 uh that uh this a d is equal to absent E 742 00:33:31,000 --> 00:33:34,566 so divide d this please uh divide d by this absent 743 00:33:34,566 --> 00:33:36,900 not into absent R by this absent R 744 00:33:36,900 --> 00:33:38,533 because it is other than 745 00:33:38,533 --> 00:33:40,700 this medium is other than the free space 746 00:33:41,166 --> 00:33:43,133 so this is your electric free linger 747 00:33:43,133 --> 00:33:45,466 uh at this uh second point 748 00:33:46,800 --> 00:33:49,066 this is your assignment for this uh week 749 00:33:49,066 --> 00:33:50,600 uh and uh uh 750 00:33:50,600 --> 00:33:51,933 you need to solve these uh 751 00:33:51,933 --> 00:33:55,166 two problem exercises at your own time however 752 00:33:55,166 --> 00:33:56,766 this submission is not required 753 00:33:56,766 --> 00:33:59,100 but it is stressed that uh 754 00:33:59,100 --> 00:34:01,366 in order to understand this topic more uh 755 00:34:01,366 --> 00:34:02,300 uh deeply 756 00:34:02,300 --> 00:34:05,966 you must solve these two uh uh practice exercises 757 00:34:07,133 --> 00:34:08,400 so today we we 758 00:34:08,400 --> 00:34:11,700 we had a review of electric boundary conditions 759 00:34:11,700 --> 00:34:14,266 and then we solved some of the case examples 760 00:34:14,266 --> 00:34:18,366 to further clarify these electric boundary conditions 761 00:34:19,866 --> 00:34:21,000 in the next lecture uh 762 00:34:21,000 --> 00:34:23,533 we are going to study the uh 763 00:34:23,533 --> 00:34:25,666 what is uh electric 764 00:34:25,800 --> 00:34:27,666 electrostatic boundary value problem 765 00:34:27,666 --> 00:34:29,733 and how to solve these types of the problem 766 00:34:29,733 --> 00:34:31,700 and what is the significance of uh 767 00:34:31,966 --> 00:34:33,266 electric boundary problem 768 00:34:33,266 --> 00:34:35,866 electric boundary value problems and uh 769 00:34:35,866 --> 00:34:37,366 how they are helpful in uh 770 00:34:37,366 --> 00:34:41,933 determining the electric uh uh field and the uh these 771 00:34:41,933 --> 00:34:44,933 uh electric potentials so that is the significance 772 00:34:44,933 --> 00:34:46,300 and uh for that thing 773 00:34:46,300 --> 00:34:48,766 we will also drive the warzones and 774 00:34:48,766 --> 00:34:51,566 uh laplash equations which are very famous equations 775 00:34:51,566 --> 00:34:52,566 for the electric fields 776 00:34:52,566 --> 00:34:53,766 and the magnetic fields as well 777 00:34:53,766 --> 00:34:54,566 later on 778 00:34:54,766 --> 00:34:57,300 and then we will understand what is the uniqueness 779 00:34:57,300 --> 00:35:00,300 uh theorem and there after we solve some of the uh 780 00:35:00,300 --> 00:35:01,633 case examples as well 781 00:35:04,700 --> 00:35:05,933 here I thank you all 782 00:35:05,933 --> 00:35:07,666 if you have any questions 783 00:35:07,666 --> 00:35:09,133 uh they would be entertained 784 00:35:09,133 --> 00:35:11,733 uh through your emails or online 785 00:35:11,733 --> 00:35:14,200 synchronous sessions that we have arranged 786 00:35:14,200 --> 00:35:15,400 at the department