00:02
In this question, solution here is given that region 1 is z less than 0, er is equal to 2 .5, region 2 z greater than 0, er is equal to 4, here e1 vector is equal to minus 30 ax cap plus 50 a cap of y plus 70 a cap of z v per m.
00:45
So now boundary conditions different for tangential components and normal components of e vector.
00:53
Therefore, xy is boundary, a cap of x, a cap of y becomes tangential components, tangential components of e vector and a cap of z becomes normal component.
01:37
At boundary tangential components remains same.
01:40
Therefore, e of t2 is equal to e of t1 which is equal to minus of 30 a cap x plus 50 a cap y.
01:57
But normal component vary as dn1 minus dn2 is equal to rho s.
02:15
Therefore, surface xy has no charge.
02:27
Rho s is equal to 0.
02:31
Therefore, dn1 is equal to dn2.
02:37
Therefore, d is equal to epsilon multiplied by e.
02:45
This implies here epsilon r1 multiplied by en1 is equal to epsilon r2 multiplied by en2.
02:59
So, we get here this en2 is equal to epsilon r1 divided by epsilon r2 by en1.
03:11
This is equal to 2 .5 divided by 4 multiplied by 70 a cap z...