00:01
The other equations, del ey by del x minus del ex by del y equals i omega b z.
00:15
Second, del e z by del y minus ik ey equals i omega bx.
00:25
Third equation ike e x minus del e z by del x equals i omega bx.
00:34
The fourth one to prove is del b .y by del x minus del b x by dl bx by dly equals minus i omega by c square ez.
00:49
The first equation del b z by minus ik by equals minus i omega by c square into ex.
01:00
And the last equation to prove is i k b x minus del b z by del b x equals minus i omega y c squared into e y.
01:17
From these equations obtain the equation e x equals 1 by omega by c whole square minus k square into k del e z by x plus omega del b x plus omega del b z by del y e y equals i by omega b c whole square minus k square into k del e z by del y minus omega del b z by del x by del x by del x b x equal i by omega b c whole square minus k square into k using these equations show the equation del square by del x square plus del square by del y square plus omega by c whole square minus k square into e z equals zero and del square by del x square plus del square by del x square plus omega by c whole square minus k square into b z equals zero and del square by del x square plus del square by del x square plus omega by c whole square minus k square from the equations, first to say an ex, ey, b, z, find the equation del square by del x square plus del square by del x square plus omega by c square minus a square into ez equals to zero, and the same for this equation.
03:04
Now, we know that the electric field in the space is given by e0 x, comma, y, e raised to the power i, a z minus omega t and the magnetic field is given by b not x y e raised to the power i he raised to the power i a z minus omega t and e not and b not are given by e not is e xx cap plus e y y cap plus e z z cap and v0 is equal to vx x cap plus, vy, y cap plus vz z cap.
04:06
Now the divergence of the electric field in the determinant form is given by del by del x, del by del x, del by del, del, del, or ex, e, e, z.
04:28
In the equation form, it will be del cross, e, e, z.
04:33
In the x direction is equal to del e z by del y minus del e y by del z then we can say it is equal to del e not x by del y minus i k e not y multiplied by e raised to the power i k z minus omega t now del cross e is also equals to minus of del b by del t which is equals to i omega b not he raised to the power i k z minus omega t now del cross b is equals to 1 by c square del e y del t which can be written as minus i omega by c square into e not, it is to the power i, kz minus omega t.
05:41
By comparing the del cross e with del cross ex, then we have the equation del ez by del t minus ik ey equals i omega bx.
06:11
Now finding del cross e z in the y, direction you can write it as del e x by del z minus del e z by del x which is equals to i k e not x minus del e0 x by del x into e raised to the power i k z minus omega t then i k e e x minus del e x minus del x equals to i own.
06:55
The second equation is true...
