00:03
So the potential, pi of x is equal to 1 over 4 pi epsilon not, integral over surface 1, sigma 1, x prime, g, xxx prime, d8 prime.
00:22
So we have the surface charge, the greens function, and the area element.
00:28
Now, the electrostatic energy w is epsilon not over 2, the integral, volume integral, magnitude of the electric field squared dv we can also write a half the volume that's using e is equal to minus the gradient of the potential row x phi x dv so over s1 row becomes sigma and d a half of the integral over surface 1 sigma 1 x and phi is 1 over 4 pi epsilon not sigma 1 x prime g x d a prime d a and so this proves the given expression which is going to be 1 over 8 pi epsilon not is the energy stored you have the integral s 1 double integral or surface now sigma 1x greens function for x and x prime and sigma 1 x prime times d a d a prime so we just proved that so let sigma x equal to sigma 1x plus some small variation delta sigma x now, the numerator is going to be the double integral of surface 1, sigma x, g x, x, x prime, sigma x prime, d a d a prime, we substitute the variational principle.
03:10
We see that it's going to be the double integral of sigma 1 x, x, g, x, x, prime...