## Surface Integral Considered as a Potential Function 2

To find
$\int_s \frac{\alpha}{\sqrt{(x-x_0)^2 +(Y-Y_0)^2 +(Z-Z_0)^2}}dS$
, the surface being a sphere not containing the point
$(x_0 , y_0 , z_0)$
we can consider the integral as a potential function. As long as the distribution of charge over the surface is spherically symmetrical, we can treat the charge distribution over the surface as a point charge at the center of the sphere.
Continuing the potential analygy, take the charge per unit area as 1. If the radius of the sphere is
$r$
then the charge equals the surface area equals
$4 \pi r^2$
.
$\int_s \frac{\alpha}{\sqrt{(x-x_0)^2 +(Y-Y_0)^2 +(Z-Z_0)^2)}}dS = \frac{4 \pi r^2}{\sqrt{x_0^2 +y_0^2+z_0^2}}$