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  
\[4 \pi r^2\]
 .
\[\int_s \frac{\alpha}{\sqrt{(x-x_0)^2 +(Y-Y_0)^2 +(Z-Z_0)^2)}}dS = \frac{4 \pi r^2}{\sqrt{x_0^2 +y_0^2+z_0^2}} \]

Add comment

Security code
Refresh