- #1
Thaakisfox
- 263
- 0
Suppose [tex]\phi[/tex] is a scalar function: [tex]R^n\to R[/tex], and it satisfies the Poisson equation:
[tex]\nabla^2 \phi=-\dfrac{\rho}{\varepsilon_0}[/tex]
Now I want to calculate the following integral:
[tex]\int \phi \nabla^2 \phi \,dV[/tex]
So using Greens first identity I get:
[tex]\int \phi \nabla^2 \phi \,dV = \oint_S \phi \nabla \phi d\vec A -\int |\nabla \phi|^2 dV [/tex]
Where S is some closed surface.
Now when we are calculating the electrostatic potential energy, we have to calculate exactly this integral, but we know the end, that:
[tex]W=-\dfrac{\varepsilon_0}{2}\int \phi \nabla^2 \phi \,dV =\dfrac{\varepsilon_0}{2}\int |\nabla \phi|^2 dV [/tex]
So what I am wondering is why will, this be true:
[tex]\oint_S \phi \nabla \phi d\vec A=0[/tex]
If [tex]\phi[/tex] satisfies the Poisson equation??
[tex]\nabla^2 \phi=-\dfrac{\rho}{\varepsilon_0}[/tex]
Now I want to calculate the following integral:
[tex]\int \phi \nabla^2 \phi \,dV[/tex]
So using Greens first identity I get:
[tex]\int \phi \nabla^2 \phi \,dV = \oint_S \phi \nabla \phi d\vec A -\int |\nabla \phi|^2 dV [/tex]
Where S is some closed surface.
Now when we are calculating the electrostatic potential energy, we have to calculate exactly this integral, but we know the end, that:
[tex]W=-\dfrac{\varepsilon_0}{2}\int \phi \nabla^2 \phi \,dV =\dfrac{\varepsilon_0}{2}\int |\nabla \phi|^2 dV [/tex]
So what I am wondering is why will, this be true:
[tex]\oint_S \phi \nabla \phi d\vec A=0[/tex]
If [tex]\phi[/tex] satisfies the Poisson equation??