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## Main Question or Discussion Point

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??