TF DerivationOfCoulombForce

Poisson's Equation

[math]\nabla^2 \phi(\vec{\xi}) = - \frac{\rho}{\epsilon_0} =- \frac{e}{\epsilon_0} \delta(\vec{\xi})[/math]

Fourier Transform of Poisson's Equation

[math]\frac{1}{(2 \pi)^{3/2}} \int e^{-i \vec{k} \cdot \vec{\xi}} \nabla^2 \phi(\vec{\xi})dV = - \frac{1}{(2 \pi)^{3/2}} \frac{e}{\epsilon_0} \int e^{-i \vec{k} \cdot \vec{\xi}}\delta(\vec{\xi}) dV [/math]

[math]\frac{1}{(2 \pi)^{3/2}} \int e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \cdot (\vec{\nabla} \phi(\vec{\xi}))dV = - \frac{e}{(2 \pi)^{3/2}\epsilon_0} (1)[/math]

Product rule for dervatives

[math]\frac{1}{(2 \pi)^{3/2}} \int \left \{ \vec{\nabla} \cdot ( e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi ) - (\vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}}) \cdot (\vec{\nabla} \phi) \right \} dV = - \frac{e}{(2 \pi)^{3/2}\epsilon_0} (1)[/math]

Gauss' Theorem:

[math]\int \vec{\nabla} \cdot ( e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi ) dV = \oint_S e^{-i \vec{k}\cdot \vec{\xi}} \vec{\nabla}\cdot d\vec{A}[/math]

Definition of derivative:

[math](\vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}}) \cdot (\vec{\nabla} \phi ) = \vec{\nabla} \cdot (\phi \vec{\nabla} e^{-i \vec{k}}) - \phi {\nabla}^2 e^{-i \vec{k} \cdot \vec{\xi}}[/math]

Substituting

[math]\frac{1}{(2 \pi)^{3/2} } \left \{ \int e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi \cdot d\vec{A} - \int \vec{\nabla} \cdot (\phi \vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}} ) dV + \int \phi {\nabla}^2 e^{-i \vec{k} \cdot \vec{\xi}} dV \right \} = \frac{-e}{2 \pi)^{3/2} \epsilon_0}[/math]

Gauss' Low:

[math]\int \vec{\nabla}\cdot (\phi \vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}} ) dV = \int \phi \vec{\nabla} e^{-i k \xi } \cdot d\vec{A}[/math]

[math]\frac{1}{(2 \pi)^{3/2} } \left \{ \int e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi - \phi \vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}} \right \}[/math]