TF DerivationOfCoulombForce

Poisson's Equation

$\nabla^2 \phi(\vec{\xi}) = - \frac{\rho}{\epsilon_0} =- \frac{e}{\epsilon_0} \delta(\vec{\xi})$

Fourier Transform of Poisson's Equation

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

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

Product rule for dervatives

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

Gauss' Theorem:

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

Definition of derivative:

$(\vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}}) \cdot (\vec{\nabla} \phi ) = \vec{\nabla} \cdot (\phi \vec{\nabla} e^{-i \vec{k}}) - \phi {\nabla}^2 e^{-i \vec{k} \cdot \vec{\xi}}$

Substituting

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

Gauss' Low:

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