Derivation of Poisson's Equation and Laplace's Equation

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SUMMARY

This discussion focuses on the derivation of Poisson's Equation and Laplace's Equation from Maxwell's equations, specifically in electrostatic systems. The relevant Maxwell's equations are identified as \(\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon}\) and \(\vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}\). In a static scenario where the magnetic field does not change, it is established that \(\vec{\nabla} \times \vec{E} = 0\), allowing the electric field \(\vec{E}\) to be expressed as the gradient of a scalar potential \(\vec{E} = - \vec{\nabla} \varphi\). This leads to the formulation of Poisson's Equation as \(\nabla^2 \varphi = - \frac{\rho}{\epsilon}\) and identifies Laplace's Equation as the case where \(\rho = 0\).

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MadMike1986
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Hi,

Can someone point me in the right direction to a derivation of Poisson's Equation and of Laplace's Equation, (from Maxwell's equations I think) both in a vacuum and in material media?

How does one get from Maxwell's equations to Poisson's and Laplace's?
 
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So the two relevant maxwell's equations are:
\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon}
\vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}

For an electrostatic system, there is no changing B field so,
\vec{\nabla} \times \vec{E} = 0
Which implies E can be written as the gradient of a scalar potential,
\vec{E} = - \vec{\nabla} \varphi

Combining this fact with the first equation,
\vec{\nabla} \cdot \vec{\nabla} \varphi = \nabla^2 \varphi = - \frac{\rho}{\epsilon}

And of course Laplace's equation is the special case where rho is zero.

Cheers!
 
Ah, thank you very much. That's not so bad after all.
 

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