Revision as of 21:34, 6 December 2020 by Denneyl (Talk | contribs)

The Laplace operator has many applications in the physical sciences, one of which being in electric potentials. An electric field, $ E $, is defined as a vector field that describes the force of electricity per unit charge on any charge in the field. Take, for example, an electric field created by a point charge at $ (0,0) $. By Coulomb's law:

$ E = \frac{F}{q} \\ F = \large\frac{Qq}{4\pi\epsilon_{0}r^{2}} \\ E = \large\frac{Qq}{4\pi\epsilon_{0}r^{2}} \cdot \frac{1}{q} \\ E = \large\frac{Q}{4\pi\epsilon_{0}r^{2}} $

where $ Q $ is the charge of the point charge, $ q $ is the charge of a charge in the field, $ r $ is the distance of the charge from the point charge, and $ \epsilon_0 $ is vacuum permittivity, a physical constant approximately equal to $ 8.8 $ x $ 10^{-12} $ Farads per meter. If the point charge is 1 Coulomb, a 1 Coulomb charge 1 meter away from the point charge would have an electric force of $ \frac{1}{4\pi\epsilon_0} $ Newtons in the direction of the point charge.

Voltage, or electric potential, is how much work it would take to move an electric charge from one point in an electric field to another. When expressed as a scalar function relative to the origin, we get the following relation:

$ E = - \nabla V $

where $ V $ is the voltage function. Electric fields also have another relation, shown as follows:

$ \nabla \cdot E = \frac{\rho}{\epsilon_0} $

where $ \rho $ is the density of charge in space. Taking both relations into account, we get:

$ \nabla \cdot (-\nabla V) = \frac{\rho}{\epsilon_0} \\ \Delta V = -\Large\frac{\rho}{\epsilon_0} $

Going back to the intuition behind the Laplacian, $ \Delta V $ is low at local maxima of $ V $. Since $ \epsilon_0 $ is a constant, we can conclude from the inverse relationship between $ \Delta V $ and $ \rho $ that $ \rho $ is high at these local minima.

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