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→Coulomb's Law: Fill it in!
The first equation is just [[C. A. Coulomb|Coulomb]]'s law of electrostatics, manipulated very elegantly (as usual) by [[Michael Faraday|Faraday]] and [[Gauss]]. Coulomb's law simply says that the electric force between two charged particles acts in the direction of the line between them, is attracting if they have like charges and repelling if unlike, is proportional to the product of the charges, and is inversely proportional to the square of the distance between them:
The first equation is just [[C. A. Coulomb|Coulomb]]'s law of electrostatics, manipulated very elegantly (as usual) by [[Michael Faraday|Faraday]] and [[Gauss]]. Coulomb's law simply says that the electric force between two charged particles acts in the direction of the line between them, is attracting if they have like charges and repelling if unlike, is proportional to the product of the charges, and is inversely proportional to the square of the distance between them:
:<math>F = \frac{q_1 q_2}{4 \pi \epsilon\ d^2}</math>
:<math>F = \frac{q_1 q_2}{4 \pi \epsilon\ d^2}</math>
In this case, the constant defining the strength of the electric force is <math>4 \pi \epsilon</math> in the denominator. More about that presently.
In [[International_System_of_Units|SI units]] the charges are measured in [[International_System_of_Units#Coulomb|Coulombs]], the force in [[International_System_of_Units#Newton|Newtons]], the distance in [[International_System_of_Units#Meter|Meters]], and the value of <math>\epsilon</math> is <math>8.854 \times 10^{-12}</math> Coulombs<sup>2</sup> per Newton meter<sup>2</sup>, or [[International_System_of_Units#Farad|Farads]] per meter.
Michael Faraday reformulated the electric and magnetic forces in terms of ''fields'' He said that what was really happening was that each charge was creating and electric field (called E) that acted on the other charge. The field created by the charge q<sub>1</sub>, as observed at distance d, is
:<math>E = \frac{q_1}{4 \pi \epsilon\ d^2}</math>
and points directly outward from that charge, in all directions. The force felt by charge q<sub>2</sub> is
:<math>F = q_2 E</math>
Now consider a sphere of radius d with the charge at the center. If <math>\rho</math> is the charge density in Coulombs per cubic meter (Maxwell's equations are in terms of densities), the total charge in some volume is the integral, over that volume, of <math>\rho</math>.
So we have
:<math>q = \int_V \rho\, \mathrm{d}V</math>
Now the field at the surface of the sphere is
:<math>\frac{q_1}{4 \pi \epsilon\ d^2}</math>, or <math>\frac{1}{4 \pi \epsilon\ d^2} \int_V \rho\, \mathrm{d}V</math>
That field is directly outward, perpendicular to the sphere's surface, and is uniform over the surface. The integral of the field over the surface is <math>4 \pi d^2</math> times that (the surface area of the sphere is <math>4 \pi d^2</math>; this is why we have the pesky factor of <math>4 \pi</math> in various formulas; remember that d is the distance, and hence is the sphere's ''radius'', not its diameter), so
:<math>\int_V \frac{\rho}{\epsilon}\, \mathrm{d}V = \oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A}</math>
But, by Gauss's Theorem,
:<math>\oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A} = \int_V \nabla \cdot \mathbf{E}\ \mathrm{d}V</math>
So
:<math>\int_V \nabla \cdot \mathbf{E}\ \mathrm{d}V = \int_V \frac{\rho}{\epsilon}\, \mathrm{d}V</math>
Since this is true for any volume, we have
:<math>\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon}</math>
Now D = <math>\epsilon\ E</math> in the straightforward case (more about that later), so
:<math>\nabla \cdot \mathbf{D} = \rho</math>
==Other Formulations==
==Other Formulations==