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What is Coulomb's law

  1. Jul 24, 2014 #1

    Coulomb's law is an inverse-square law stating that the force vector between two stationary charges is a constant times the unit vector between them and times the product of the magnitudes of the charges divided by the square of the distance between them: [itex]\mathbf{F}_{12}\ \propto\ Q_1Q_2\,\mathbf{\hat{r}}_{12}/r_{12}^2[/itex]

    That constant (Coulomb's constant) is the same in any material, and is [itex]1/4\,\pi\,\varepsilon_0[/itex], where [itex]\varepsilon_0[/itex] is the electric constant (the permittivity of the vacuum, with dimensions of charge²/force.area, and measured in units of farads/metre), and [itex]4\,\pi[/itex] is the ratio between the surface area of a sphere and the radius squared.

    If the charges have the same sign, then [itex]Q_1Q_2[/itex] is positive, and the force vector points outward (the force is repulsive); if they have opposite signs, then [itex]Q_1Q_2[/itex] is negative, and the force vector points inward (the force is attractive).

    Gauss' law (one of Maxwell's equations) may be derived from Coulomb's law.


    Force on stationary charge 2 from stationary charge 1:

    [tex]\mathbf{F}_{12}\ =\ \frac{Q_1\,Q_2}{4\,\pi\,\varepsilon_0\,r_{12}^2}\,\mathbf{\hat{r}}_{12}[/tex]

    Electric field of charge 1 at position of charge 2 (from Lorentz force equation):

    [tex]\mathbf{E}_{12}\ =\ \frac{\mathbf{F}_{12}}{Q_2}\ =\ \frac{Q_1}{4\,\pi\,\varepsilon_0\,r_{12}^2}\,\mathbf{\hat{r}}_{12}[/tex]

    Since this is independent of the magnitude of charge 2, it may be rewritten:

    [tex]\mathbf{E}_1(\mathbf{r})\ =\ \frac{Q_1}{4\,\pi\,\varepsilon_0\,r^2}\,\mathbf{\hat{r}}\ =\ \frac{Q_1}{\varepsilon_0\,A(r)}\,\mathbf{\hat{r}}[/tex]

    where [itex]A(r)[/itex] is the surface area of the sphere [itex]S(r)[/itex] through [itex]\mathbf{r}[/itex] with charge 1 at its centre.

    Obviously, the divergence of [itex]\mathbf{E}_1[/itex] at any point other than the position of charge 1 is zero (differential form of Gauss' law for zero charge density [itex]\rho[/itex]):

    [tex]\nabla\cdot\mathbf{E}_1\ =\ 0\ \ \text{if}\ \ \rho\ =\ 0[/tex]

    And the flux of [itex]\mathbf{E}_1[/itex] through the sphere [itex]S(r)[/itex] is [itex]Q_1/\varepsilon_0[/itex]:

    [tex]\oint_{S(r)}\,\mathbf{E}_1\cdot(\mathbf{\hat{r}}\,dA)\ =\ \frac{Q_1}{\varepsilon_0 \,A(r)}\,\oint_{S(r)}\,\mathbf{\hat{r}}\cdot\mathbf{\hat{r}}\,dA\ =\ \frac{Q_1}{\varepsilon_0 \,A(r)}\,\oint_{S(r)}dA\ =\ \frac{Q_1}{\varepsilon_0}[/tex]

    and so, from Stoke's theorem, the flux of [itex]\mathbf{E}_1[/itex] through any closed surface S containing charge 1 is [itex]Q_1/\varepsilon_0[/itex] and through any other closed surface is zero:

    [tex]\oint_S \, \mathbf{E}_1 \cdot(\mathbf{\hat{n}} \, dA)\ =\ \left\{\begin{array}{cc}
    Q_1/\varepsilon_0 & \text{if S contains charge 1}\\
    0 & \text{if S does not contain charge 1}\end{array}\right.[/tex]

    Extended explanation

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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