Coulomb's Law

Key Questions

  • Question #d2176

    \vec{F}_{12} = 1/(4\pi\epsilon_o)(q_1q_2)/|\vec{r}_1-\vec{r}_2|^2=-\vec{F}_21

    Cosmic Defect · 2 · Dec 17 2014
  • Can Coulomb's law be used to derive Gauss's law? How?

    Answer:

    I shall outline a better method. Not just by assuming the surface is spherical with area 4pir^2.

    So, to start with I shall prove it for the field due to a single point charge. (At O)
    Since superposition principle applies to electric fields, this can be generalised to any number of charges quite easily.

    There it is, given below.

    Explanation:

    Consider a charged particle q placed at O and is being surrounded and enclosed by a closed surface S. (Of arbitrary size and shape).

    Take an arbitrary surface element dS on the surface at a distance r from O.

    Let the normal on dS make an angle theta with the electric field of q passing through dS.

    Thus, vec E*dvecS = EdS Cos theta which is equal to the flux through the small area.

    Now, by Coulomb's law,

    E = q/(4piepsilon_0r^2)

    Which gives, dphi = q/(4piepsilon_0)*dScos theta/(r^2)

    Now, dS Cos theta/(r^2) is the solid angle which can be denoted as, domega

    Thus, the flux through dS is,

    dphi = q/(4piepsilon_0)*domega

    Integrating over the entire surface,

    int int_S vecE*dvecS = q/(4piepsilon_0) int domega

    But, the net solid angle subtended by a closed surface at an internal point is always 4pi.

    int int_S vecE*dvecS = q/(4piepsilon_0)*4pi

    Thus,
    int int_S vecE*dvecS = q/epsilon_0

    Which is Gauss law in integral form.

    One may derive the differential form from Coulomb's law as well by taking the Divergence of the electric field and then proceeding with vector calculus methods.

    The differential form looks something like this, nabla*vec E = rho/epsilon_0 where rho is the charge density.

    Aritra G. · 8 · Mar 25 2017
  • What is the Coulomb's law?

    Answer:

    F=k(Q_1,Q_2)/r^2

    Explanation:

    assume there are two charges Q_1, Q_2

    the force between them can be found through this relation

    F=k(Q_1*Q_2)/r^2

    Where k is Coulomb's constant
    and r is the distance between the two charges

    and the force is attraction if the two charges have different sign
    (+ve and -ve)
    and repulsion if the two charges have the same sign

    Yahia M. · 1 · Apr 21 2018

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Electrical Energy and Current