Gauss law in differential form derivation

Last time I have derived and discussed the gauss law for electrostatics. Let us today derive and discuss the gauss law for electrostatics in differential form.

STATEMENT:-Differential form of Gauss law states that the divergence of electric field E at any point in space is equal to 1/ε0 times the volume charge density,ρ, at that point.

Del.E=ρ/ε0

Where ρ is the volume charge density (charge per unit volume) and ε0 the permittivity of free space.It is one of the Maxwell’s equation.

Derivation or Proof .Consider a region of continuous charge distribution with varrying volume density of charge ρ(charge per unit volume).In this region,consider a volume V enclosed by the surface S.if dV is an infinitesimal small volume element enclosed by the surface dS,then according to Gauss’s law for a continuous charge distribution

∫E.dS=1/ε0∫ρdV                                                    (1)

According to Gauss-divergence theorem

∫E.dS=∫(del.E)dV                                                (2)

By comparing equations(1) and (2),we get

∫(del.E)dV=1/ε0∫ρdV

Or                               ∫[del.E-ρ/ε0] dV=0

As the volume under consideration is arbitrary, therefore,

Del.E-ρ/ε0=0

Or                       del.E=ρ/ε0 (3)

Equation (3) is the differential form of Gauss’s law in differential form or a continuous charge distribution in M.K.S system. The physical meaning of this differential form of Gauss’s law is that it relates the electric field at a point in space to the charge distribution ρ at that point in space.

In C.G.S Gaussian system Gauss’s law can be expressed as

Del.E=4πρ

You can also see the gauss law for electrostatics derivation or proof.

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One Response to Gauss law in differential form derivation

  1. Nasir shah says:

    it’s very easy

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