differential form of gauss's law is....
Answers
Explanation:
It states that the divergence of the electric field at any point is just a measure of the charge density there.
it is the differential form of gauss law
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Answer:
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