Simply Lecture

The Maxwell first equation in electrostatics is called the Gauss law in electrostatics. Statement:  It states that the total electric flux \(\psi_E\) passing through a closed hypothetical surface is equal to \(\frac{1}{\epsilon_0}\) enclosed by the surface. Integral Form: $$\phi_E = \int E.ds = \frac{q}{\epsilon_0}$$ $$\int D.ds = q$$ where, $$D = \epsilon_0 E = displacement-vector$$ Let the change be distributed over a volume v and \(\rho\) be the volume charge density. Hence, $$q = \int \rho dv$$ Therefore; $$\int D.ds = \int_{v} \rho dv$$ .........(1) Equ(1) is the integral form of Maxwell first law Differential form: Apply Gauss divergence theorem to the L.H.S of equ(1) from surface integral to volume integral. $$\int D.ds = \int (\nabla.D)dv$$ Substituting this equation to equ(1) $$\int(\nabla.D)dv = \int_{v} \rho dv$$ As two volume integrals are equal only if their integrands are equal. Thus; $$\nabla.D = \rho v$$ ............(2) Equ(2) is the dif...