Question

Define electric displacement vector D vector and deduce Gauss's law for dielectrics

Answer 1

Electric displacement vector D  is the charge per unit area that would be displaced across a layer of conductor placed across an electric field.  It is also known as flux density.

Now we know that according to gauss law,

$$\Phi=\int E_0dA=\frac{q}{\epsilon_0}\\E_0=\frac{q}{A\epsilon_0}$$

Where Φ is the flux , E₀ is the electric field when no diectric is placed , A is the surface area, q is the charge and €_o is permeability in vacuum.

Now let E be the electric filed when a dielectric is placed such that ,

$$\epsilon=k\epsilon_0$$

Where k is the dielectric constant and € is the permiability in the dielectric.

Now

$$E=\frac{q}{A\epsilon}=\frac{q}{kA\epsilon_0}$$