Question

PART B
5. a) Define electric displacement vector D and deduce Gauss's law for dielectrics.​

Answer 1

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Explanation:

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Answer 2

ELECTRIC DISPLACEMENT is the charge per unit area that would be displaced across a layer of conductor placed across an electric field.

.) It is the VECTOR QUANTITY

.) It is denoted by 'D'

.) SI unit of ELECTRIC DISPLACEMENT is COULOMB per unit AREA ie C/m^2.

.) It is also known as FLUX DENSITY

.) Now we know that acc to gauss law,

Φ = § Eo.A = q/€o

Or,

Eo = q/A€o --------(1)

Where Φ is the flux , Eo is the electric field when no diectric is placed ,A is the surface area, q is the charge and €o is permeability in air/vacuum.

.) Now let E be the electric filed when a dielectric is placed such that , K = €/€o

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

.) Now,

E = q/A€ -----------(2)

Now dividing (1) and (2) , we get

E/Eo = €o / €

=> E/Eo = 1/K

Hence we get ,

E = Eo / K

=> E = q / KA€o