APPLICATION OF GAUSS LAW

CAPACITANCE OF A PARALLEL PLATE CAPACITOR WHEN DIELECTRIC SLAB IS INTRODUCED BETWEEN THE SLAB

LET ,E₀ is applied electric field

                        E₀  =  q/ є₀A     -----------------------(1)

                        E = Induced electric field

             Potential difference across the capacitor

                      V = E  ×t + E₀ (d-t)

            From definition of dielectric constant

                     K = E₀ / E

                Or  E = E₀/K

            On substituting the value of E

                V =   E₀t/K +  E₀ (d-t)

                 V   = E₀ [t/k + (d-t)]              -------------------(2)

          As     c= q/V             ----------------------(3)

           Now from equation (1)  in (2)

                 V = q/ є₀A  [t/k + (d-t)]            ----------------(4)            

Now, Equation (4) in (3)

               c = q/ (q/ є₀A  [t/k + (d-t)])           

                c = є₀A/[t/k + (d-t)]           

 If  t = d

       So  c = k є₀A/d

When battery is removed & dielectric slab is introduced the charge will be same.