Question

When a dielectric slab of thickness 6cm is introduced between the plates of parallel plate condenser,it is found that the distance between the plates has to be increased by 4cm to restore the capacity to original value. The dielectric constant of the slab is ?

Answer 1

     for a parallel plate capacitor
             C = ε K   A / d
It is possible that we have air/vacuum and then a medium in between the parallel plates.  Initially there is a medium of dielectric constant K and of thickness d.

    Let  K1 be the dielectric constant of the slab. 
     let d1 be the thickness of slab = 6cm.

Formula for capacity with multiple media in between the plates is
             C =  ε A / [ d1/K1 + d2 / K ]
          total gap =  d1 + d2 = d + 4 cm          =>  d2 = d + 4 cm - 6 cm = d - 2
 
         d / K =  d1/ K1 + d2 / K
         d/ K = 6 / K1  + (d-2)/K

          d - (d-2)  = 6 K / K1
                K1 = 3 K
  If there was air originally in between the plates,  K = 1,  then answer is 3.

Answer 2

Answer:

Let initial capacitance be C

C= εo A/d

A and d are in cm

Final capacitance is same as initial capacitance.

d/εo A = d+4−6/ εo A + 6/KεoA

Solving, we get:

K=3