Question

A dielectric slab of thickness is placed between the plates of a parallel plate capacitor. If the distance between plates is reduced by , the capacity of the capacitor remains the same. Find the dielectric constant of the medium.

A


B


C


D


Medium

Solution

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Correct option is D)

 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.

Solve any question of Electrostatic Potential and Capacitance with:-

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