Question

68. A parallel plate air capacitor has a capacitance 18 uF. If
the distance between the plates is tripled and a dielectric
medium is introduced, the capacitance becomes 72 uF.
The dielectric constant of the medium is​

Answer 1

Answer:

Explanation: c = (A∈0/d) = 18

c' = [(A∈0k)/d'] = 72

∴ (c1/c) = (kd/d')

∴ (72/18) = [(k ∙ d)/(3d)]

------ (d' = 3d)

∴ 4 = (k/3)

k = 12 ←

Answer 2

Answer:

Dielectric constant of medium placed between parallel plate capacitor is 12.

Explanation:

capacitance of parallel plate capacitor is denoted by

 $c=\frac{\epsilon A}{d}$

where  ∈= permittivity      

      A = area of plates      

      d = distance between them

capacitance in presence of dielectric is  

$C' =\frac{k \epsilon A }{d'}$  

where k= dielectric constant

Given :   C =18μf  ,   C' = 72μf        

         d' = 3d

so we have  

 $C = \frac{\epsilon A}{d}\\18 = \frac{\epsilon A}{d}$        ---- (i)

also for dielectric

$C' =\frac{k\epsilon A}{d'}\\\\\\ 72 = \frac{k\epsilon A}{d'}$    --------(ii)

dividing both equations

$\frac{18}{72} =\frac{\epsilon A d'}{k \epsilon A d}$

$\frac{1}{4}=\frac{3d}{kd} \\\\k= 12$

hence dielectric constant of the medium is 12 .