Question

If a slab of insulating material (conceptual). $4 \times 10^{-3} m$ thick is introduced between the plates of a parallel plate capacitor, the separation between the plates has to be increased by $3.5 \times 10^{-3} m$ to restore the capacity to original value. The dielectric constant of the material will be

• A. 6
• B. 8
• C. 10
• D. 12

Answer 1

Answer: (B)

If $k$ be the dielectric constant, the distance increased due to introduced dielectric is $x = t -\frac{ t }{ k } \Rightarrow x = t \left(1-\frac{1}{ k }\right)$
Where, $t \rightarrow$ thickness of the dielectric.
According to the question,
$ x =3.5 \times 10^{-3} m $
$ t \left(1-\frac{1}{ k }\right)=3.5 \times 10^{-3} $
$1-\frac{1}{ k }=\frac{3.5 \times 10}{4 \times 10}=\frac{3.5}{4} $
$\frac{1}{ k }=1-\frac{35}{4}=\frac{0.5}{4}=0.125$
$k =8$