Question

If a slab of insulating material $4\times {10}^{-5}m$ thick is introduced between the plates of a parallel plate capacitor, the distance between the plates has to be increased by $3.5\times {10}^{-5}m$ to restore the capacity to original value. Then the dielectric constant of the material of slab is :

• A. $8$
• B. $6$
• C. $12$
• D. $10$

Answer 1

The correct option is A $8$

If $K$ be the dielectric constant, the distance increased due to introduce dielectric is $x=t-\frac{t}{K}=t(1-\frac{1}{K})$

where $t=$ thickness of dielectric.
According to question, $x=3.5\times {10}^{-5}$
$\Rightarrow t(1-\frac{1}{K})=3.5\times {10}^{-5}$

$\Rightarrow 1-\frac{1}{K}=\frac{3.5\times {10}^{-5}}{4\times {10}^{-5}}=0.875$

$\Rightarrow \frac{1}{K}=1-0.875=0.125$

$\Rightarrow K=8$