Q. The plates of a parallel plate capacitor are charged upto 200 V. A dielectric slab of thickness 4 mm is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by 3.2 mm. The dielectric constant of the dielectric slab is
A) 1
B) 4
C) 5
D) 6
Explanation
(c) $\frac{\varepsilon_{0}A}{d}\ = \ \frac{\varepsilon_{0}A}{d^{'} - t + \frac{t}{K}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d = d^{'} - t + \frac{1}{K}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d^{'} - d = t\left( 1 - \frac{1}{K} \right)$
Here, $d^{'} - d = 3.2\ mm,\ t = 4mm$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.2 = 4\left( 1 - \frac{1}{K} \right)1 - \frac{1}{K}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \frac{3.2}{4} = 1 - \frac{1}{K}$
$\Rightarrow \ \ \ \ \ \ 1 - \frac{1}{K} = \frac{4}{5}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K = 5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.