Q. In a capacitor of capacitance $20\mu\ F,$ the distance between the plates is $2mm.$ If a dielectric of width $1\ mm$ and dielectric constant 2 is inserted between the plates, then the new capacitance is
A) $2\mu\ F$
B) $15.5\mu\ F$
C) $26.6\mu\ F$
D) $32\mu\ F$
Explanation
(c) $C = \frac{\varepsilon_{0}A}{d}$ and $C^{'} = \frac{\varepsilon_{0}A}{\left( d - t + \frac{t}{K} \right)} \Rightarrow \frac{C}{C'} = \frac{\left( d - t + \frac{t}{K} \right)}{d}$
$\Rightarrow \frac{20}{C'} = \frac{\left( 2 \times 10^{- 3} - 1 \times 10^{- 3} + \frac{1 \times 10^{- 3}}{2} \right)}{2 \times 10^{- 3}} \Rightarrow C^{'} = 26.6\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.