Q. The capacitance of an air capacitor is $15\mu F$ the separation between the parallel plates is $6mm.$ A copper plate of $3mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes
A) $5\mu F$
B) $7.5\mu F$
C) $22.5\mu F$
D) $30\mu F$
Explanation
(d) By using $C_{air} = \frac{\varepsilon_{0}A}{d},\ C_{medium} = \frac{\varepsilon_{0}A}{d - t + \frac{t}{K}}$
For $K = \infty\ \ C_{medium} = \frac{\varepsilon_{0}A}{d - t}$
$\Rightarrow \frac{C_{m}}{C_{a}} = \frac{d}{d - t} \Rightarrow \frac{C_{m}}{15} = \frac{6}{6 - 3} \Rightarrow C_{m} = 30\mu\ C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.