Q. The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4\ cm$ is $2\mu F.$ The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is
A) $11.2\mu F$
B) $15.6\mu F$
C) $19.2\mu F$
D) $22.4\mu F$
Explanation
(a) $C = \frac{\varepsilon_{0}KA}{d} \Rightarrow \frac{C_{1}}{C_{2}} = \frac{K_{1}}{K_{2}} \times \frac{d_{2}}{d_{1}}$
$\frac{2}{C_{2}} = \frac{1}{2.8} \times \frac{(0.4/2)}{(0.4)} \Rightarrow C_{2} = 11.2\mu\ F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.