Q. A parallel plate capacitor with air between the plates has a capacitance of $9\ pF.$ The separation between its plates is $'d^{'}.$ The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_{1} = 3$ and thickness $d/3$ while the other one has dielectric constant $k_{2} = 6$ and thickness $2d/3.$ Capacitance of the capacitor is now
A) $45\ pF$
B) $40.5\ pF$
C) $20.25\ pF$
D) $1.8\ pF$
Explanation
(b) [image:1_image334.emf]
$C_{air} = \frac{\varepsilon_{0A}}{d} = 9,\frac{1}{C_{med}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{d_{1}}{k_{1}\varepsilon_{0}A} + \frac{d_{2}}{k_{2}\varepsilon_{0}A}$
$\Rightarrow C_{med} = \frac{k_{1}k_{2}\varepsilon_{0}A}{k_{1}d_{2} + k_{2}d_{1}}$
$= \frac{3 \times 6 \times \varepsilon_{0}A}{3 \times 2d/3 + 6 \times d/3} = \frac{18}{4} \times 9 = 40.5\ pF$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.