A parallel plate capacitor with air between the plates has a capacitance of 12 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 $\frac{d}{3}$ while the other one has dielectric constant $k_{2} = 6$ and thickness $\frac{2 d}{3} .$ Capacitance of the

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

54 pF

36 pF

18 pF

72 pF

answer is C.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$The air filled capacitance , C = \frac{A ϵ_{0}}{d} = 12 pF$

According to question there are two capacitor with capacitances C₁ and C₂ and they are in series.

$Here, C_{1} = \frac{A k_{1} ϵ_{0}}{(d / 3)} = 9 \frac{A ϵ_{0}}{d} = 9 C as k_{1} = 3$

$and C_{2} = \frac{{Ak}_{2} ϵ_{0}}{(2 d / 3)} = 9 \frac{A ϵ_{0}}{d} = 9 C as k_{2} = 6$

$The equivalent capacitance, C_{eq} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{(9 C) (9 C)}{9 C + 9 C} = \frac{9}{2} C = (9 / 2) 12 = 54 pF$

$12 \times 3 + \frac{12 \times 3}{2} = 54 pF$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test