A container has a base of 50 cm × 5 cm and height 50 cm, as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm × 50 cm. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant 3 at a uniform rate of 250cm3s−1. What is the value of the capacitance of the container after 10 seconds? [Given: Permittivity of free space ϵ0=9×10−12C2N−1m−2, the effects of the non-conducting walls on the capacitance are negligible]
(A)27 pF
(B)63 pF✓ Correct
(C)81 pF
(D)135 pF
Answer: (B)
Solution
After 10 seconds, volume of liquid filled: V=250×10=2500 cm³.
Height of liquid: h=50×52500=10 cm.
The capacitor has two regions in series:
Lower part (with dielectric, height 10 cm): Cd=dAϵ0k=5×10−250×10−2×10×10−2×ϵ0×3=3ϵ0
Upper part (air, height 40 cm): Ca=dAϵ0=5×10−250×10−2×40×10−2×ϵ0=4ϵ0
Total capacitance (parallel combination): C=Cd+Ca=7ϵ0=7×9×10−12=63 pF.
JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option
