A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K=2. The level of liquid is d3 initially. Suppose the liquid level decreases at a constant speed υ, the time constant as a function of time τ is 6ε0RPd+q v t then (p+q) is

C1=∈0A2d3+Vt ; C2=K∈0Ad3−VtCeq=ϵ0KAd3-Vt·ϵ0 A2 d3+Vtϵ0KAd3-Vt+ϵ0 A2 d3+Vt=ϵ0KAK2 d3+Vt+d3-Vt=2ϵ04d3+2Vt+d3-Vt=2ϵ05 d3+Vt=6ϵ05 d+3Vtp = 5 , q = 3p+q = 8

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant $K = 2$ . The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed $υ$ , the time constant as a function of time $τ$ is $\frac{6 ε_{0} R}{P d + q v t}$ then $(p + q)$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 8.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$C_{1} = \frac{\in_{0} A}{(\frac{2 d}{3} + V t)}; C_{2} = \frac{K \in_{0} A}{(\frac{d}{3} - V t)}$

$C_{eq} = \frac{\frac{ϵ_{0} KA}{\frac{d}{3} - Vt} \cdot \frac{ϵ_{0} A}{\frac{2 d}{3} + Vt}}{\frac{ϵ_{0} KA}{\frac{d}{3} - Vt} + \frac{ϵ_{0} A}{\frac{2 d}{3} + Vt}} = \frac{ϵ_{0} KA}{K (\frac{2 d}{3} + Vt) + \frac{d}{3} - Vt}$