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

Book Online Demo

Check Your IQ

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

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

answer is 8.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

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