A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of di-electric constant K=2. The level of the liquid is d/3 initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time t is?

Question

Chetan Mandayam Nayakar · Answer

let level of liquid be 'h'. Capacitance Ceff =((C1*C2)/(C1+C2), where C1=(Ko/(d-h)) and C2=2/d,at t=0,h=d/3,

C=∫(from h=d/3 to ((d/3)-vt))(dCeff/dt) dt dCeff/dt=(1/(C1+C2)²)((C1+C2)d(C1*C2)/dt -(C1*C2)d(C1+C2)/dt)

Samarth · Answer

Level of liquid = --t+d/3 C1 =E/(vt+2d/3) C2=E/(-vt+d/3) Ceff=C1×C2/(C1+C2) Ceff=6E/(5d-3vt) T=R(Ceff) T=6ER/(5d-3vt)

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A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of di-electric constant K=2. The level of the liquid is d/3 initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time t is?

A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of di-electric constant K=2. The level of the liquid is d/3 initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time t is?

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