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A parallel plate capacitor C with plates...

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 `d//3` initially. Suppose the liquid level decreases at a constant speed v, the time constant as a function of time t is-

A

(a) `(6epislon_0R)/(5d+3vt)`

B

(b) `((15d+9vt)epsilon_0R)/(2d^2-3dvt-9v^2t^2)`

C

(c) `(6epsilon_0R)/(5d-3vt)`

D

(d) `((15d-9vt)epsilon_0R)/(2d^2-3dvt-9v^2t^2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Let the level of liquid at an instant of time 't' be x. Then
`V=-(dx)/(dt)impliesdx=-vdt`
`implies int_(d//3)^xdx=-vint_0^tdt`
`impliesx-d/3=vt`
`impliesx=d/3-vt` ltbrrgt
Also the capacitance can be conisdered as an equivalent of two capacitances in series such that
`(1)/(C_(eq))=(1)/(C_1)+(1)/(C_2)`
`implies(1)/(C_(eq))=(1)/((in_0A)/(d-x))+(1)/((in_0AK)/(x))=(d-x)/(in_0A)+(x)/(in_0AK)`
`:.` `C_(eq)=(in_0AK)/(Kd+x(1-K))`
But `A=1`, `K=2` and `x=d/3-vt`
`:.` `C_(eq)=(in_0xx1xx2)/(2d+[d/3-vt](1-2))=(6in_0)/(5d+3vt)`
`:.` Time constant `tau=RC_(eq)=(6Rin_0)/(5d+3vt)`
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