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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. `(6epsiolon_(0)R)/(5d+3Vt)`B. `((15d+9Vt)epsilon_(0)R)/(2d^(2)-3dVt-9V^(2)t^(3))`C. `(6epsilon_(0)R)/(5d-3Vt)`D. `((15d+9Vt)epsilon_(0)R)/(2d^(2)-3dVt-9V^(2)t^(3))` |
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Answer» Correct Answer - A Let the fall in the lavel in time `t` be `x`. Then `x=vt`. At that time the thickness of air in the capacitor is `d_(1) = ((2d)/(3)+vt)` and the thickness of liquid column is `d_(2) = ((d)/(3)-vt)`. At that time the equivalent capacitance is. `(1)/(C) =(1)/(C_(1)) +(1)/(C_(2)) =(((2d)/(3)+vt))/(in_(0)A) +(((d)/(3)-vt))/(Kin_(0)A)` put `A = 1` and `K = 2`, we get `C = (6in_(0))/(5d+3vt)` `tau = RC = (6in_(0)C )/(5d +3vt)` |
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