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IIT-JEE 2008 · Paper 2, Physics question 30

A parallel plate capacitor C with plates of unit area and separation dd is filled with a liquid of dielectric constant K=2K=2. The level of liquid is d3\frac d3 initially. Suppose the liquid level decreases at a constant speed VV, the time constant as a function of time tt is

A horizontal parallel-plate capacitor C (plate separation d) with liquid filling the lower part to a height d/3, connected in series with a resistor R and a battery.
Choose one option
Answer and worked solution

Official answer: A

Capacitance with liquid height h (unit plate area): a dielectric layer of thickness hh in series with air of thickness d−hd-h:
C=ε0(d−h)+hK=ε0d−h2.C=\frac{\varepsilon_0}{(d-h)+\frac hK}=\frac{\varepsilon_0}{d-\frac h2}.

Level at time t. h=d3−Vth=\frac d3-Vt, so d−h2=5d6+Vt2=5d+3Vt6d-\frac h2=\frac{5d}6+\frac{Vt}2=\frac{5d+3Vt}6.

Time constant. τ=RC=6ε0R5d+3Vt\tau=RC=\dfrac{6\varepsilon_0R}{5d+3Vt}.

(A)

Topics:DielectricsRC circuits
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