Question
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 $$v,$$ 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 dielectric constant $$K = 2.$$ The level of liquid is $$\frac{d}{3}$$ initially. Suppose the liquid level decreases at a constant speed $$v,$$ the time constant as a function of time $$t$$ is-
A.
$$\frac{{6{\varepsilon _0}R}}{{5d + 3vt}}$$
B.
$$\frac{{\left( {15d + 9vt} \right){\varepsilon _0}R}}{{2{d^2} - 3dvt - 9{v^2}{t^2}}}$$
C.
$$\frac{{6{\varepsilon _0}R}}{{5d - 3vt}}$$
D.
$$\frac{{\left( {15d - 9vt} \right){\varepsilon _0}R}}{{2{d^2} - 3dvt - 9{v^2}{t^2}}}$$
Answer :
$$\frac{{6{\varepsilon _0}R}}{{5d + 3vt}}$$
Solution :
Let the level pf liquid at an instant of time $$'t'$$ be $$x.$$ Then
$$\eqalign{ & v = - \frac{{dx}}{{dt}} \Rightarrow dx = - vdt \cr & \Rightarrow \int\limits_{\frac{d}{3}}^x {dx = - v\int\limits_0^t {dt} } \cr & \Rightarrow x - \frac{d}{3} = - vt \cr & \Rightarrow x = \frac{d}{3} - vt \cr} $$
Also the capacitance can be considered as an equivalent of two capacitances in series such that
$$\eqalign{ & \frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} \cr & \Rightarrow \frac{1}{{{C_{eq}}}} = \frac{1}{{\frac{{{ \in _0}A}}{{d - x}}}} + \frac{1}{{\frac{{{ \in _O}AK}}{x}}} = \frac{{d - x}}{{{ \in _0}A}} + \frac{x}{{{ \in _o}AK}} \cr & \therefore {C_{eq}} = \frac{{{ \in _0}AK}}{{Kd + x\left( {1 - K} \right)}} \cr & {\text{But }}A = 1,K = 2{\text{ and }}x = \frac{d}{3} - vt \cr & \therefore {C_{eq}} = \frac{{{ \in _0} \times 1 \times 2}}{{2d + \left[ {\frac{d}{3} - vt} \right]\left( {1 - 2} \right)}} = \frac{{6{ \in _0}}}{{5d + 3vt}} \cr & \therefore {\text{Time constant }}\tau = R{C_{eq}} = \frac{{6R{ \in _0}}}{{5d + 3vt}} \cr} $$
Let the level pf liquid at an instant of time $$'t'$$ be $$x.$$ Then
$$\eqalign{ & v = - \frac{{dx}}{{dt}} \Rightarrow dx = - vdt \cr & \Rightarrow \int\limits_{\frac{d}{3}}^x {dx = - v\int\limits_0^t {dt} } \cr & \Rightarrow x - \frac{d}{3} = - vt \cr & \Rightarrow x = \frac{d}{3} - vt \cr} $$
Also the capacitance can be considered as an equivalent of two capacitances in series such that
$$\eqalign{ & \frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} \cr & \Rightarrow \frac{1}{{{C_{eq}}}} = \frac{1}{{\frac{{{ \in _0}A}}{{d - x}}}} + \frac{1}{{\frac{{{ \in _O}AK}}{x}}} = \frac{{d - x}}{{{ \in _0}A}} + \frac{x}{{{ \in _o}AK}} \cr & \therefore {C_{eq}} = \frac{{{ \in _0}AK}}{{Kd + x\left( {1 - K} \right)}} \cr & {\text{But }}A = 1,K = 2{\text{ and }}x = \frac{d}{3} - vt \cr & \therefore {C_{eq}} = \frac{{{ \in _0} \times 1 \times 2}}{{2d + \left[ {\frac{d}{3} - vt} \right]\left( {1 - 2} \right)}} = \frac{{6{ \in _0}}}{{5d + 3vt}} \cr & \therefore {\text{Time constant }}\tau = R{C_{eq}} = \frac{{6R{ \in _0}}}{{5d + 3vt}} \cr} $$
