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WBJEEA parallel plate capacitor \(C\) with plates of unit area and separation \(d\)...
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
Solution:
After time \(t\), thickness of liquid will remain \(\left(\frac{d}{3}-v t\right)\).
Now, time constant as function of time
\(\)
\begin{aligned}
\tau_c & =C R \\
& \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} \quad \quad \text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \\
& =\frac{6 \varepsilon_0 R}{5 d+3 v t}
\end{aligned}
\(\)
\(\therefore\) correct option is (a).
Now, time constant as function of time
\(\)
\begin{aligned}
\tau_c & =C R \\
& \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} \quad \quad \text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \\
& =\frac{6 \varepsilon_0 R}{5 d+3 v t}
\end{aligned}
\(\)
\(\therefore\) correct option is (a).
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