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JEE Advanced · Physics · 18. Capacitance

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

  1. A \(\frac{6 \varepsilon_0 R}{5 d+3 v t}\)
  2. B \(\frac{(15 d+9 v t) \varepsilon_0 R}{2 d^2-3 d v t-9 v^2 t^2}\)
  3. C \(\frac{6 \varepsilon_0 R}{5 d-3 v t}\)
  4. D \(\frac{(15 d-9 v t) \varepsilon_0 R}{2 d^2+3 d v t-9 v^2 t^2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{6 \varepsilon_0 R}{5 d+3 v t}\)

Step-by-step Solution

Detailed explanation

After time \(t\), thickness of liquid will remain \(\left(\frac{d}{3}-v t\right)\).
Now, time constant as function of time
\(\tau_c =C R \)
\( \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}}~\text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \)
\( =\frac{6 \varepsilon_0 R}{5 d+3 v t}\)
\(\therefore\) correct option is (a).
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