Electrostatics Ques 111
- 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
$(2008,3 M)$
(a) $\frac{6 \varepsilon_{0} R}{5 d+3 v t}$
(b) $\frac{(15 d+9 v t) \varepsilon_{0} R}{2 d^{2}-3 d v t-9 v^{2} t^{2}}$
(c) $\frac{6 \varepsilon_{0} R}{5 d-3 v t}$
(d) $\frac{(15 d-9 v t) \varepsilon_{0} R}{2 d^{2}+3 d v t-9 v^{2} t^{2}}$
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Answer:
Correct Answer: 111.(a)
Solution:
Formula:
After time $t$, thickness of liquid will remain $\frac{d}{3}-v t$.
Now, time constant as a function of time
$ \tau_{c}=C R=\frac{\varepsilon_{0} \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} $
$ Applying $ C=\frac{\varepsilon_{0} A}{d-\frac{t}{k}} $ $
$ =\frac{6 \varepsilon_{0} R}{5 d+3 v t} $
$\therefore$ Correct option is (a).
BETA
