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A parallel plate capacitor \mathrm{C} with plates of unit area and separation \mathrm{d} is filled with a liquid of dielectric constant \mathrm{k=2}.The level of liquid is \mathrm{d / 3} initally. Suppose the liquid level decreases at a constant speed v, the time constant as a function of time \mathrm{t} is

 

Option: 1

\mathrm{\frac{6 \epsilon_{0} R}{5 d+3 v t}}


Option: 2

\mathrm{\frac{(15 d+9 v t) \epsilon _{0} R^{2}}{2 d^{2}-3 d v t-9 v^{2} t^{2}}}


Option: 3

\mathrm{\frac{16\epsilon _{0} R}{15 d-3 v t}}


Option: 4

\mathrm{\frac{(15 d-9 v t) \epsilon_{0} R}{12 d^{2}+3 d v t-9 v^{2} t^{2}}}


Answers (1)

best_answer

After time \mathrm{t}, thickness of liquid will remain \mathrm{\left\{\frac{d}{3}-v t\right\}}

Now, time constant  as a function of time:-

\mathrm{\tau_{c}=C R=\frac{\epsilon_{0}(1) \cdot R}{\left(\alpha-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}}}
\mathrm{\tau_{c}=\frac{6 \epsilon _{0} R}{5 d+3 v t} \quad\left\{\text { Apply } c=\frac{\epsilon _{0} A}{d-t+\frac{t}{k}}\right.}

 

Posted by

Ritika Kankaria

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