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. $\frac{6 \varepsilon_{6} R}{5 d+3 V t}$
• B. $\frac{(15 d +9 Vt ) \varepsilon_{0} R }{2 d ^{2}-3 d V t -9 V ^{2} t ^{2}}$
• C. $\frac{6 \varepsilon_{0} R}{5 d-3 V t}$
• D. $\frac{(15 d -9 Vt ) \varepsilon_{0} R }{2 d ^{2}+3 d Vt -9 V ^{2} t ^{2}}$

Answer 1

Answer: (A)

$C_{\text {equivielet }}=\frac{\frac{2 \varepsilon_{0}}{\frac{ d }{3}- vt } \cdot \frac{\varepsilon_{0}}{\frac{2 d }{3}+ vt }}{\frac{2 \varepsilon_{0}}{\frac{ d }{3}- vt }+\frac{\varepsilon_{0}}{\frac{2 d }{3}+ vt }}$
$\therefore \tau= C _{\text {equivalent }} R$