A parallel plate capacitor C with plates of u | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The capacitor can be modeled as two capacitors in series: one filled with dielectric $(K=2)$ and one with air $(K=1)$.Let the thickness of the die

Vedclass · Accepted Answer

$\frac{6 \varepsilon_0 R}{5 d+3 Vt}$

Vedclass · Answer

(A) The capacitor can be modeled as two capacitors in series: one filled with dielectric $(K=2)$ and one with air $(K=1)$.Let the thickness of the dielectric be $x(t) = \frac{d}{3} - Vt$ and the thickness of the air gap be $y(t) = d - x(t) = \frac{2d}{3} + Vt$.The capacitance of the dielectric part is $C_1 = \frac{K \varepsilon_0 A}{x(t)} = \frac{2 \varepsilon_0}{(\frac{d}{3} - Vt)}$.The capacitance of the air part is $C_2 = \frac{\varepsilon_0 A}{y(t)} = \frac{\varepsilon_0}{(\frac{2d}{3} + Vt)}$.Since they are in series,the equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.$C_{eq} = \frac{\frac{2 \varepsilon_0}{\frac{d}{3} - Vt} \cdot \frac{\varepsilon_0}{\frac{2d}{3} + Vt}}{\frac{2 \varepsilon_0}{\frac{d}{3} - Vt} + \frac{\varepsilon_0}{\frac{2d}{3} + Vt}} = \frac{2 \varepsilon_0^2 / [(\frac{d}{3} - Vt)(\frac{2d}{3} + Vt)]}{\varepsilon_0 [\frac{2(\frac{2d}{3} + Vt) + (\frac{d}{3} - Vt)}{(\frac{d}{3} - Vt)(\frac{2d}{3} + Vt)}]} = \frac{2 \varepsilon_0}{\frac{4d}{3} + 2Vt + \frac{d}{3} - Vt} = \frac{2 \varepsilon_0}{\frac{5d}{3} + Vt} = \frac{6 \varepsilon_0}{5d + 3Vt}$.The time constant is $	au = C_{eq} R = \frac{6 \varepsilon_0 R}{5d + 3Vt}$.

$A$ parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The initial level of the liquid is $\frac{d}{3}$. Suppose the liquid level decreases at a constant speed $V$. Find the time constant $\tau$ as a function of time $t$.

Similar Questions

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become:

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

$A$ capacitor of capacitance $C = 10 \ \mu F$ is connected to a $12 \ V$ battery. When a dielectric slab of dielectric constant $K = 5$ is inserted between its plates,how much additional charge (in $\mu C$) will flow from the battery to the capacitor?

Vedclass Test Series

Exam Paper Generator

Online Exam Module