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Electrostatics

Section: Physics

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Question : 59 of 59

Marks: +1, -0

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

[JEE Adv 2008 P2]

$\frac{6\varepsilon_0 R}{5d+3Vt}$
$\frac{(15d+9Vt)\varepsilon_0 R}{2d^2 - 3dVt - 9V^2 t^2}$
$\frac{6\varepsilon_0 R}{5d - 3Vt}$
$\frac{(15d-9Vt)\varepsilon_0 R}{2d^2 + 3dVt - 9V^2 t^2}$

Solution:

The effective capacitance

C_{\text{eff}}

=

\frac{A\varepsilon_0}{d - y + \frac{y}{\varepsilon_r}}

where A is the area of the capacitor plate, d is the plate separation, y is the thickness of dielectric and

\varepsilon_r

is the dielectric constant. It is given that rate of change of dielectric liquid is

\frac{dy}{dt}

= - V ⇒ y = -- Vt +

C_1

At t = 0, we have

y =

\frac{d}{3}

⇒

C_1

=

\frac{d}{3}

Therefore,

C_{\text{eff}}

=

\frac{\varepsilon_0}{d + Vt - \frac{d}{3} + \frac{-Vt + \frac{d}{3}}{\varepsilon_r}}

=

\frac{\varepsilon_0\varepsilon_r}{\frac{d(2\varepsilon_r+1)+3V(\varepsilon_r-1)}{3}}

=

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

The time constant is

τ =

R C_{\text{eff}}

=

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

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