QJEEAdv08Ph7858: JEE Advanced 2008 Physics - Class 12 JEEAdv08Ph7858 Solution: A parallel plate capacitor C with plates of unit area and…

Time constant ( $\tau$ ) = RC $_{eq}$

Where R = Resistance

C $_{eq}$ = equivalent capacitance

And, $v = {{ - dx} \over {dt}} \Rightarrow dx = - vdt$

$\Rightarrow \int_{d/3}^x {dx = - v\int_0^t {dt} }$

$\Rightarrow [x]_{d/3}^x = - v[t]_0^t$

$x - {d \over 3} = - vt$

$x = {d \over 3} - vt$ ..... (i)

And also, ${1 \over {{C_{eq}}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}} = {1 \over {{{{\varepsilon _0}A} \over {(d - x)}}}} + {1 \over {{{{\varepsilon _0}Ak} \over x}}}$

${1 \over {{C_{eq}}}} = {{d - x} \over {{\varepsilon _0}A}} + {x \over {{\varepsilon _0}Ak}}$

$= {1 \over {{\varepsilon _0}A}}\left[ {d - x + {x \over k}} \right]$

$= {1 \over {{\varepsilon _0}A}}\left[ {{{(d - x)k + x} \over k}} \right]$

${1 \over {{C_{eq}}}} = {1 \over {{\varepsilon _0}Ak}}[dk - xk + x]$

$\therefore$ ${C_{eq}} = {{{\varepsilon _0}A} \over {kd + x(1 - k)}}$ .... (ii)

Given, $A = 1,k = 2,x = {d \over 3} - vt$

Hence, ${C_{eq}} = {{{\varepsilon _0} \times 1} \over {2d + \left[ {{d \over 3} - vt} \right](1 - 2)}}$

$= {{6{\varepsilon _0}} \over {5d + 3vt}}$

Therefore, $\tau = R{C_{eq}} = {{6R{\varepsilon _0}} \over {5d + 3vt}}$