Question

A capacitor (Area of plates = $A$) initially given a charge $Q_0$ is connected across a resistor $R$ at $t=0$. The separation between the plates changes according to $d=\frac{d_0}{(1+t)}(0\le t<1)$. A small bulb is connected across the plates of the capacitor which lights when potential difference across the plates of the capacitor reaches $V_0$. Then

• A. The variation of charge with time is $q=Q_0(1+t{)}^{-\frac{d_0}{R{\epsilon }_{0}A}}$.
• B. The variation of charge with time is $q=Q_0(1+t{)}^{-\frac{d_0}{R{\epsilon }_{0}A}}$.
• C. Time when the bulb will light is $\{{\left(\frac{Q_0{d}_0}{{\epsilon }_{0}AV}\right)}^{\left(\frac{R{\epsilon }_{0}A}{R{\epsilon }_{0}A+d_0}\right)}-1\}$
• D. Time when the bulb will light is $\{{\left(\frac{Q_0{d}_0}{{\epsilon }_{0}AV}\right)}^{\left(\frac{R{\epsilon }_{0}A}{R{\epsilon }_{0}A+2d_0}\right)}-1\}$.

Answer 1

Answer: (A), (C)

The correct options are
A The variation of charge with time is $q=Q_0(1+t{)}^{-\frac{d_0}{R{\epsilon }_{0}A}}$.
C Time when the bulb will light is $\{{\left(\frac{Q_0{d}_0}{{\epsilon }_{0}AV}\right)}^{\left(\frac{R{\epsilon }_{0}A}{R{\epsilon }_{0}A+d_0}\right)}-1\}$


Capacitance at $t=0,C_0=\frac{{\epsilon }_{0}A}{d_0},C=C_0(1+t)$
Using Kirchoff's law
$\frac{q}{C}-Ri=0$
(i) $\frac{q}{C_0(1+t)}+R\frac{dq}{dt}=0$
$\frac{dq}{q}=-\frac{1}{R{C}_0}\frac{dt}{(1+t)}$
$lnq{|}_{Q_0}^q=-\frac{1}{R{C}_0}ln(1+t){|}_0^t$
$ln\left(\frac{q}{Q_0}\right)=-\frac{1}{R{C}_0}ln(1+t)$
$ln\left(\frac{q}{Q_0}\right)=ln(1+t{)}^{-\frac{1}{R{C}_0}}$
$q=Q_0(1+t{)}^{-\frac{1}{R{C}_0}}$
Subtituting $C_0=\frac{{\epsilon }_{0}A}{d_0}$ we get,
$q=Q_0(1+t{)}^{-\frac{d_0}{R{\epsilon }_{0}A}}$
$V=\frac{q}{C}=\frac{Q_0(1+t{)}^{-\frac{1}{RC}}}{C_0(1+t)}=\frac{Q_0}{C_0}(1+t{)}^{-(\frac{1}{RC}+1)}$
which gives $t=\{{\left(\frac{Q_0}{C_{0}V}\right)}^{\left(\frac{R{C}_0}{R{C}_0+1}\right)}-1\}$

Subtituting $C_0$ we get,
$t=\{{\left(\frac{Q_0{d}_0}{{\epsilon }_{0}AV}\right)}^{\left(\frac{R{\epsilon }_{0}A}{R{\epsilon }_{0}A+d_0}\right)}-1\}$