Question

A capacitor initially given a charge $Q_o$ is connected across a resistor $R$ at $t=0$. The separation between the plates changes according to, $d=\frac{d_o}{1+t}(0\le t<1)$.. The variation of charge with time is $Q_o{(x+t)}^{-1/R{C}_o}$. The value of $x$ is ______.

Answer 1

Capacitance at $t=0$,

$C_o=\frac{{ϵ}_{o}A}{d_o}$

Also, at any time $t$, capacitance,

$C=\frac{{ϵ}_{o}A}{d_o/(1+t)}=C_o(1+t)$

Using Kirchhoff's Law,

$\frac{q}{C}-RI=0$

$\Rightarrow \frac{q}{C_o(1+t)}+R\frac{dq}{dt}=0$

$\Rightarrow \frac{dq}{q}=-\frac{1}{R{C}_o}\frac{dt}{(1+t)}$

$\Rightarrow {\left[\ln q\right]}_{Q_o}^q=-\frac{1}{R{C}_o}{\left[\ln \right(1+t\left)\right]}_0^t$

$\Rightarrow \ln \left(q/Q_o\right)=-\frac{1}{R{C}_o}\ln (1+t)$

$\Rightarrow \ln \left(q/Q_o\right)=\ln (1+t{)}^{-1/R{C}_o}$

$\Rightarrow q=Q_o(1+t{)}^{-1/R{C}_o}=Q_o(x+t{)}^{-1/R{C}_o}$

$\therefore x=1$