Treacherous Cylinder

Suppose that the charge contained on the surface of inner cylinder is $Q(t)$

Gauss Law in a dielectric system gives:

$\large \oint_{}^{}D.dA=Q_{enc}$

$\large D=\frac{Q}{2\pi rl}$

$\large E=\frac{D}{\varepsilon}=\frac{D}{k\varepsilon_{0}}$

$\large E=\frac{Q}{2\pi \varepsilon_{0} krl}$

Ohm's Law

$\large J=\frac{E}{\rho}= \frac{Q}{2\pi \varepsilon_{0} \rho krl}$

$\large I=\int J. dA= \frac{Q}{2\pi \varepsilon_{0} \rho krl}2\pi rl=\frac{Q}{\rho\varepsilon_{0}k}$

$\large \frac{-dQ}{dt}=\frac{Q}{\rho\varepsilon_{0}k}$

Rearranging and integrating gives you:

$\large Q= Q_{0}e^{\frac{-t}{\rho\varepsilon_{0}k}}$

When $\large t=\rho\varepsilon_{0}k$ , $Q=\frac{Q_{0}}{e}$

Finally, the charge flown is:

$\large Q_{flown}=Q_{0}(1-\frac{1}{e})=0.63212Q_{0}$

$\large S=0.63212$

Relevant wiki: RC Circuits (Direct Current)

$\text{Charge flown from a capacitor having resistance } R_{eq} \text{ and capacitance } C_{eq} \text{ as a function of time is given by:}$ $\boxed{q=Q_0\left(1-e^{{\dfrac{-t}{R_{eq} C_{eq}}}} \right)} \quad \cdots(1)$

$\text{ ; where }$ $\large R_{eq}C_{eq}= \tau$ $\text{ is the time constant.}$

$\text{To find } R_{eq} \text{ let us assume a small element } dx \text{, which is } x \text{ distance away from the inner plate of radius } a$ .

$\text{The resistance of this elementary portion is}$ $\boxed{ dR =\dfrac{\rho.dx}{2\pi xl}}$

$\text{As these resistances are in series we can directly integrate } dR \text{ to find the net resistance of the capacitor.}$ $R_{eq}=\int _{ a }^{ b }{ \dfrac{\rho.dx}{2\pi xl} } = \dfrac{\rho\left.(\ln{x}) \right|^{b}_{a}}{2\pi l} \\ \Rightarrow \boxed{R_{eq}=\dfrac{\rho \ln{\dfrac {b}{a}}}{2\pi l}}$ $\text{Now, equivalent capacitance of a cylindrical capacitor is given by}$ $\boxed{C_{eq} = {\dfrac{2\pi k\epsilon_0 l}{\ln{\dfrac{b}{a}}}}}$

$\text{Hence, }$ $\tau = \dfrac{ {2\pi l} k\epsilon_0 }{\ln{\dfrac{b}{a}}}\times\dfrac{\rho \ln{\dfrac {b}{a}}}{{2\pi l}}=\boxed{k\rho \epsilon_0}$ $\text{Putting this value of } \tau \text{ and } t \text{ (given in the question) in equation } (1)\text{, we get:}$ $q= Q_0\left(1-{e^{{\dfrac{-\rho k \epsilon_0}{\rho k \epsilon_0}}}} \right) = Q_0\left( 1-e^{-1} \right) = Q_0\left(1-0.36787 \right) =\boxed{Q_0\left(0.63212\right)} \\ \Rightarrow \boxed{S=0.63212}$

$\text{For capacitance of a cylindrical capacitor:}$

• $\text{Take an arbitrary radius } r \text{ in between the plates of the capacitor}$
• $\text{Using Gauss's law find the electric field at that radius } r \text{ assuming that the field is radially outward.}$
• $\text{Then using the relation } E= -\dfrac{dV}{dr} \text{ find the potential difference between }a\text{ and }b \text{, using integration.}$
• $\text{Put this result in the formula } C=\dfrac{Q}{V} \text{ to get the value of capacitance.}$

NOTE

$\rightarrow \text{Resistances are in series because potential is varying with radius and is not constant across each elementary resistance }$ $\text{and in series combinations, potential isn't conserved.}$

$\rightarrow dR = {{\dfrac{\rho.dx}{2\pi xl}}}$ $\text{ is obtained from the formula } R={\dfrac{\rho\ l}{A}}.$ $2 \pi xl=A \text{ and } dx=l \text{ because the electric field in the capacitor is radially outward. }$