Flowing charges

First, let us calculate the resistance of the spherical shell.

Considering a spherical shell as an element at a distance of $r$ from the centre and having thickness $dr$ and resistance $dR$ :

$dR = \dfrac{1}{\sigma}\dfrac{dr}{4\pi r^2}$

As all of elements are in series, $\int dR = R$

$\therefore R = \int\limits_{r_1}^{r_2}\frac{1}{\sigma}\frac{dr}{4\pi r^2} = \dfrac{1}{4\pi\sigma}\Bigg(\dfrac{1}{r_1} - \dfrac{1}{r_2}\Bigg)$

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Let, at any time $t$ , charge $q$ be induced on the outer surface. Since the shell was electrically neutral, it follows that charge $-q$ must be induced on the inner surface.

Now, let's find the electric field at a distance $r$ from the centre, using Gauss' Law , it can be easily calculated to be:

$E(r) = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_0-q}{r^2}$

$\implies -\dfrac{dV}{dr} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_0-q}{r^2} \Bigg(\because -\dfrac{dV}{dr} = E\Bigg)$

As $q$ is independent of $r$ , this can be easily integrated using the fact that $\int dV = V_{out} - V_{in} = -\Delta V =$ potential difference between the two surfaces.

Since the outer surface is getting positively charged, the flow of conventional current is from the inner surface to the outer surface, which means that the inner surface is at higher potential which is why I'm writing $V_{out} - V_{in}$ as $-\Delta V$ to keep the signs consistent later on while using Ohm's Law.

$\implies \Delta V = \dfrac{q_0 - q}{4\pi\epsilon_0}\Bigg(\dfrac{1}{r_1}-\dfrac{1}{r_2}\Bigg)$

Using Ohm's Law , $\Delta V = IR$

$\implies I\cdot\dfrac{1}{4\pi\sigma}\Bigg(\dfrac{1}{r_1} - \dfrac{1}{r_2}\Bigg) = \dfrac{q_0 - q}{4\pi\epsilon_0}\Bigg(\dfrac{1}{r_1}-\dfrac{1}{r_2}\Bigg)$

$\implies I = (q_0-q)\cdot\dfrac{\sigma}{\epsilon_0}$

$\implies \dfrac{dq}{dt} = (q_0-q)\cdot\dfrac{\sigma}{\epsilon_0}$

$\implies \dfrac{dq}{q_0-q} =\dfrac{\sigma}{\epsilon_0}\cdot dt$

Integrating both sides to find a general expression for charge at time $t$ gives us:

$-\ln(q_0 - q) = \dfrac{\sigma}{\epsilon_0}\cdot t + C$

Using the fact that at $t=0$ , $q = 0$ $\implies$ $C = -\ln(q_0)$

Taking anti-log and re-arranging gives us:

$q = q_0\Bigg(1 - e^{-\dfrac{\sigma}{\epsilon_0}t} \Bigg)$

Differentiating both sides with respect to time gives us:

$I = q_0\cdot\dfrac{\sigma}{\epsilon_0}\cdot e^{-\dfrac{\sigma}{\epsilon_0}t}$

Substituting all values according to question gives:

$I = 3\cdot e^{-\ln(2)} = \dfrac{3}{2} = \boxed{1.5}$

The system of shells work as a capacitor.