Calculating surface charge density as a function of time 
An infinitely large conducting sheet of thickness $d$ and of conductivity $\eta$ is uncharged. At an instant $t=0$, an external uniform and electric field $E$ is switched on. Due to this, the conducting sheet starts developing surface charge density $\sigma$. Now how do we find $\sigma(t)$?
For $\sigma(t)$, I think first we have to find the elctric field which is induced inside the sheet as a function of time. So how do we find it? Also, I wonder whether will the electric field just outside the sheet vary with time or will it remain constant?
The answer given is $\sigma(t)=\epsilon_{o}E\left(1-e^{\dfrac{-\eta t}{\epsilon_{o}}}\right)$
 A: Let at time $t$ charge on a area $A$ of the sheet be $Q$. charge density on one face will be $\sigma_{t}$ and on the other face will be -$\sigma_{t}$. the net field inside the conducting sheet will be $$E_{tot}=E-\frac{Q}{A\epsilon_{o}}=E-\frac{\sigma_{t}}{\epsilon_{o}}$$Current Density, $J=\eta E_{tot}$
Due to this current a charge $JAdt$ flows from one surface to the other in a time interval dt. the increase in charge density, $d\sigma_{t}=\frac{JAdt}{A}=Jdt$
$$\implies d\sigma_{t}=\eta\bigg(E-\frac{\sigma_{t}}{\epsilon_{o}}\bigg)dt$$
$$\implies \int\limits_{0}^{\sigma_{t}}\dfrac{d\sigma_{t}}{\bigg(E-\frac{\sigma_{t}}{\epsilon_{o}}\bigg)}=\int\limits_{o}^{t}\eta dt$$
$$\implies \sigma_{t}=\epsilon_{o}E\left(1-e^{\dfrac{-\eta t}{\epsilon_{o}}}\right)$$
A: For a conductor placed in electric field we have a vector form of Ohm's law-
$$\vec{j} = \eta\vec{E}$$
where $\vec{j}$ is current density and $\eta$ is conductivity of conductor (which is constant at given temperature), $\vec{E}$ is net electric field inside  conductor.
note that, for $\vec{E}$, you need net electric field inside the conductor, which would be $E - \cfrac{\sigma}{\epsilon_0}$($\vec{\sigma}$ is induced surface charge density at any instant), and since there would be a current inside the conductor due to this net field ($\vec{E}$), for any elemental charge that would flow due to this $\vec{E}$ will be $dq$, and the additional induced surface charge density would be $d\sigma$, so $\vec{j} = \cfrac{d\sigma}{dt}$. rest part is easy calculation using calculus , you can easily find induced surface charge density as a function of $t$.
