This one's charged!

Lets derive the form, I can see, many did it with that form, But none posted how to derive,

So Here we go,

Lets think of an R-C Circuit where Resistance connected is $R$ and Capacitance is $C$ .

Now say, the charge developed at any time instant $t$ on the parallel plate capacitor before reaching steady point be q.

So We know the Emf Given by battery is $\epsilon$ .

So we can say that due to the capacitor the there will be an back emf and hence the net emf at any time instant will be

$\epsilon - \dfrac{q}{C} = i R$ .

$\implies \dfrac{C \times \epsilon - q}{C} = \dfrac{dq}{dt} R$

$\implies C \times \epsilon- q = RC \dfrac{dq}{dt}$

$\implies \dfrac{dt}{RC} = \dfrac{dq}{C \epsilon - q}$

$\implies \displaystyle{\int _{0}^{t} \dfrac{dt}{RC} = \int_{0}^{q_{0}}\dfrac{dq}{C \epsilon - q}}$

After solving we will get

$\displaystyle{q=q_{0} (1-e^{\tiny\dfrac{-t}{RC}}})$

Putting the values we get $q$ at any time instant, so after inputting $t=2s$ we get the answer.

Kindly add in the question the units in which the answer must be expressed.(i.e-coulomb,millicoulomb)

In an RC circuit, the charge that has flowed through the resistor by time $t$ is given by $Q_0 (1-e^{-t/ \tau})$ , where $Q_O$ is the initial charge and $\tau=RC=5*10^{6}*2*10^{-6}=10$ . Therefore, our answer is $(1*10^{-3}) (1-e^{-2/10}) \approx 0.00018 C$ .