In the following circuit the switch S is closed at t = 0 . The...

In given circuit capacitors ${C}_{1}$ and ${C}_{2}$ are connected in series , so equivalent capacitance will be

$\dfrac{1}{{C}_{eq}} = \dfrac{1}{{C}_{1}} + \dfrac{1}{{C}_{2}} ⇒ {C}_{eq} = \dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}$

at $t = 0$, the switch is closed and capacitors start charged, at time $t$

charge on the capacitor is given by

$q = {q}_{0}\left[1 - {e}^{- t / τ}\right]$

where ${q}_{0}$ is the maximum charge on the capacitor at steady state and $τ$ is the time constant.

and ${q}_{0} = {C}_{eq}E$ and $τ = R{C}_{eq}$

Then from the above equation

$q = {C}_{eq}E\left[1 - {e}^{- t / R{C}_{eq}}\right]$

$∴ q = {C}_{eq}E\left[1 - exp\left(\dfrac{t}{R{C}_{eq}}\right)\right]$
Both capacitor will have same charge as they are connected in series