At t = 0 , a capacitor having capacitance C is connected through a...

The charge on the capacitor at time $t$ is given by $q = C\mathcal{E}(1 - e^{-t/RC})$.

The current in the circuit is $i = \dfrac{dq}{dt} = \dfrac{\mathcal{E}}{R} e^{-t/RC}$.

The rate of energy storage in the capacitor is $P = \dfrac{dU}{dt} = \dfrac{d}{dt} \left( \dfrac{q^2}{2C} \right) = \dfrac{q}{C} i$.

Substituting the expressions for $q$ and $i$ gives:

$P = \dfrac{C\mathcal{E}(1 - e^{-t/RC})}{C} \dfrac{\mathcal{E}}{R} e^{-t/RC} = \dfrac{\mathcal{E}^2}{R} (e^{-t/RC} - e^{-2t/RC})$.

To find the maximum rate of energy storage, let $x = e^{-t/RC}$. Then $P = \dfrac{\mathcal{E}^2}{R} (x - x^2)$.

For maximum $P$, $\dfrac{dP}{dx} = 0 \Rightarrow 1 - 2x = 0 \Rightarrow x = \dfrac{1}{2}$.

Thus, $e^{-t/RC} = \dfrac{1}{2} \Rightarrow t = CR \ln 2$.

The maximum rate of energy storage is P_\max = \dfrac{\mathcal{E}^2}{R} \left( \dfrac{1}{2} - \dfrac{1}{4} \right) = \dfrac{\mathcal{E}^2}{4R}.