Through resistanceless leads, a capacitor of capacitance 12.0 is...

The energy stored in a capacitor is given by $U = \dfrac{q^2}{2C}$.

The rate at which the energy is increasing is $\dfrac{dU}{dt} = \dfrac{q}{C} \dfrac{dq}{dt} = V_c I$.

For an RC circuit during charging, the voltage across the capacitor and the current are given by:

$V_c = E(1 - e^{-t/RC})$

$I = \dfrac{E}{R} e^{-t/RC}$

The time constant of the circuit is $\tau = RC = (1.00 \Omega)(12.0 \mu\text{F}) = 12.0 \mu\text{s}$.

At the given time $t = 12.0 \mu\text{s}$, we have $t = \tau$. Substituting this into the equations:

$V_c = E(1 - e^{-1})$

$I = \dfrac{E}{R} e^{-1}$

The rate of increase of energy is:

$\dfrac{dU}{dt} = E(1 - e^{-1}) \times \dfrac{E}{R} e^{-1} = \dfrac{E^2}{R} (e^{-1} - e^{-2})$

Substituting the given values $E = 6.00 \text{ V} and$R = 1.00 \Omega$:<br><br>$\dfrac{dU}{dt} = \dfrac{(6.00)^2}{1.00} \left(\dfrac{1}{e} - \dfrac{1}{e^2}\right)

Using $e \approx 2.718$:

\dfrac{dU}{dt} =$36.0 \times (0.3679 - 0.1353) = 36.0 \times 0.2326 = 8.37 \text{ W}$.