A parallel-plate capacitor features a plate area of 20 cm^2 , a...

The capacitance of the parallel-plate capacitor with the dielectric slab is given by:

$C = \dfrac{K \epsilon_0 A}{d}$

Substituting the given values:

$C = \dfrac{5.0 \times 8.85 \times 10^{-12} \times 20 \times 10^{-4}}{1.0 \times 10^{-3}} = 88.5 \times 10^{-12} \text{ F}$

The time constant of the RC circuit is:

$\tau = RC = (100 \times 10^3) \times (88.5 \times 10^{-12}) = 8.85 \times 10^{-6} \text{ s} = 8.85 \mu\text{s}$

The given time is $t = 8.9 \mu\text{s}$, which is approximately equal to one time constant $\tau$. The voltage across the capacitor at time $t$ is:

$V = V_0 (1 - e^{-t/\tau})$

For $t \approx \tau$:

$V \approx 6.0 (1 - e^{-1}) \approx 6.0 (1 - 0.37) = 3.78 \text{ V}$

The energy stored in the capacitor at this time is:

$U = \dfrac{1}{2} C V^2$

$U = \dfrac{1}{2} \times 88.5 \times 10^{-12} \times (3.78)^2$

$U \approx 6.3 \times 10^{-10} \text{ J}$

Answer: $6.3 \times 10^{-10} \text{ J}$