A parallel-plate capacitor features a plate area of 100 cm^2 and a...

Given plate area $A = 100 \text{ cm}^2 = 10^{-2} \text{ m}^2$ and separation $d = 1.0 \text{ cm} = 10^{-2} \text{ m}$.

The space between the plates is completely filled with two dielectric slabs. The thickness of the glass plate is $t_1 = 6.0 \text{ mm} = 6 \times 10^{-3} \text{ m}$ with dielectric constant $K_1 = 6.0$.

The thickness of the ebonite plate is $t_2 = d - t_1 = 1.0 \text{ cm} - 0.6 \text{ cm} = 0.4 \text{ cm} = 4 \times 10^{-3} \text{ m}$ with dielectric constant $K_2 = 4.0$.

The equivalent capacitance of the parallel-plate capacitor with two dielectric slabs in series is given by:

$C = \dfrac{\varepsilon_0 A}{\dfrac{t_1}{K_1} + \dfrac{t_2}{K_2}}$

Substituting the given values:

$\dfrac{t_1}{K_1} = \dfrac{6 \times 10^{-3}}{6} = 10^{-3} \text{ m}$

$\dfrac{t_2}{K_2} = \dfrac{4 \times 10^{-3}}{4} = 10^{-3} \text{ m}$

$\dfrac{t_1}{K_1} + \dfrac{t_2}{K_2} = 10^{-3} + 10^{-3} = 2 \times 10^{-3} \text{ m}$

Now, calculating the capacitance:

$C = \dfrac{8.85 \times 10^{-12} \times 10^{-2}}{2 \times 10^{-3}}$

$C = 4.425 \times 10^{-11} \text{ F}$

$C = 44.25 \text{ pF} \approx 44 \text{ pF}$