A parallel plate capacitor has square plates of side \(L=\) \(10.0 \mathrm{~cm}\) and a distance \(d=1.00 \mathrm{~cm}\) between the plates. Of the space between the plates, \(\frac{1}{3}\) is filled with a dielectric with dielectric constant \(\kappa_{1}=20.0 .\) The remaining \(\frac{4}{5}\) of the space is filled with a different dielectric with \(\kappa_{2}=5.00 .\) Find the capacitance of the capacitor.

Recall that the basic formula for the capacitance of a parallel plate capacitor without any dielectric is given by: \(C_0 = \frac{\epsilon_0 A}{d}\), where \(\epsilon_0\) is the vacuum permittivity, \(A\) is the area of the plates, and \(d\) is the distance between the plates.

Given the side length of the square plates \(L=10.0\,\mathrm{cm}\), we can calculate the area of the plates: \(A = L^2 = (10.0\,\mathrm{cm})^2 = 100\,\mathrm{cm}^2\). Convert this into SI units, we get \(A = 100\,\mathrm{cm}^2 \times (0.01\,\mathrm{m/cm})^2 = 0.01\,\mathrm{m}^2\).

First, we calculate the capacitance for each dielectric separately, taking into account their portions of the total distance between the plates: \(d_1 = \frac{1}{3}d\) and \(d_2 = \frac{4}{5}d\). We know that the capacitance of the capacitors with dielectrics is given by \(C_i = \kappa_iC_{0_i}\), where \(\kappa_i\) is the dielectric constant of the dielectric, and \(C_{0_i}\) is the capacitance of the capacitor without the dielectric. For dielectric 1, its capacitance is given by: \(C_1 = \kappa_{1} \frac{\epsilon_0 A}{d_1} = 20.0 \times \frac{\epsilon_0 A}{\frac{1}{3}d}\). For dielectric 2, its capacitance is given by: \(C_2 = \kappa_{2} \frac{\epsilon_0 A}{d_2} = 5.00 \times \frac{\epsilon_0 A}{\frac{4}{5}d}\).

The capacitors formed by the two dielectrics are effectively in series, so their overall capacitance can be found using the formula for capacitors in series: \(\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}\). Now, we can plug in the expressions for \(C_1\) and \(C_2\) we calculated earlier: $$ \frac{1}{C} = \frac{1}{20.0\frac{\epsilon_0 A}{\frac{1}{3}d}} + \frac{1}{5.00\frac{\epsilon_0 A}{\frac{4}{5}d}} $$

Simplify the equation in step 4 and solve for the overall capacitance, \(C\). As \(\epsilon_0\) and \(A\) are common terms in both terms, we can factor them out: $$ \frac{1}{C} = \frac{\epsilon_0 A}{20.0 \times \frac{1}{3}d} + \frac{\epsilon_0 A}{5.00 \times \frac{4}{5}d} $$ $$ \frac{1}{C} = \epsilon_0 A \left( \frac{1}{20.0 \times \frac{1}{3}d} + \frac{1}{5.00 \times \frac{4}{5}d} \right) $$ Now, find the reciprocal of the expression to get the capacitance \(C\): $$ C = \frac{1}{\epsilon_0 A \left( \frac{1}{20.0 \times \frac{1}{3}d} + \frac{1}{5.00 \times \frac{4}{5}d} \right)} $$ Plug in the values of \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\), \(A = 0.01 \mathrm{m}^2\), and \(d = 0.01 \mathrm{m}\), and compute the capacitance \(C\): $$ C = \frac{1}{(8.85 \times 10^{-12} \mathrm{F/m})(0.01\,\mathrm{m^2}) \left( \frac{1}{20.0 \times \frac{1}{3}(0.01\,\mathrm{ m})} + \frac{1}{5.00 \times \frac{4}{5}(0.01\,\mathrm{ m})} \right)} $$ Evaluating this expression, we obtain \(C \approx 2.91 \times 10^{-10}\mathrm{F}\). So, the capacitance of the parallel plate capacitor with the two given dielectrics is approximately \(2.91 \times 10^{-10}\mathrm{F}\).