Question

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, \(\frac{1}{5}\) 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.

Short answer

Answer: To calculate the capacitance of the capacitor filled with two different dielectrics, follow these steps: 1. Calculate the area of each section occupied by the dielectrics using the total area of the plates (A=L×L). 2. Calculate the capacitance of each section using the formula \(C = \kappa \epsilon_0 \frac{A}{d}\). 3. Determine the equivalent capacitance of the capacitor by treating the dielectric sections as being in series. 4. Solve for the equivalent capacitance, \(C_\mathrm{eq}\). By carrying out these steps, you will find the capacitance of the parallel plate capacitor filled with two different dielectrics.

Step-by-step solution

Identify the formula to find capacitance

We know that the formula for capacitance with a dielectric is given by: \(C = \kappa \epsilon_0 \frac{A}{d}\) Where \(C\) is the capacitance, \(\kappa\) is the dielectric constant, \(\epsilon_0\) is the vacuum permittivity \((8.85 \times 10^{-12} \mathrm{F\:m^{-1})\), \(A\) is the area of the plates, and \(d\) is the distance between the plates.

Find the area of each section of the dielectric

The total area of the plates \(A\) is given by \(A=L \times L\), where \(L=10.0 \mathrm{~cm} = 0.1 \mathrm{~m}\). However, we have two different dielectrics present. The first dielectric occupies \(\dfrac{1}{5}\) of the space, and the second dielectric occupies \(\dfrac{4}{5}\) of the space. Therefore, the area of the sections occupied by each dielectric is: \(A_1=\dfrac{1}{5}A\) and \(A_2=\dfrac{4}{5}A\)

Calculate the capacitance for each section

Using the capacitance formula from Step 1, we need to calculate the capacitances of each section: \(C_1 = \kappa_{1} \epsilon_0 \frac{A_1}{d}\) and \(C_2 = \kappa_{2} \epsilon_0 \frac{A_2}{d}\) Plug in the given values and solve for \(C_1\) and \(C_2\).

Calculate the equivalent capacitance

Since the two dielectrics occupy different parts of the same space, they are essentially in series. To calculate the total capacitance for a series configuration, we use the following formula: \(\dfrac{1}{C_\mathrm{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}\) Using the calculated values for \(C_1\) and \(C_2\) from Step 3, we can find the equivalent capacitance, \(C_\mathrm{eq}\), of the capacitor.

Final answer

With the calculated value of \(C_\mathrm{eq}\), we have found the capacitance of the capacitor filled with two different dielectrics.