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, \(\frac{1}{3}\) 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.

Short answer

The capacitance of the parallel plate capacitor is approximately \(2.91 \times 10^{-10}\mathrm{F}\).

Step-by-step solution

Understand Basic Capacitance Formula

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.

Calculate the area of 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 \(A = 100\,\mathrm{cm}^2 \times (0.01\,\mathrm{m/cm})^2 = 0.01\,\mathrm{m}^2\).

Calculate the capacitance of each dielectric separately

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: \(C_2 = \kappa_{2} \frac{\epsilon_0 A}{d_2} = 5.00 \times \frac{\epsilon_0 A}{\frac{4}{5}d}\).

Calculate overall capacitance for the capacitors in series

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: $$ \frac{1}{C} = \frac{1}{20.0\frac{\epsilon_0 A}{\frac{1}{3}d}} + \frac{1}{5.00\frac{\epsilon_0 A}{\frac{4}{5}d}} $$

Solve for the overall capacitance

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 \(2.91 \times 10^{-10}\mathrm{F}\).

Key concepts

Parallel Plate Capacitor

A Parallel Plate Capacitor is a fundamental electrical component that consists of two conductive plates separated by a small distance. The purpose of such a capacitor is to store electrical energy in an electric field created between these plates. The basic principle is that the plates can hold opposite charges, creating a potential difference due to the separation.
Meaning, the larger the area (\(A\)) of the plates and the smaller the distance (\(d\)) between them, the greater the capacitance (\(C_0\)), which is given by the formula:

• \(C_0 = \frac{\epsilon_0 A}{d}\)

Here, \(\epsilon_0\) is the vacuum permittivity. By understanding this basic structure, students can infer how changes in each variable affect the overall capacitance.

Dielectric Constant

The Dielectric Constant, also known as relative permittivity, is a key property of an insulating material placed between the plates of a capacitor. It quantifies the ability of the material to increase the capacitance beyond what would be present if there was only a vacuum between the plates.
The modified capacitance is calculated as:

• \(C = \kappa \cdot C_{0}\)

Where \(\kappa\) is the dielectric constant of the material. This means that a higher dielectric constant will result in a higher stored charge for the same voltage applied, effectively enhancing the capacitor's ability to store energy.

Capacitors in Series

When dealing with multiple capacitors in an electric circuit, it's important to understand how they interact with each other. In a series configuration, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances:

• \(\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots\)

For capacitors with different dielectrics in between, like in our original exercise, each section can be thought of as a separate capacitor in series with the others.
It’s essential knowledge for students learning how to calculate the equivalent capacitance of multiple sections or units arranged in this manner.

Vacuum Permittivity

Vacuum Permittivity, symbolized as \(\epsilon_0\), is a constant that represents the ability of a vacuum to permit electric field lines. This constant fundamentally affects how electric fields interact in a vacuum and is crucial when calculating the basic capacitance of a capacitor using the formula:

• \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\)

In problems like our exercise, \(\epsilon_0\) remains a constant reference point when assessing how additional materials (dielectrics) affect the overall capacitance.