Question

A parallel-plate capacitor is filled with a nonuniform dielectric characterized by \(\epsilon_{r}=2+2 \times 10^{6} x^{2}\), where \(x\) is the distance from one plate in meters. If \(S=0.02 \mathrm{~m}^{2}\) and \(d=1 \mathrm{~mm}\), find \(C\).

Short answer

Based on the given problem of a parallel-plate capacitor filled with a nonuniform dielectric, we have calculated the capacitance of the capacitor to be approximately \(3.54 \times 10^{-11} \mathrm{F}\).

Step-by-step solution

Calculate the relative permittivity

Given, the relative permittivity is defined by the formula \(\epsilon_{r} = 2 + 2 \times 10^{6}x^{2}\). The total permittivity, \(\epsilon\), can be found by multiplying the relative permittivity by the vacuum permittivity, \(\epsilon_{0}\). So, \(\epsilon = \epsilon_{0} \epsilon_{r} = \epsilon_{0}(2 + 2 \times 10^{6}x^{2})\).

Calculate capacitance for infinitesimally small distances

We'll now find the capacitance for an infinitesimally small distance \(dx\) within the capacitor. Using the formula \(C = \frac{\epsilon A}{d}\), the capacitance for a small distance \(dx\) will be \(dC = \frac{\epsilon S}{dx} = \frac{\epsilon_{0}(2 + 2 \times 10^{6}x^{2})S}{dx}\).

Integrate to find the total capacitance

To find the total capacitance of the capacitor, we'll integrate the \(dC\) expression over the entire distance between the plates, from 0 to \(d=1\) mm, or 0 to \(10^{-3}\) meters. This will give us: \(C = \int_{0}^{10^{-3}} \frac{\epsilon_{0}(2 + 2 \times 10^6 x^2)S}{dx}\).

Perform the integration

To perform the integration, we can first distribute the constants outside of the integral: \(C = \epsilon_{0}S\int_{0}^{10^{-3}} \frac{2 + 2 \times 10^{6}x^{2}}{dx}\). Now, integrating with respect to \(x\), we get: \(C = \epsilon_{0}S\left[2x + \frac{2}{3} \times 10^{6}x^{3}\right]_{0}^{10^{-3}}\).

Evaluate the integral

Evaluating the integral at the limits of 0 and \(10^{-3}\), we get: \(C = \epsilon_{0}S\left[2(10^{-3}) + \frac{2}{3} \times 10^{6}(10^{-3})^{3}\right]\).

Calculate the capacitance

Using the value of vacuum permittivity, \(\epsilon_{0} = 8.854 \times 10^{-12} \mathrm{F/m}\), and the given plate area, \(S = 0.02 \mathrm{m^2}\), we can calculate the capacitance: \(C = (8.854 \times 10^{-12})(0.02)\left[2(10^{-3}) + \frac{2}{3} \times 10^{6}(10^{-3})^{3}\right]\). After evaluating this expression, we find that the capacitance of the parallel-plate capacitor is approximately \(3.54 \times 10^{-11} \mathrm{F}\).

Key concepts

Parallel-Plate Capacitor

A parallel-plate capacitor is a fundamental component in electromagnetics. It consists of two conducting plates separated by an insulator or dielectric. The key goal of a capacitor is to store electrical energy. The capacitance, or the ability of a capacitor to store charge, depends on the area of the plates and the distance between them.

• For a given area, larger plate areas allow for more charge storage.
• Smaller separation between the plates increases capacitance, as electric fields can influence charges more effectively.

Understanding these principles helps in designing efficient capacitors for various applications, from small electronic devices to large circuits.

Nonuniform Dielectric

In this scenario, the parallel-plate capacitor is filled with a nonuniform dielectric. Unlike uniform dielectrics with a constant permittivity, nonuniform dielectrics have variable permittivity. Here, the relative permittivity, \(\epsilon_{r} = 2 + 2 \times 10^{6} x^{2}\), varies with distance \(x\) from one plate.

• This variation impacts how the electric field distributes across the capacitor.
• The relative permittivity describes how easily a material allows electric fields to pass through.

Nonuniform dielectrics require more complex calculations, often involving integration, to determine total capacitance.

Integration in Electromagnetics

Integration is a crucial mathematical tool in electromagnetics, especially for calculating capacitance or finding electric fields in varying conditions. In the given problem, we calculate total capacitance by integrating the differential capacitance \(dC\) with the changing permittivity.

• This involves integrating the expression \(\int_{0}^{d} \epsilon_{0}(2 + 2 \times 10^6 x^2)S \, dx\), where \(d\) is the separation between the plates.
• The integral calculates the effect of the varying dielectric over the entire distance.

Mastery of integration is essential for solving such real-world physics problems.

Vacuum Permittivity

Vacuum permittivity, denoted as \(\epsilon_{0}\), is a constant that describes how electric fields behave in a vacuum. Its value, \(8.854 \times 10^{-12} \mathrm{F/m}\), is fundamental in determining capacitance. In the problem, we use \(\epsilon_{0}\) to compute the total permittivity of the dielectric.

• The permittivity of a material determines its ability to allow electric fields to pass through.
• For capacitors, \(\epsilon_{0}\) forms the base constant multiplied by relative permittivity to find total permittivity.

Understanding \(\epsilon_{0}\) is key to solving problems involving dielectrics and capacitance calculations.