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 \(3.54 \times 10^{-11} \mathrm{F}\).