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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

Expert verified
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

01

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, \(\epsilon = \epsilon_{0} \epsilon_{r} = \epsilon_{0}(2 + 2 \times 10^{6}x^{2})\).
02

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 \(dC = \frac{\epsilon S}{dx} = \frac{\epsilon_{0}(2 + 2 \times 10^{6}x^{2})S}{dx}\).
03

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: \(C = \int_{0}^{10^{-3}} \frac{\epsilon_{0}(2 + 2 \times 10^6 x^2)S}{dx}\).
04

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: \(C = \epsilon_{0}S\left[2x + \frac{2}{3} \times 10^{6}x^{3}\right]_{0}^{10^{-3}}\).
05

Evaluate the integral

Evaluating the integral at the limits of 0 and \(10^{-3}\), we get: \(C = \epsilon_{0}S\left[2(10^{-3}) + \frac{2}{3} \times 10^{6}(10^{-3})^{3}\right]\).
06

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

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Most popular questions from this chapter

An air-filled parallel-plate capacitor with plate separation \(d\) and plate area \(A\) is connected to a battery that applies a voltage \(V_{0}\) between plates. With the battery left connected, the plates are moved apart to a distance of \(10 d\). Determine by what factor each of the following quantities changes: \((a) V_{0} ;(b) C ;(c) E ;(d) D ;(e) Q ;(f) \rho_{S} ;(g) W_{E}\)

Let \(S=100 \mathrm{~mm}^{2}, d=3 \mathrm{~mm}\), and \(\epsilon_{r}=12\) for a parallel-plate capacitor. (a) Calculate the capacitance. (b) After connecting a 6-V battery across the capacitor, calculate \(E, D, Q\), and the total stored electrostatic energy. (c) With the source still connected, the dielectric is carefully withdrawn from between the plates. With the dielectric gone, recalculate \(E, D, Q\), and the energy stored in the capacitor. \((d)\) If the charge and energy found in part \((c)\) are less than the values found in part \((b)\) (which you should have discovered), what became of the missing charge and energy?

A parallel-plate capacitor has plates located at \(z=0\) and \(z=d\). The region between plates is filled with a material that contains volume charge of uniform density \(\rho_{0} \mathrm{C} / \mathrm{m}^{3}\) and has permittivity \(\epsilon\). Both plates are held at ground potential. ( \(a\) ) Determine the potential field between plates. (b) Determine the electric field intensity \(\mathbf{E}\) between plates. \((c)\) Repeat parts \((a)\) and \((b)\) for the case of the plate at \(z=d\) raised to potential \(V_{0}\), with the \(z=0\) plate grounded.

Coaxial conducting cylinders are located at \(\rho=0.5 \mathrm{~cm}\) and \(\rho=1.2 \mathrm{~cm}\). The region between the cylinders is filled with a homogeneous perfect dielectric. If the inner cylinder is at \(100 \mathrm{~V}\) and the outer at \(0 \mathrm{~V}\), find (a) the location of the \(20 \mathrm{~V}\) equipotential surface; \((b) E_{\rho \max } ;(c) \epsilon_{r}\) if the charge per meter length on the inner cylinder is \(20 \mathrm{nC} / \mathrm{m}\).

Construct a curvilinear-square map of the potential field between two parallel circular cylinders, one of \(4 \mathrm{~cm}\) radius inside another of \(8 \mathrm{~cm}\) radius. The two axes are displaced by \(2.5 \mathrm{~cm}\). These dimensions are suitable for the drawing. As a check on the accuracy, compute the capacitance per meter from the sketch and from the exact expression: $$C=\frac{2 \pi \epsilon}{\cosh ^{-1}\left[\left(a^{2}+b^{2}-D^{2}\right) /(2 a b)\right]}$$ where \(a\) and \(b\) are the conductor radii and \(D\) is the axis separation.

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