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A parallel-plate capacitor is filled with a nonuniform dielectric characterized by ϵr=2+2×106x2, where x is the distance from one plate in meters. If S=0.02 m2 and d=1 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 3.54×1011F.

Step by step solution

01

Calculate the relative permittivity

Given, the relative permittivity is defined by the formula ϵr=2+2×106x2. The total permittivity, ϵ, can be found by multiplying the relative permittivity by the vacuum permittivity, ϵ0. So, ϵ=ϵ0ϵr=ϵ0(2+2×106x2).
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=ϵAd, the capacitance for a small distance dx will be dC=ϵSdx=ϵ0(2+2×106x2)Sdx.
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 103 meters. This will give us: C=0103ϵ0(2+2×106x2)Sdx.
04

Perform the integration

To perform the integration, we can first distribute the constants outside of the integral: C=ϵ0S01032+2×106x2dx. Now, integrating with respect to x, we get: C=ϵ0S[2x+23×106x3]0103.
05

Evaluate the integral

Evaluating the integral at the limits of 0 and 103, we get: C=ϵ0S[2(103)+23×106(103)3].
06

Calculate the capacitance

Using the value of vacuum permittivity, ϵ0=8.854×1012F/m, and the given plate area, S=0.02m2, we can calculate the capacitance: C=(8.854×1012)(0.02)[2(103)+23×106(103)3]. After evaluating this expression, we find that the capacitance of the parallel-plate capacitor is approximately 3.54×1011F.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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, ϵr=2+2×106x2, 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 0dϵ0(2+2×106x2)Sdx, 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 ϵ0, is a constant that describes how electric fields behave in a vacuum. Its value, 8.854×1012F/m, is fundamental in determining capacitance. In the problem, we use ϵ0 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, ϵ0 forms the base constant multiplied by relative permittivity to find total permittivity.
Understanding ϵ0 is key to solving problems involving dielectrics and capacitance calculations.

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Most popular questions from this chapter

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 ρ0C/m3 and has permittivity ϵ. Both plates are held at ground potential. ( a ) Determine the potential field between plates. (b) Determine the electric field intensity E between plates. (c) Repeat parts (a) and (b) for the case of the plate at z=d raised to potential V0, with the z=0 plate grounded.

Construct a curvilinear-square map of the potential field between two parallel circular cylinders, one of 4 cm radius inside another of 8 cm radius. The two axes are displaced by 2.5 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=2πϵcosh1[(a2+b2D2)/(2ab)] where a and b are the conductor radii and D is the axis separation.

Concentric conducting spheres are located at r=5 mm and r=20 mm The region between the spheres is filled with a perfect dielectric. If the inner sphere is at 100 V and the outer sphere is at 0 V(a) Find the location of the 20 V equipotential surface. (b) Find Er,max(c) Find ϵr if the surface charge density on the inner sphere is 1.0μC/m2.

Coaxial conducting cylinders are located at ρ=0.5 cm and ρ=1.2 cm. The region between the cylinders is filled with a homogeneous perfect dielectric. If the inner cylinder is at 100 V and the outer at 0 V, find (a) the location of the 20 V equipotential surface; (b)Eρmax;(c)ϵr if the charge per meter length on the inner cylinder is 20nC/m.

Let ϵr1=2.5 for \(0

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