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Ch. 24 - Capacitance, Dielectrics, Electric Energy, Storage
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 23, Problem 19b

Small distances can be measured using a capacitor whose plate separation 𝓍 is variable. Consider an air-filled parallel-plate capacitor with fixed plate area A = 25 mm² separated by a variable distance 𝓍. Assume this capacitor is attached to a capacitance-measuring instrument which can measure capacitance C in the range 1.0 pF to 1000.0 pF with an accuracy of ∆C = 0.1 pF. Define ∆𝓍 to be the accuracy (magnitude) to which 𝓍 can be determined, and determine a formula for ∆𝓍.

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Start by recalling the formula for the capacitance of a parallel-plate capacitor: C=εx, where c is the capacitance, e is permittivity and x distance

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance

Capacitance is the ability of a capacitor to store electrical charge per unit voltage. It is defined by the formula C = ε(A/x), where C is the capacitance, ε is the permittivity of the dielectric material (air in this case), A is the area of the plates, and x is the separation between them. Understanding this relationship is crucial for determining how changes in plate separation affect the measured capacitance.
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Measurement Accuracy

Measurement accuracy refers to the degree to which a measured value reflects the true value. In this context, the accuracy of the capacitance measuring instrument is given as ∆C = 0.1 pF, which indicates the smallest change in capacitance that can be reliably detected. This concept is essential for determining how precisely the plate separation x can be inferred from the measured capacitance.
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Error Propagation

Error propagation is the process of determining the uncertainty in a calculated result based on the uncertainties in the measurements used to obtain it. In this case, the uncertainty in the capacitance measurement (∆C) will affect the calculated uncertainty in the plate separation (∆x). Understanding how to propagate these errors is key to deriving a formula for ∆x based on the given ∆C.
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Related Practice
Textbook Question

To get an idea how big a farad is, suppose you want to make a 1-F air-filled parallel-plate capacitor for a circuit you are building. To make it a reasonable size, suppose you limit the plate area to 1.0 cm². What would the gap have to be between the plates? Is this practically achievable?

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Textbook Question

Suppose in Fig. 24–27 that C₁ = C₃ = 8.0μF, C₂ = C₄ = 16μF, and Q₃ = 21μC. Determine the voltage Vba across the combination.

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Textbook Question

(II) A 7.7-μF capacitor is charged by a 185-V battery (Fig. 24–21a) and then is disconnected from the battery. When this capacitor (C₁) is then connected (Fig. 24–21b) to a second (initially uncharged) capacitor, C₂, the final voltage on each capacitor is 15 V. What is the value of C₂? [Hint: Charge is conserved.]

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Textbook Question

Two capacitors connected in parallel produce an equivalent capacitance of 32.9-μF, but when connected in series the equivalent capacitance is only 5.5 μF. What is the individual capacitance of each capacitor?

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Textbook Question

In an electrostatic air cleaner (“precipitator”), the strong nonuniform electric field in the central region of a cylindrical capacitor (with outer and inner cylindrical radii Rₐ and R₆ ) is used to create ionized air molecules for use in charging dust and soot particles (Fig. 24–22). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength Eₛ ≈ 3.0 x 10⁶ N/C, air molecules will dissociate into positively charged ions and free electrons. In a precipitator, the region within which air is ionized (the corona discharge region) occupies a cylindrical volume of radius R that is typically five times that of the inner cylinder. Assume a particular precipitator is constructed with R₆ = 0.10 mm and Rₐ = 10.0 cm. In order to create a corona discharge region with radius R = 5.0 R₆, what potential difference V should be applied between the precipitator’s inner and outer conducting cylinders? [Besides dissociating air, the charged inner cylinder repels the resulting positive ions from the corona discharge region, where they are put to use in charging dust particles, which are then “collected” on the negatively charged outer cylinder.]

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