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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 34c

Suppose in Fig. 24–27 that C₁ = C₃ = 8.0μF, C₂ = C₄ = 16μF, and Q₃ = 21μC. Determine the voltage Vba across the combination.
Circuit diagram showing capacitors C₁, C₂, C₃, and C₄ arranged in a combination with labeled terminals a and b.

Verified step by step guidance
1
Identify the configuration of the capacitors in the circuit. Based on the problem, capacitors C₁ and C₂ are in series, and C₃ and C₄ are in series. These two series combinations are then connected in parallel. This is a mixed series-parallel circuit.
Calculate the equivalent capacitance of the series combination of C₁ and C₂. For capacitors in series, the formula is: 1C=1C1+1C2. Substitute C₁ = 8.0 μF and C₂ = 16 μF into the formula.
Similarly, calculate the equivalent capacitance of the series combination of C₃ and C₄ using the same formula: 1C=1C3+1C4. Substitute C₃ = 8.0 μF and C₄ = 16 μF.
Add the equivalent capacitances of the two series combinations to find the total equivalent capacitance of the circuit. For capacitors in parallel, the formula is: C=Ceq1+Ceq2, where C_eq1 and C_eq2 are the equivalent capacitances of the two series combinations.
Use the relationship between charge, capacitance, and voltage: Q=CV. Given Q₃ = 21 μC, determine the voltage V₆ₐ across the entire combination by dividing the total charge by the total equivalent capacitance: V=QC.

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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 system to store an electric charge per unit voltage. It is measured in farads (F) and is defined by the formula C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage across the capacitor. In this problem, understanding the values of the capacitors (C₁, C₂, C₃, C₄) is essential for calculating the total capacitance in the circuit.
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Capacitors & Capacitance (Intro)

Voltage in Capacitor Networks

In a network of capacitors, the total voltage across the combination can be determined by analyzing how the capacitors are arranged—either in series or parallel. For capacitors in series, the total capacitance decreases, and the voltage across each capacitor adds up to the total voltage. In contrast, for capacitors in parallel, the total capacitance increases, and the voltage across each capacitor remains the same as the source voltage.
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Charge Conservation

Charge conservation is a fundamental principle stating that the total electric charge in an isolated system remains constant. In capacitor circuits, this means that the charge stored in capacitors can be redistributed but not created or destroyed. In this problem, knowing the charge Q₃ allows us to relate it to the voltages and capacitances in the circuit, which is crucial for finding the voltage V₆ₐ across the combination.
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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

What is the capacitance of a pair of circular plates with a radius of 5.0 cm separated by 2.3 mm of mica?

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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

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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Textbook Question

A cylindrical capacitor (Example 24–2) has Rₐ = 3.5 mm and R₆.= 0.50 mm. The two conductors have a potential difference of 625 V, with the inner conductor at the higher potential. Calculate the energy stored in a 1.0-m length of the 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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