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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 36

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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1
Step 1: Understand the problem. Two capacitors are connected in parallel and series, and their equivalent capacitances are given. The goal is to find the individual capacitances of each capacitor, denoted as \( C_1 \) and \( C_2 \).
Step 2: Recall the formulas for equivalent capacitance. For capacitors in parallel, the equivalent capacitance is \( C_{eq} = C_1 + C_2 \). For capacitors in series, the equivalent capacitance is given by \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \).
Step 3: Use the parallel formula to set up the first equation. Substitute the given parallel equivalent capacitance: \( C_1 + C_2 = 32.9 \, \mu\text{F} \).
Step 4: Use the series formula to set up the second equation. Substitute the given series equivalent capacitance: \( \frac{1}{5.5} = \frac{1}{C_1} + \frac{1}{C_2} \). Rewrite this equation as \( \frac{1}{C_1} + \frac{1}{C_2} = 0.1818 \, \text{(approx)} \).
Step 5: Solve the system of equations. Use the two equations \( C_1 + C_2 = 32.9 \, \mu\text{F} \) and \( \frac{1}{C_1} + \frac{1}{C_2} = 0.1818 \) to find the values of \( C_1 \) and \( C_2 \). This involves algebraic manipulation or substitution to solve for the individual capacitances.

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Key Concepts

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

Capacitance in Parallel

When capacitors are connected in parallel, the total or equivalent capacitance is the sum of the individual capacitances. This means that if two capacitors, C1 and C2, are connected in parallel, the equivalent capacitance (Ceq) can be calculated using the formula Ceq = C1 + C2. This configuration allows for increased capacitance, as the charge storage capability is enhanced.
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Capacitance in Series

In a series connection, the equivalent capacitance is found using the formula 1/Ceq = 1/C1 + 1/C2. This means that the reciprocal of the total capacitance is equal to the sum of the reciprocals of the individual capacitances. Series connections result in a lower equivalent capacitance compared to the individual capacitors, as the charge must pass through each capacitor sequentially.
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Solving for Individual Capacitances

To find the individual capacitances of capacitors connected in both parallel and series, one can set up a system of equations based on the known equivalent capacitances. By using the equations for capacitance in parallel and series, one can solve for the unknown values of C1 and C2. This often involves algebraic manipulation and substitution to isolate the variables.
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Related Practice
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

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

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

A 3500-pF air-gap capacitor is connected to an 18-V battery. If a piece of mica fills the space between the plates, how much charge will flow from the battery?

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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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