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

Giancoli Douglas 5th edition

Ch. 24 - Capacitance, Dielectrics, Electric Energy, Storage

Problem 57

Chapter 23, Problem 57

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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Identify the given values: The capacitance of the air-gap capacitor is \( C = 3500 \, \text{pF} = 3.5 \times 10^{-9} \, \text{F} \), the voltage across the capacitor is \( V = 18 \; \text{V} \), and the dielectric material introduced is mica, which has a dielectric constant \( \kappa \approx 7 \).

Recall the formula for capacitance with a dielectric: \( C' = \kappa C \), where \( C' \) is the new capacitance after the dielectric is introduced. Substitute the given values to calculate \( C' \).

Use the relationship between charge, capacitance, and voltage: \( Q = C' V \), where \( Q \) is the charge stored on the capacitor. Substitute \( C' \) and \( V \) into this formula to find the charge.

Perform the unit conversions if necessary to ensure consistency (e.g., converting capacitance to farads and charge to coulombs).

Conclude that the charge \( Q \) represents the amount of charge that flows from the battery to the capacitor when the mica is introduced.

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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 charge per unit voltage, measured in farads (F). It is defined by the formula C = Q/V, where C is capacitance, Q is charge, and V is voltage. In this scenario, the initial capacitance of the air-gap capacitor is given as 3500 pF (picoFarads), which indicates how much charge it can hold at a specific voltage.

Dielectric Material

A dielectric material, such as mica, is an insulating substance that increases a capacitor's capacitance when placed between its plates. The presence of a dielectric reduces the electric field within the capacitor, allowing it to store more charge for the same voltage. The dielectric constant of mica is significantly higher than that of air, which means it will enhance the capacitor's ability to hold charge when inserted.

Charge Flow

Charge flow refers to the movement of electric charge, typically measured in coulombs (C), from the battery to the capacitor. When the capacitor is connected to the battery, the voltage causes charge to accumulate on the plates until the voltage across the capacitor equals the battery voltage. The total charge can be calculated using the modified capacitance with the dielectric, following the formula Q = C * V.

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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

A parallel-plate capacitor with plate area 2.0 cm² and air-gap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected. What is the charge on the capacitor?

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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

The capacitor shown in Fig. 24–34 is connected to an 80.0-V battery. Calculate (and sketch) the electric field everywhere between the capacitor plates. Find both the free charge on each capacitor plate and the induced charge on the faces of the glass dielectric plate.

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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 parallel-plate capacitor has square plates 12 cm on a side separated by 0.10 mm of plastic with a dielectric constant of K = 3.8. The plates are connected to a battery, causing them to become oppositely charged. Since the oppositely charged plates attract each other, they exert a pressure on the dielectric. If this pressure is 40.0 Pa, what is the battery voltage?

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