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 53

Chapter 23, Problem 53

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

Verified step by step guidance

Start by recalling the formula for the capacitance of a parallel plate capacitor:

C = \frac{ε}{d} \cdot A

, where

ε

is the permittivity of the material between the plates,

d

is the separation between the plates, and

A

is the area of one plate.

Determine the permittivity of mica. The permittivity

ε

is given by

ε = ε

₀⋅κ, where

ε

₀ is the permittivity of free space (

8.85×10

^-12 F/m) and

κ

is the dielectric constant of mica (approximately 7).

Calculate the area of one circular plate using the formula for the area of a circle:

A = π \cdot r

², where

r

is the radius of the plate. Convert the radius from centimeters to meters before substituting into the formula.

Convert the separation between the plates from millimeters to meters. This is necessary to ensure all units are consistent in the SI system.

Substitute the values for

ε

A

, and

d

into the capacitance formula to calculate the capacitance. Ensure that all calculations are performed with consistent units.

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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 electric charge per unit voltage. It is defined as the ratio of the electric charge (Q) stored on the plates to the potential difference (V) between them, expressed as C = Q/V. The unit of capacitance is the farad (F), and it is influenced by the physical characteristics of the capacitor, including the area of the plates and the distance between them.

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates separated by an insulating material, known as a dielectric. The capacitance of such a capacitor can be calculated using the formula C = (ε₀ * ε_r * A) / d, where ε₀ is the permittivity of free space, ε_r is the relative permittivity of the dielectric, A is the area of one plate, and d is the separation between the plates. This configuration allows for efficient charge storage.

Dielectric Material

A dielectric material is an insulating substance that can be polarized by an electric field, which enhances the capacitance of a capacitor. Mica, in this case, is a type of dielectric with a high relative permittivity (ε_r), which increases the capacitor's ability to store charge compared to a vacuum. The presence of a dielectric reduces the electric field strength between the plates, allowing for greater charge accumulation.

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Intro To Dielectrics

Related Practice

Textbook Question

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

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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 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

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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