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Chapter 24: Problem 11

(1) To make a \(0.40-\mu\) F capacitor, what area must the plates have if they are to be separated by a 2.8 -mm air gap?

Short Answer

Expert verified
The plate area should be approximately 126.48 m虏.

Step by step solution

01

Understand the Capacitance Formula

The formula for the capacitance of a parallel plate capacitor is given by \( C = \frac{\varepsilon_0 \cdot A}{d} \), where \( C \) is the capacitance, \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ F/m} \) (the permittivity of free space), \( A \) is the area of the plates, and \( d \) is the separation between the plates. We need to find \( A \).
02

Convert Units for Consistency

The given capacitance \( C = 0.40 \text{ }\mu\text{F} = 0.40 \times 10^{-6} \text{ F} \). The separation \( d = 2.8 \text{ mm} = 2.8 \times 10^{-3} \text{ m} \).
03

Rearrange the Formula to Solve for Area

Rearrange the formula to solve for \( A \): \[ A = \frac{C \cdot d}{\varepsilon_0} \]
04

Plug in the Values

Substitute the known values into the rearranged formula:\[ A = \frac{0.40 \times 10^{-6} \text{ F} \times 2.8 \times 10^{-3} \text{ m}}{8.854 \times 10^{-12} \text{ F/m}} \]
05

Calculate the Area

Perform the calculation: \[ A = \frac{1.12 \times 10^{-9}}{8.854 \times 10^{-12}} \approx 126.48 \text{ m}^2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance
Capacitance is a measure of a capacitor's ability to store an electric charge. Simply put, it's like a "container" for electric charge, determining how much electric energy can be stored. Capacitance is denoted by the symbol \( C \) and is measured in Farads (F). The greater the capacitance, the more charge a capacitor can store at a given voltage.
Capacitance depends on several factors:
  • The surface area of the plates: Larger areas allow for more charge to be stored.
  • The distance between the plates: Closer plates can store more charge.
  • The material between the plates: Different materials affect how charge is stored.
Understanding these factors helps in determining how to construct a practical capacitor for specific applications. Knowing how they work is crucial for designing and troubleshooting circuits in electronics.
Parallel Plate Capacitor
A parallel plate capacitor is a type of capacitor where two conductive plates are placed parallel to each other. This is a simple and common design used in various electronic devices. The space between the plates can be filled with air, a vacuum, or some insulating material, known as a dielectric, to enhance capacitance.
The ability of a parallel plate capacitor to store charge is calculated using the formula: \( C = \frac{\varepsilon_0 \cdot A}{d} \). Here,
  • \( \varepsilon_0 \) refers to the permittivity of free space, a constant value that represents how much electric flux is allowed through a vacuum.
  • \( A \) is the area of one of the plates. Larger areas provide more space to store charge.
  • \( d \) is the distance between the plates. Smaller distances increase the capacitance.
This simple structure allows for great versatility and is central to the functioning of modern electronics.
Permittivity of Free Space
Permittivity of free space, denoted by \( \varepsilon_0 \), is a fundamental physical constant describing how electric fields interact in a vacuum. It is used in calculating capacitance and appears in equations involving electric fields and forces. The constant has a value of approximately \( 8.854 \times 10^{-12} \text{ F/m} \).
The concept of permittivity is crucial in understanding how capacitors interact with electric fields:
  • It determines how much electric field can "pass through" a certain space.
  • Higher permittivity allows more electric field lines within a given space, enhancing the ability to store charge.
  • In equations for capacitance, permittivity helps in determining the influence of the dielectric material between the plates.
Overall, permittivity helps us comprehend how materials affect and store electric charges, making it key for accurate design and analysis of electronic components.

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Most popular questions from this chapter

The charge on a capacitor increases by \(26 \mu \mathrm{C}\) when the voltage across it increases from \(28 \mathrm{~V}\) to \(78 \mathrm{~V}\). What is the capacitance of the capacitor?

(II) Two capacitors, \(C_{1}=3200 \mathrm{pF}\) and \(C_{2}=1800 \mathrm{pF},\) are connected in series to a \(12.0-\mathrm{V}\) battery. The capacitors are later disconnected from the battery and connected directly to each other, positive plate to positive plate, and negative plate to negative plate. What then will be the charge on each capacitor?

(II) Two identical capacitors are connected in parallel and each acquires a charge \(Q_{0}\) when connected to a source of voltage \(V_{0} .\) The voltage source is disconnected and then a dielectric \((K=3.2)\) is inserted to fill the space between the plates of one of the capacitors. Determine \((a)\) the charge now on each capacitor, and \((b)\) the voltage now across each capacitor.

In a dynamic random access memory } & \text { (DRAM) }\end{array}\( computer chip, each memory cell chiefly consists of a capacitor for charge storage. Each of these cells represents a single binary-bit value of 1 when its 35 -fF capacitor \)\left(1 \mathrm{fF}=10^{-15} \mathrm{~F}\right)\( is charged at \)1.5 \mathrm{~V},\( or 0 when uncharged at \)0 \mathrm{~V}\(. \)(a)\( When it is fully charged, how many excess electrons are on a cell capacitor's negative plate? (b) After charge has been placed on a cell capacitor's plate, it slowly "leaks" off (through a variety of mechanisms) at a constant rate of \)0.30 \mathrm{fC} / \mathrm{s} .\( How long does it take for the potential difference across this capacitor to decrease by \)1.0 \%$ from its fully charged value? (Because of this leakage effect, the charge on a DRAM capacitor is "refreshed" many times per second.)

Two capacitors, \(C_{1}=3200 \mathrm{pF}\) and \(C_{2}=1800 \mathrm{pF},\) are connected in series to a 12.0-V battery. The capacitors are later disconnected from the battery and connected directly to each other, positive plate to positive plate, and negative plate to negative plate. What then will be the charge on each capacitor?
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