/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 The plates of a parallel-plate c... [FREE SOLUTION] | 91Ó°ÊÓ

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The plates of a parallel-plate capacitor are made of circular discs of radii \(5.0 \mathrm{~cm}\) each. If the separation between the plates is \(1 \cdot 0 \mathrm{~mm}\), what is the capacitance?

Short Answer

Expert verified
The capacitance is approximately 0.695 pF.

Step by step solution

01

Understand the Formula

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

Convert Given Quantities to SI Units

Convert the quantities into SI units:- The radius \( r \) is given as \( 5.0 \mathrm{~cm} = 0.05 \mathrm{~m} \).- The separation \( d \) is given as \( 1.0 \mathrm{~mm} = 0.001 \mathrm{~m} \).
03

Calculate the Area of the Plates

The area \( A \) of a circular disc is given by\[ A = \pi r^2 \]Substituting the radius \( r = 0.05 \mathrm{~m} \), we get:\[ A = \pi (0.05)^2 = \pi \times 0.0025 \approx 0.00785 \mathrm{~m}^2 \].
04

Substitute Values into the Capacitance Formula

Substitute the values into the formula for capacitance:\[ C = \frac{\varepsilon_0 A}{d} = \frac{8.854 \times 10^{-12} \times 0.00785}{0.001} \approx 6.95 \times 10^{-13} \mathrm{~F} \].
05

Express the Capacitance in Picofarads

Convert the capacitance to picofarads (pF), since 1 pF = \(10^{-12}\) F:\[ C \approx 6.95 \times 10^{-13} \mathrm{~F} = 0.695 \mathrm{~pF} \].

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Key Concepts

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

Understanding the Capacitance Formula
The capacitance formula for a parallel-plate capacitor is an essential tool in understanding how capacitors work. It is given by \( C = \frac{\varepsilon_0 A}{d} \), where:
  • \( C \) represents the capacitance, measured in farads (F).
  • \( \varepsilon_0 \) denotes the permittivity of free space, a constant value crucial in determining how the electric field behaves between the plates.
  • \( A \) is the area of one of the plates, measured in square meters (\( \text{m}^2 \)).
  • \( d \) is the distance between the two plates, measured in meters (m).
This formula tells us that the capacitance is directly proportional to the area of the plates and the permittivity of free space but inversely proportional to the distance between the plates. Consequently, larger plates or closer spacing between them results in higher capacitance. Understanding this relationship is critical when designing circuits and electronic devices.
Exploring the Permittivity of Free Space
Permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental constant in electromagnetism. It defines how much electric field is "permitted" to pass through the vacuum of free space. Its value is \( 8.854 \times 10^{-12} \text{ F/m} \). This quantity is vital because it helps us understand how electric fields interact with materials.

Several factors make \( \varepsilon_0 \) important, including:
  • Its role in determining the energy stored in the capacitor.
  • The impact on the force between charged particles in a vacuum.
Understanding \( \varepsilon_0 \) allows us to predict and control how devices like capacitors, insulators, and even transmitters behave in different conditions. Its consistent value ensures that the laws of electrical physics hold true across various applications.
The Importance of SI Units Conversion
Converting quantities to the International System of Units (SI) is crucial for ensuring consistency and accuracy, particularly in scientific calculations. This system is widely used because it provides a standard measurement framework that is universally understood.

In the context of our capacitor problem:
  • The radius given as \( 5.0 \text{ cm} \) is converted to meters by multiplying by 0.01, making it \( 0.05 \text{ m} \).
  • The separation of \( 1.0 \text{ mm} \) is converted to meters by multiplying by 0.001, resulting in \( 0.001 \text{ m} \).
Conversions ensure that calculations are done in the same unit system, which avoids errors and makes the results dependable. This is particularly important when dealing with formulas that involve constants, such as \( \varepsilon_0 \), which are defined in specific units. Sticking to SI units in scientific contexts ensures clarity and keeps the findings consistent with global scientific standards.

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