/*! 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 25 The plates of an empty parallel-... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The plates of an empty parallel-plate capacitor of capacitance \(5.0 \mathrm{pF}\) are \(2.0 \mathrm{mm}\) apart. What is the area of each plate?

Short Answer

Expert verified
The area of each plate of the parallel-plate capacitor is approximately \(1.13 \times 10^{-3} \text{ m}^2\).

Step by step solution

01

Write down the given information

We are given the following information: Capacitance: \(C = 5.0 \text{ pF} = 5.0 \times 10^{-12} \text{ F}\) Distance between the plates: \(d = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}\)
02

Write down the formula for the capacitance of a parallel-plate capacitor

The formula for the capacitance of a parallel-plate capacitor is: \(C = \frac{\varepsilon_0 \cdot A}{d}\) Where: C is the capacitance, \(\varepsilon_0\) is the vacuum permittivity (\(\varepsilon_0 \approx 8.85 \times 10^{-12} \text{ F/m}\)), A is the area of each plate, and d is the distance between the plates.
03

Rearrange the formula to solve for the area A

We want to find the area A of each plate, so we need to rearrange the formula to solve for A: \(A = \frac{C \cdot d}{\varepsilon_0}\)
04

Substitute the given values into the formula and solve for A

Now, we'll substitute the given values into the formula and solve for A: \(A = \frac{(5.0 \times 10^{-12} \text{ F}) \cdot (2.0 \times 10^{-3} \text{ m})}{(8.85 \times 10^{-12} \text{ F/m})}\) \(A \approx 1.13 \times 10^{-3} \text{ m}^2\)
05

Write the final answer

The area of each plate of the parallel-plate capacitor is approximately \(1.13 \times 10^{-3} \text{ m}^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Capacitance Calculation
The concept of capacitance is fundamental when studying capacitors, especially parallel-plate capacitors. Capacitance is a measure of a capacitor's ability to store electric charge per unit voltage that is applied to it. The formula to calculate the capacitance of a parallel-plate capacitor is given by:
\(C = \frac{\varepsilon_0 \cdot A}{d}\)
where \(C\) represents the capacitance in farads (F), \(\varepsilon_0\) is the vacuum permittivity or the permittivity of free space, \(A\) is the area of one of the plates in square meters (m²), and \(d\) is the distance between the two plates in meters (m).

In the context of our exercise, knowing the capacitance and the distance between the plates allows us to calculate the area of the plates. This ability to determine one unknown parameter when others are known makes the capacitance calculation a powerful tool in circuit design and analysis.

Practical Applications

Calculating capacitance is not only essential for answering academic problems but also plays a vital role in designing electronic components. It helps engineers control the charge storage and discharge rates in electronic circuits, affecting everything from timing to signal processing.
Vacuum Permittivity
Vacuum permittivity, denoted by \(\varepsilon_0\), is a physical constant that describes how an electric field affects and is affected by the vacuum of space. It is also known as the electric constant and has a value of approximately \(8.85 \times 10^{-12} \text{ F/m}\). It plays a crucial role in electrodynamics, as it appears in Coulomb's law, which dictates the force of attraction or repulsion between two point charges.

Vacuum permittivity is part of our capacitance calculation for parallel plate capacitors. Although it might seem abstract because it involves the concept of a vacuum, it's actually instrumental in determining how the electric field behaves in various media. Whenever a capacitor is not filled with a material (dielectric), we use vacuum permittivity in our calculations. However, if a dielectric material is present, the permittivity of that material would be used instead.
Electric Field in Capacitors
The electric field in a capacitor is a uniform field created by the separation of charges on the two plates of the capacitor. For a parallel-plate capacitor, the electric field \(E\) between the plates is directly proportional to the surface charge density \(\sigma\) and is inversely proportional to the vacuum permittivity \(\varepsilon_0\). The relationship is described by:
\(E = \frac{\sigma}{\varepsilon_0}\)
where \(\sigma\) is the charge per unit area on the plates. The electric field is crucial for a capacitor's function and affects characteristics such as voltage, energy storage, and the forces on charges within the field.

Uniform Electric Field

One of the most interesting aspects of parallel-plate capacitors is that the electric field between the plates is considered uniform. This assumption holds true when the dimensions of the plates are significantly larger than the distance between them, preventing fringe effects. The uniform field greatly simplifies the analysis of capacitors and is a reason why parallel-plate capacitors are extensively studied in basic physics courses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(4.00-\mu \mathrm{F}\) capacitor and a \(6.00-\mu \mathrm{F}\) capacitor are connected in parallel across a 600 -V supply line. (a) Find the charge on each capacitor and voltage across each. (b) The charged capacitors are disconnected from the line and from each other. They are then reconnected to each other with terminals of unlike sign together. Find the final charge on each capacitor and the voltage across each.

Three capacitors, with capacitances of \(C_{1}=2.0 \mu \mathrm{F}\) \(C_{2}=3.0 \mu \mathrm{F}, \quad\) and \(\quad C_{3}=6.0 \mu \mathrm{F}, \quad\) respectively, \(\quad\) are connected in parallel. A \(500-\mathrm{V}\) potential difference is applied across the combination. Determine the voltage across each capacitor and the charge on each capacitor.

A parallel-plate capacitor is made of two square plates \(25 \mathrm{cm}\) on a side and \(1.0 \mathrm{mm}\) apart. The capacitor is connected to a \(50.0-\mathrm{V}\) battery. With the battery still connected, the plates are pulled apart to a separation of 2.00 mm. What are the energies stored in the capacitor before and after the plates are pulled farther apart? Why does the energy decrease even though work is done in separating the plates?

A spherical capacitor is formed from two concentric spherical conducting spheres separated by vacuum. The inner sphere has radius \(12.5 \mathrm{cm}\) and the outer sphere has radius \(14.8 \mathrm{cm} .\) A potential difference of \(120 \mathrm{V}\) is applied to the capacitor. (a) What is the capacitance of the capacitor? (b) What is the magnitude of the electrical field at \(r=12.6 \mathrm{cm},\) just outside the inner sphere? (c) What is the magnitude of the electrical field at \(r=14.7 \mathrm{cm},\) just inside the outer sphere? (d) For a parallel-plate capacitor the electrical field is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

Some cell walls in the human body have a layer of negative charge on the inside surface. Suppose that the surface charge densities are \(\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) the cell wall is \(5.0 \times 10^{-9} \mathrm{m}\) thick, and the cell wall material has a dielectric constant of \(\kappa=5.4 .\) (a) Find the magnitude of the electric field in the wall between two charge layers. (b) Find the potential difference between the inside and the outside of the cell. Which is at higher potential? (c) A typical cell in the human body has volume \(10^{-16} \mathrm{m}^{3}\) Estimate the total electrical field energy stored in the wall of a cell of this size when assuming that the cell is spherical. (Hint: Calculate the volume of the cell wall.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.