/*! 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 63 Two parallel plates \(10 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two parallel plates \(10 \mathrm{cm}\) on a side are given equal and opposite charges of magnitude \(5.0 \times 10^{-9} \mathrm{C}\). The plates are \(1.5 \mathrm{mm}\) apart. What is the potential difference between the plates?

Short Answer

Expert verified
The potential difference between the parallel plates is \(847.5 \mathrm{V}\).

Step by step solution

01

Calculate the electric field between the plates

The electric field between two parallel plates with equal and opposite charges can be calculated using the formula: \(E = \dfrac{Q}{A\epsilon_0}\) where \(E\) is the electric field, \(Q\) is the charge on each plate, \(A\) is the area of the plates, and \(\epsilon_0\) is the vacuum permittivity constant \((8.85 \times 10^{-12} \mathrm{C^2/Nm^2})\). First, let's calculate the area of the plates: \(A = (10 \mathrm{cm})^2 = (0.1 \mathrm{m})^2 = 0.01 \mathrm{m^2}\) Now, let's plug in the values: \(E = \dfrac{5.0 \times 10^{-9} \mathrm{C}}{0.01 \mathrm{m^2}(8.85 \times 10^{-12} \mathrm{C^2/Nm^2})} = 5.65 \times 10^5 \mathrm{N/C}\) So, the electric field between the plates is \(5.65 \times 10^5 \mathrm{N/C}\).
02

Calculate the potential difference between the plates

To find the potential difference between the plates, we can use the relationship between electric field and potential difference: \(V = E \cdot d\) where \(V\) is the potential difference, \(E\) is the electric field, and \(d\) is the distance between the plates. The distance between the plates is given as \(1.5 \mathrm{mm}\), which is equal to \(0.0015 \mathrm{m}\). Now, let's plug in the values: \(V = (5.65 \times 10^5 \mathrm{N/C}) (0.0015 \mathrm{m}) = 847.5 \mathrm{V}\) So, the potential difference between the plates is \(847.5 \mathrm{V}\).

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.

Electric Field
The concept of the electric field is central to understanding how charges interact with their environment. The electric field is a vector quantity, meaning it has both magnitude and direction. It influences how charged objects will move when placed within the field.
The electric field (E) between two charged parallel plates is uniform, which means it has the same strength and direction everywhere between the plates. It can be calculated using the formula:
  • \( E = \dfrac{Q}{A \epsilon_0} \)
  • \( Q \) is the magnitude of the charge on each plate.
  • \( A \) is the area of the plates.
  • \( \epsilon_0 \) is the vacuum permittivity, which we'll explore further later.
The uniform electric field between parallel plates is one of its most significant properties, simplifying calculations and allowing us to make predictions about forces, potential differences, and motions of charged particles.
Parallel Plates
Parallel plates are a common setup for creating a uniform electric field. They consist of two flat, conductive plates that are very close to each other and are charged with equal but opposite charges. This configuration helps us study the properties of electric fields in a controlled environment.
In our example problem, we see two plates each being 10 cm on a side, and they are separated by just 1.5 mm. This small distance coupled with the larger area, ensures a uniform electric field can be created between the plates.
This kind of setup is often used in capacitors, devices that store electrical energy. The key features of parallel plates include:
  • The potential difference is directly proportional to the separation of the plates.
  • The area of the plates affects the amount of charge they can hold.
  • They allow simple calculations that are crucial for understanding basic principles in electromagnetism.
Vacuum Permittivity
Vacuum permittivity, denoted as \( \epsilon_0 \), is a physical constant that describes how electric field lines pass through free space. It is vital for understanding how electromagnetic forces work in a vacuum and is crucial in calculations involving electric fields.
The value of vacuum permittivity is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2 \). This constant appears in many fundamental equations in physics such as Coulomb's law and, as mentioned earlier, in the formula for the electric field between parallel plates.
The significance of vacuum permittivity lies in its role in determining how easily electric fields can establish and propagate through space. It essentially acts as a measure of "resistance" that a material (in this case, a vacuum) provides to the formation of an electric field.
  • Essential for calculating electric fields in a vacuum.
  • Widely used in formulas relating to electrical force, field, and potential.
Potential Difference
The potential difference, often referred to as voltage (V), is the work needed to move a charge from one point in an electric field to another. It is what drives electric current in circuits and is a critical concept in the study of electromagnetism.
Between two parallel plates, the potential difference is a result of the electric field established by the charges on the plates. We can calculate the potential difference using:
  • \( V = E \cdot d \)
  • \( E \) is the uniform electric field between the plates.
  • \( d \) is the distance between the plates.
In our calculation example, with \( E = 5.65 \times 10^5 \text{ N/C} \) and \( d = 0.0015 \text{ m} \), we find that the potential difference \( V \) equals to 847.5 volts.
The potential difference is critical as it defines how easy or difficult it is for charges to move between two points. This concept is not only important in physics but also in everyday technology like batteries and power supplies.

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

If the potential due to a point charge is \(5.00 \times 10^{2} \mathrm{V}\) at a distance of \(15.0 \mathrm{m},\) what are the sign and magnitude of the charge?

A metallic sphere of radius 2.0 cm is charged with \(+5.0-\mu \mathrm{C}\) charge, which spreads on the surface of the sphere uniformly. The metallic sphere stands on an insulated stand and is surrounded by a larger metallic spherical shell, of inner radius 5.0 cm and outer radius 6.0 cm. Now, a charge of \(-5.0-\mu \mathrm{C}\) is placed on the inside of the spherical shell, which spreads out uniformly on the inside surface of the shell. If potential is zero at infinity, what is the potential of (a) the spherical shell, (b) the sphere, (c) the space between the two, (d) inside the sphere, and (e) outside the shell?

Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. What is the voltage across an 8.00 -nm-thick membrane if the electric field strength across it is \(5.50 \mathrm{MV} / \mathrm{m}\) ? You may assume a uniform electric field.

Earth has a net charge that produces an electric field of approximately \(150 \mathrm{N} / \mathrm{C}\) downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center? (b) What acceleration will the field produce on a free electron near Earth's surface? (c) What mass object with a single extra electron will have its weight supported by this field?

To form a helium atom, an alpha particle that contains two protons and two neutrons is fixed at one location, and two electrons are brought in from far away, one at a time. The first electron is placed at \(0.600 \times 10^{-10} \mathrm{m}\) from the alpha particle and held there while the second electron is brought to \(0.600 \times 10^{-10} \mathrm{m}\) from the alpha particle on the other side from the first electron. See the final configuration below. (a) How much work is done in each step? (b) What is the electrostatic energy of the alpha particle and two electrons in the final configuration?

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.