/*! 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 12 As a space shuttle moves through... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 12

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, its potential is typically changed by \(-1.0 \mathrm{~V}\) during one revolution. By assuming that the shuttle is a sphere of radius \(10 \mathrm{~m}\), estimate the amount of charge it collects.

Short Answer

Expert verified
The amount of charge collected by the shuttle is approximately \(-1.11 \times 10^{-9} \mathrm{~C}\).

Step by step solution

01

- Understand the Problem

The problem involves calculating the amount of charge collected by a space shuttle, modeled as a sphere, when its potential changes by \(-1.0 \mathrm{~V}\). Given the shuttle's radius (\(R = 10 \mathrm{~m}\)), we need to find the amount of charge.
02

- Use the Potential Formula

The potential \(V\) of a sphere of radius \(R\) carrying a charge \(Q\) is given by: \[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \]
03

- Solve for Charge \(Q\)

Rearrange the formula to solve for \(Q\): \[ Q = V \cdot 4 \pi \epsilon_0 \cdot R \]
04

- Substitute Known Values

Substitute \(V = -1.0 \mathrm{~V}\), \(R = 10 \mathrm{~m}\), and \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{~F/m}\) into the equation: \[ Q = (-1.0 \mathrm{~V}) \cdot 4 \pi \cdot (8.85 \times 10^{-12} \mathrm{~F/m}) \cdot 10 \mathrm{~m} \]
05

- Simplify the Expression

Compute the charge \(Q\): \[ Q = -1.0 \cdot 4 \pi \cdot 8.85 \times 10^{-12} \cdot 10 \approx -1.11 \times 10^{-9} \mathrm{~C} \]

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 91影视!

Key Concepts

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

potential of a charged sphere
The potential of a charged sphere is a fundamental concept in electrostatics, particularly relevant in the space environment. For a sphere with radius \( R \) and charge \( Q \), the potential \( V \) at its surface can be calculated using: \[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \].
This formula reflects how the potential is directly proportional to the charge and inversely proportional to the radius. Here, \( \epsilon_0 \) is the permittivity of free space (8.85 x 10-12 F/m).
For example, if we have a sphere like the space shuttle with a radius of 10 meters, and it collects charge until its potential changes by -1.0 V, we can substitute these values into the formula to find the collected charge.
Rearranging the initial equation to solve for \( Q \) results in \[ Q = V \cdot 4\pi \epsilon_0 \cdot R \].
This calculation shows us how to determine the quantity of charge based on the potential change.
space shuttle ionosphere interaction
The interaction between the space shuttle and the ionosphere is an intriguing aspect of electrostatics in the space environment. As the shuttle moves through the ionosphere, it encounters a dilute ionized gas, affecting its electrical properties.
Typically, the potential of the shuttle changes due to the collection of charged particles, primarily electrons and ions.
One of the most important factors in this interaction is the shuttle's motion through the ionosphere. This motion causes it to sweep up charged particles, leading to the shuttle acquiring a net charge.
As seen in our example, the shuttle is modeled as a sphere with a radius of 10 meters, and its potential decreases by -1.0 V with each revolution.
This information allows us to estimate the amount of charge accumulated using the potential formula for a sphere, highlighting the interconnectedness of motion, charge collection, and potential change.
Coulomb's law
Coulomb's law is a cornerstone of electrostatics, governing the force between two charges. It states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) is directly proportional to the product of the charges and inversely proportional to the square of the distance \( r \) between them: \[ F = k \frac{q_1 q_2}{r^2} \].
Here, \( k \) is Coulomb's constant, approximately equal to \( 8.99 \times 10^9 \mathrm{N \cdot m^2/C^2} \).
This law explains the attraction or repulsion between charged particles.
In the context of the space shuttle, understanding how these forces work helps clarify why the shuttle can accumulate charge as it traverses the ionosphere. The electric forces between the shuttle and the ionized gas particles play a significant role in charge dynamics.
By combining the knowledge of Coulomb's law with the potential of a charged sphere, we get a comprehensive understanding of charge collection and distribution in the space environment.

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

An electron is projected with an initial speed of \(3.2 \times 10^{5} \mathrm{~m} / \mathrm{s}\) directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to twice the initial value?

The electric field inside a nonconducting sphere of radius \(R\), with charge spread uniformly throughout its volume, is radially directed and has magnitude $$E(r)=|\vec{E}(r)|=\frac{|q| r}{4 \pi \varepsilon_{0} R^{3}}.$$ Here \(q\) (positive or negative) is the total charge within the sphere, and \(r\) is the distance from the sphere's center. (a) Taking \(V=0\) at the center of the sphere, find the electric potential \(V(r)\) inside the sphere. (b) What is the difference in electric potential between a point on the surface and the sphere's center? (c) If \(q\) is positive, which of those two points is at the higher potential?

What is the excess charge on a conducting sphere of radius \(r=0.15 \mathrm{~m}\) if the potential of the sphere is \(1500 \mathrm{~V}\) and \(V=0\) at infinity?

Two tiny metal spheres \(A\) and \(B\) of mass \(m_{A}=5.00 \mathrm{~g}\) and \(m_{B}=10.0 \mathrm{~g}\) have equal positive charges \(q=5.00 \mu \mathrm{C}\). The spheres are connected by a massless nonconducting string of length \(d=1.00 \mathrm{~m}\), which is much greater than the radii of the spheres. (a) What is the electric potential energy of the system? (b) Suppose you cut the string. At that instant, what is the acceleration of each sphere? (c) A long time after you cut the string, what is the speed of each sphere?

Consider two widely separated conducting spheres, 1 and 2 , the second having twice the diameter of the first. The smaller sphere initially has a positive charge \(q\), and the larger one is initially uncharged. You now connect the spheres with a long thin wire. (a) How are the final potentials \(V_{1}\) and \(V_{2}\) of the spheres related? (b) What are the final charges \(q_{1}\) and \(q_{2}\) on the spheres, in terms of \(q ?(\mathrm{c})\) What is the ratio of the final surface charge density of sphere 1 to that of sphere \(2 ?\)
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Medical Physics

Read Explanation

Nuclear Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Work Energy and Power

Read Explanation

Scientific Method Physics

Read Explanation
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.