Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 36

How far must two electrons be placed on the Earth's surface for there to be an electrostatic force between them equal to the weight of one of the electrons?

Short Answer

Expert verified
Answer: The two electrons should be placed approximately \(5.47 * 10^{-11}\) meters apart.

Step by step solution

01

Write down the given information

The charge of an electron (e): \(1.6 * 10^{-19}\) C The mass of an electron (m): \(9.11 * 10^{-31}\) kg Coulomb's constant (k): \(8.988 * 10^{9} Nm^2/C^2\) Acceleration due to gravity (g): \(9.81 m/s^2\)
02

Compute the weight of one electron

To find the weight of one electron, we use the gravitational force formula: Weight (W) = mass (m) × acceleration due to gravity (g) \(W = m \times g\)
03

Write the formula for the electrostatic force between two electrons

The electrostatic force between two electrons with charges q1 and q2 at a distance r apart is given by the formula: \(F = k \frac{q1 * q2}{r^2}\) In our case, q1 = q2 = charge of an electron (e).
04

Set the weight equal to the electrostatic force and solve for r

We want to find the distance (r) between the electrons when the weight of one electron equals the electrostatic force between them: \(W = F\) \(m \times g = k \frac{e^2}{r^2}\) Now, let's solve for r: \(r^2 = k \frac{e^2}{m \times g}\) \(r = \sqrt{k \frac{e^2}{m \times g}}\) Now substitute the values of k, e, m, and g into the equation and calculate r: \(r = \sqrt{ (8.988 * 10^{9}) \frac{(1.6 * 10^{-19})^2}{(9.11 * 10^{-31}) \times (9.81)}}\) \(r \approx 5.47 * 10^{-11} m\) The two electrons must be placed approximately \(5.47 * 10^{-11}\) meters apart on the Earth's surface for the electrostatic force between them to be equal to the weight of one of the electrons.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Find the net force on a \(2.0-C\) charge at the origin of an \(x y\) -coordinate system if there is a \(+5.0-C\) charge at \((3 \mathrm{~m}, 0)\) and \(a-3.0-C\) charge at \((0,4 \mathrm{~m})\)

Four point charges are placed at the following \(x y\) coordinates: \(Q_{1}=-1 \mathrm{mC},\) at \((-3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{2}=-1 \mathrm{mC},\) at \((+3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{3}=+1.024 \mathrm{mC},\) at \((0 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{4}=+2 \mathrm{mC},\) at \((0 \mathrm{~cm},-4 \mathrm{~cm})\) Calculate the net force on charge \(Q_{4}\) due to charges \(Q_{1}, Q_{2}\) and \(Q_{3}\).

The Earth is constantly being bombarded by cosmic rays, which consist mostly of protons. These protons are incident on the Earth's atmosphere from all directions at a rate of 1245.0 protons per square meter per second. Assuming that the depth of Earth's atmosphere is \(120 \mathrm{~km},\) what is the total charge incident on the atmosphere in \(5.00 \mathrm{~min}\) ? Assume that the radius of the surface of the Earth is \(6378 \mathrm{~km}\).

In solid sodium chloride (table salt), chloride ions have one more electron than they have protons, and sodium ions have one more proton than they have electrons. These ions are separated by about \(0.28 \mathrm{nm} .\) Calculate the electrostatic force between a sodium ion and a chloride ion.

Two balls have the same mass of \(0.681 \mathrm{~kg}\) and identical charges of \(18.0 \mu \mathrm{C} .\) They hang from the ceiling on strings of identical length as shown in the figure. If the angle with respect to the vertical of the strings is \(20.0^{\circ}\), what is the length of the strings?
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Energy Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Translational Dynamics

Read Explanation

Torque and Rotational Motion

Read Explanation

Solid State 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.

Sign-up for free