/*! 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 68 How close must two electrons be ... [FREE SOLUTION] | 91Ó°ÊÓ

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How close must two electrons be if the electric force between them is equal to the weight of either at the Earth's surface?

Short Answer

Expert verified
The distance between the electrons must be approximately \(5.09 \times 10^{-11}\) meters.

Step by step solution

01

Identify the Forces

To solve the problem, we need to equate the electric force between two electrons to the gravitational force (weight) of a single electron at the Earth's surface. The electric force can be calculated using Coulomb's law, and the weight can be calculated using the formula for gravitational force. The formulas are as follows:Coulomb's law for electric force: \[ F_e = \frac{k imes e^2}{r^2} \]where \( F_e \) is the electric force, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) \), \( e \) is the elementary charge \( (1.60 \times 10^{-19} \, \text{C}) \), and \( r \) is the distance between the electrons.Gravitational force (weight):\[ F_g = m imes g \]where \( F_g \) is the gravitational force, \( m \) is the mass of an electron \( (9.11 \times 10^{-31} \, \text{kg}) \), and \( g \) is the acceleration due to gravity \( (9.81 \, \text{m/s}^2) \).
02

Set the Forces Equal

We need to find the distance \( r \) where these forces are equal, so we set the electric force \( F_e \) equal to the gravitational force \( F_g \). The equation becomes:\[ \frac{k imes e^2}{r^2} = m imes g \]
03

Isolate the Distance (r)

Solve the equation for \( r \). Start by multiplying both sides by \( r^2 \) to get rid of the denominator:\[ k imes e^2 = m imes g imes r^2 \]Then, divide both sides by \( m imes g \) to isolate \( r^2 \):\[ r^2 = \frac{k imes e^2}{m imes g} \]
04

Calculate the Distance (r)

Substitute the known values into the equation:\[ r^2 = \frac{ (8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) imes (1.60 \times 10^{-19} \, \text{C})^2 }{ 9.11 \times 10^{-31} \, \text{kg} imes 9.81 \, \text{m/s}^2 } \]Calculate the expression to find \( r^2 \). Finally, take the square root of both sides to solve for \( r \):\[ r = \sqrt{ \frac{ (8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) imes (1.60 \times 10^{-19} \, \text{C})^2 }{ 9.11 \times 10^{-31} \, \text{kg} imes 9.81 \, \text{m/s}^2 } } \]
05

Solve for Numerical Value

Solve the equation to find the numerical value of \( r \). After substituting the values and calculating:\[ r \approx 5.09 \times 10^{-11} \, \text{m} \]

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Key Concepts

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

Coulomb's law
Coulomb's law is a fundamental principle in physics that describes how electric forces interact between charged particles. It forms the basis for understanding electromagnetic interactions. The law states that the electric force between two point charges is directly proportional to the product of their charges. It is also inversely proportional to the square of the distance between them. This can be understood with the formula: \[ F_e = \frac{k \times q_1 \times q_2}{r^2} \]
  • Here, \( F_e \) is the electric force.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance between the charges.
This law enables us to compute the strength and direction of the electric force, which can be attractive or repulsive depending on the nature of the charges.
When considering electrons, which both have a negative charge, the force is always repulsive. This leads them to push away from each other at any distance. The application of Coulomb's law is evident in this exercise as it helps determine the distance at which their electromagnetic repulsion equals gravitational attraction.
gravitational force
Gravitational force is a natural phenomenon by which all objects with mass are brought toward one another. On the Earth's surface, this force is experienced as weight. The gravitational force acting on an object is given by the equation: \[ F_g = m \times g \]
  • Here, \( F_g \) is the gravitational force or the weight.
  • \( m \) is the mass of the object, such as an electron in this scenario.
  • \( g \) is the acceleration due to gravity on Earth, approximately \( 9.81 \, \text{m/s}^2 \).
Despite the fact that gravitational forces are universal and act over much larger distances compared to electric forces, they are significantly weaker. For example, gravity keeps us grounded on Earth, but it requires a significant mass to exert a force comparable to electromagnetic interactions at small scales. In our exercise, the task was to find the distance at which the electric force between two electrons equals the gravitational force on one. This demonstrates the vastly different magnitudes of these two forces.
elementary charge
The elementary charge is the smallest unit of electric charge that is relevant in physics. It is the charge of a single proton or, in the context of this exercise, an electron, albeit with opposite signs. The value of the elementary charge \( e \) is approximately \( 1.60 \times 10^{-19} \, \text{C} \).
The concept of elementary charge is crucial for understanding chemical reactions and physical processes involving charged particles. In the context of Coulomb's law, it allows us to calculate the force between particles like electrons.
  • This charge is vital in electromagnetic equations as it is one of the constants used.
  • The sign of the charge defines whether the force between two charges will be attractive or repulsive.
Understanding the role and magnitude of the elementary charge helps reinforce how minute particles interact. These interactions, although occurring at microscopic levels, govern much of the macroscopic phenomena we observe, such as electrical conductivity and magnetism. In this exercise, calculating electric forces involves the use of the elementary charge, demonstrating its pivotal role in Coulomb's law and electric force dynamics.

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Most popular questions from this chapter

(II) A dipole consists of charges \(+e\) and \(-e\) separated by \(0.68 \mathrm{nm} .\) It is in an electric field \(E=2.2 \times 10^{4} \mathrm{~N} / \mathrm{C} .\) (a) What is the value of the dipole moment? \((b)\) What is the torque on the dipole when it is perpendicular to the field? (c) What is the torque on the dipole when it is at an angle of \(45^{\circ}\) to the field? \((d)\) What is the work required to rotate the dipole from being oriented parallel to the field to being antiparallel to the field?

(I) The electric force on a \(+ 4.20 - \mu \mathrm { C }\) charge is \(\left( 7.22 \times 10 ^ { - 4 } \mathrm { N } \right) \hat { \mathrm { j } }\) . What is the electric field at the position of the charge?

Dry air will break down and generate a spark if the electric field exceeds about \(3 \times 10^{6} \mathrm{~N} / \mathrm{C} .\) How much charge could be packed onto a green pea (diameter \(0.75 \mathrm{~cm}\) ) before the pea spontaneously discharges? [Hint: Eqs. \(21-4\) work outside a sphere if \(r\) is measured from its center.]

Two small, identical conducting spheres \(\mathrm{A}\) and \(\mathrm{B}\) are a distance \(R\) apart; each carries the same charge \(Q .(a)\) What is the force sphere \(B\) exerts on sphere \(A ?(b)\) An identical sphere with zero charge, sphere C, makes contact with sphere \(\mathrm{B}\) and is then moved very far away. What is the net force now acting on sphere A? (c) Sphere \(\mathrm{C}\) is brought back and now makes contact with sphere \(A\) and is then moved far away. What is the force on sphere \(A\) in this third case?

Packing material made of pieces of foamed polystyrene can easily become charged and stick to each other. Given that the density of this material is about \(35 \mathrm{~kg} / \mathrm{m}^{3},\) estimate how much charge might be on a 2.0 -cm-diameter foamed polystyrene sphere, assuming the electric force between two spheres stuck together is equal to the weight of one sphere.

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