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Chapter 1: Problem 1

Gravity vs. electricity (a) In the domain of elementary particles, a natural unit of mass is the mass of a nucleon, that is, a proton or a neutron, the basic massive building blocks of ordinary matter. Given the nucleon mass as \(1.67 \cdot 10^{-27} \mathrm{~kg}\) and the gravitational constant G as \(6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \mathrm{s}^{2}\right)\), compare the gravitational attraction of two protons with their electrostatic repulsion. This shows why we call gravitation a very weak force. (b) The distance between the two protons in the helium nucleus could be at one instant as much as \(10^{-15} \mathrm{~m}\). How large is the force of electrical repulsion between two protons at that distance? Express it in newtons, and in pounds. Even stronger is the nuclear force that acts between any pair of hadrons (including neutrons and protons) when they are that close together.

Short Answer

Expert verified
The calculation shows that the gravitational force between two protons is extremely weak, while the electrostatic repulsion is much stronger. At a distance of \(10^{-15}\) m, the electric force between two protons is even larger. Then, to convert these force values to pounds from newtons, a simple conversion factor is used.

Step by step solution

01

Calculate the Gravitational Force between two Protons

The formula F = G * (m1 * m2) / r^2, with m1=m2 being the mass of a proton, r being the distance between them (proximity at atomic level), and G the gravitational constant, can be used. However, since the mass of a proton is very small and gravity is a weak force, the result will be negligible.
02

Calculate the Electrostatic Repulsion between two Protons

By Coulomb's law, F = k * (q1 * q2) / r^2, with q1=q2 being the charge of a proton, r being the distance between them and k being Coulomb's constant. This calculation will provide a significantly larger result explaining why gravity is considered a weak force at this level.
03

Calculate the Electric Force at a Distance of \(10^{-15} m\)

Use the same equation for the electrostatic force as in Step 2, but substitute \(10^{-15} m\) for r to compute the electric force at this distance. The result can be expressed in terms of newtons.
04

Convert Newtons to Pounds

To express the force in pounds instead of newtons, use the conversion 1 N = 0.2248 lb.

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Key Concepts

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

Gravitational Force
The gravitational force is a natural phenomenon by which all things with mass or energy are brought toward one another. Newton's universal law of gravitation explains that every point mass attracts every other point mass by a force acting along the line intersecting both points. This force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers.

Using the formula \( F = G \times \frac{m1 \times m2}{r^2} \), where \( G \) is the gravitational constant, \( m1 \) and \( m2 \) are the masses of the objects, and \( r \) is the distance between the centers of their masses, we can calculate the gravitational attraction between any two objects. In the context of elementary particles, such as protons, the gravitational force is exceedingly small because the masses involved are incredibly tiny.
Electrostatic Repulsion
In contrast to the gravitational force, electrostatic forces are the interactions that occur between electrically charged particles. When two particles have the same type of charge, either positive or positive or negative or negative, they repel each other. This is known as electrostatic repulsion.

The strength of this repulsive force can be substantial, especially at the small distances that are typical in atomic and subatomic scales. Because protons carry a positive charge, two protons will experience a strong electrostatic repulsion pushing them apart, which is what happens at the atomic level.
Coulomb's Law
To quantify the electrostatic repulsion between two charged particles, we use Coulomb's law. Coulomb's law states that the force \( F \) between two point charges \( q1 \) and \( q2 \) is directly proportional to the product of their charges and inversely proportional to the square of the distance \( r \) between them, as given by \( F = k \times \frac{q1 \times q2}{r^2} \), where \( k \) is Coulomb's constant.

This relationship means that the electrostatic force is much stronger than gravity at the scale of elementary particles because charges involved are significant at such small distances. However, unlike gravitational forces that are always attractive, electrostatic forces can be either attractive or repulsive.
Nuclear Force
While gravitational and electrostatic forces play a critical role in the interactions between particles, there's another force that is even stronger at small distances 鈥� the nuclear force. Not to be confused with nuclear power, which involves reactions that release energy, the nuclear force (also called the strong force) is responsible for holding the nuclei of atoms together.

The nuclear force acts between hadrons, which are particles like protons and neutrons, when they are extremely close to each other. Despite the powerful electrostatic repulsion that protons experience due to their like charges, the nuclear force is able to bind them together within the atomic nucleus. This force is many orders of magnitude stronger than both gravitational and electrostatic forces at the scale of nuclei but operates over a much shorter range.
Elementary Particles
Elementary particles are the smallest known building blocks of the universe. These include quarks, which make up protons and neutrons, as well as leptons, such as electrons, and bosons, like photons.

The properties of these particles, including mass and charge, determine how they interact with each other through the four fundamental forces: gravitational, electromagnetic, strong nuclear, and weak nuclear forces. The study of these particles and their interactions is a key part of quantum physics and helps us understand the composition and structure of matter at the most fundamental level.

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Most popular questions from this chapter

Hole in a shell \(*\) Figure \(1.52\) shows a spherical shell of charge, of radius \(a\) and surface density \(\sigma\), from which a small circular piece of radius \(b \ll a\) has been removed. What is the direction and magnitude of the field, at the midpoint of the aperture? There are two ways to get the answer. You can integrate over the remaining charge distribution, to sum the contributions of all elements to the field at the point in question. Or, remembering the superposition principle, you can think about the effect of replacing the piece removed, which itself is practically a little disk. Note the connection of this result with our discussion of the force on a surface charge - perhaps that is a third way in which you might arrive at the answer.

Hole in a plane : (a) A hole of radius \(R\) is cut out from a very large flat sheet with uniform charge density \(\sigma\). Let \(L\) be the line perpendicular to the sheet, passing through the center of the hole. What is the electric field at a point on \(L\), a distance \(z\) from the center of the hole? Hint: Consider the plane to consist of many concentric rings. (b) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), very close to the center of the hole, show that it undergoes oscillatory motion, and find the frequency \(\omega\) of these oscillations. What is \(\omega\) if \(m=1 \mathrm{~g},-q=-10^{-8} \mathrm{C}, \sigma=10^{-6} \mathrm{C} / \mathrm{m}^{2}\), and \(R=0.1 \mathrm{~m} ?\) (c) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), a distance \(z\) from the sheet, what is its speed when it passes through the center of the hole? What does your answer reduce to for large \(z\) (or, equivalently, small \(R\) )?

Flux through a cube (a) A point charge \(q\) is located at the center of a cube of edge \(d\). What is the value of \(\int \mathbf{E} \cdot d \mathbf{a}\) over one face of the cube? (b) The charge \(q\) is moved to one corner of the cube. Now what is the value of the flux of \(\mathbf{E}\) through each of the faces of the cube? (To make things well defined, treat the charge like a tiny sphere.)

Zero field in a sphere In Fig. \(1.51\) a sphere with radius \(R\) is centered at the origin, an, infinite cylinder with radius \(R\) has its axis along the \(z\) axis, and an infinite slab with thickness \(2 R\) lies between the planes \(z=-R\) and \(z=R\). The uniform volume densities of these objects are \(\rho_{1}, \rho_{2}\) and \(\rho_{3}\), respectively. The objects are superposed on top of each other; the densities add where the objects overlap. How should the three densities be related so that the electric field is zero everywhere throughout the volume of the sphere? Hint: Find a vector expression for the field inside each object, and then use superposition.

Fields at the surfaces Consider the electric field at a point on the surface of (a) a sphere with radius \(R\), (b) a cylinder with radius \(R\) whose length is infinite, and (c) a slab with thickness \(2 R\) whose other two dimensions are infinite. All of the objects have the same volume charge density \(\rho\). Compare the fields in the three cases, and explain physically why the sizes take the order they do.
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