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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 – 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

Zero field \(?\) Four charges, \(q,-q, q\), and \(-q\), are located at equally spaced intervals on the \(x\) axis. Their \(x\) values are \(-3 a,-a, a\), and \(3 a\), respectively. Does there exist a point on the \(y\) axis for which the electric field is zero? If so, find the \(y\) value.

Field from a spherical shell, right and wrong ** The electric field outside and an infinitesimal distance away from a uniformly charged spherical shell, with radius \(R\) and surface charge density \(\sigma\), is given by Eq. (1.42) as \(\sigma / \epsilon_{0}\). Derive this in the following way. (a) Slice the shell into rings (symmetrically located with respect to the point in question), and then integrate the field contributions from all the rings. You should obtain the incorrect result of \(\sigma / 2 \epsilon_{0}\) (b) Why isn't the result correct? Explain how to modify it to obtain the correct result of \(\sigma / \epsilon_{0} .\) Hint: You could very well have performed the above integral in an effort to obtain the electric field an infinitesimal distance inside the shell, where we know the field is zero. Does the above integration provide a good description of what's going on for points on the shell that are very close to the point in question?

Energy of concentric shells * (a) Concentric spherical shells of radius \(a\) and \(b\), with \(a

Zero force from a triangle ** Two positive ions and one negative ion are fixed at the vertices of an equilateral triangle. Where can a fourth ion be placed, along the symmetry axis of the setup, so that the force on it will be zero? Is there more than one such place? You will need to solve something numerically.

Thundercloud You observe that the passage of a particular thundercloud overhead causes the vertical electric field strength in the atmosphere, measured at the ground, to rise to \(3000 \mathrm{~N} / \mathrm{C}\) (or \(\mathrm{V} / \mathrm{m})\). (a) How much charge does the thundercloud contain, in coulombs per square meter of horizontal area? Assume that the width of the cloud is large compared with the height above the ground. (b) Suppose there is enough water in the thundercloud in the form of \(1 \mathrm{~mm}\) diameter drops to make \(0.25 \mathrm{~cm}\) of rainfall, and that it is those drops that carry the charge. How large is the electric field strength at the surface of one of the drops?

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