/*! 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 77 Imagine a sphere of radius \(a\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Imagine a sphere of radius \(a\) filled with negative charge of uniform density, the total charge being equivalent to that of two electrons. Imbed in this jelly of negative charge two protons, and assume that, in spite of their presence, the negative charge distribution remains uniform. Where must the protons be located so that the force on each of them is zero? (This is a surprisingly realistic caricature of a hydrogen molecule; the magic that keeps the electron cloud in the molecule from collapsing around the protons is explained by quantum mechanics!)

Short Answer

Expert verified
To solve the problem, we balance the electrostatic repulsion between the protons with the attractive force exerted on them by the negatively charged sphere. On doing so, we find that the protons should be located at a distance \(x\) from the center of the sphere. This \(x\) can be found as a solution of resulting cubic equation, which is derived from equating the magnitudes of two aforementioned forces.

Step by step solution

01

Determine Forces Acting on Protons

First, it should be noted that the protons experience two kinds of forces. One is from the other proton due to their like charged nature, which can be calculated using Coulomb's Law. Let's denote this as \(F_{pp}\). It pushes the protons away from each other. The other force is from the negative charge of the sphere. Since the charge distribution is uniform, the force exerted by the sphere on a proton is radially inward. This can be calculated using Gauss’s Law in electrostatics. Let's denote this force as \(F_{sp}\).
02

Calculating the Force between the Protons

The force between the two protons depends on their distance apart and the charges they carry. Using Coulomb's Law, the force is given by \(F_{pp} = k \frac{e^2}{r^2}\), where \(e\) is the charge of a proton, \(r\) is the distance between the protons and \(k\) is Coulomb's constant.
03

Calculating the Force due to Negative Sphere

Considering a proton at distance \(x\) from the center of the sphere, the sphere exerts an inward force on the proton. This is because parts of the sphere closer to the proton exert more force than parts farther away, due to Coulomb's Law. By symmetry, all these force components add up to give a resultant force pointing toward the center of the sphere. The magnitude of this force is given by \(F_{sp} = k \frac{(2e)(xe/r^3)}{x^2}\), where \(2e\) is the total charge of the sphere (equivalent to that of two electrons). Note that this force only takes into account the negative charge within a smaller sphere inscribed inside the larger sphere, the radius of smaller sphere being the distance \(x\) of the proton from the sphere's center.
04

Setting Forces Equal for Equilibrium

For equilibrium, the force exerted by the sphere of charge on a proton must balance the force between the two protons. Therefore, we set the magnitudes of the two forces equal: \(F_{pp} = F_{sp}\). Simplifying the resulting equation will yield a cubic equation in terms of \(x\). Solving that equation will give the distance \(x\), which indicates the position from the center of the sphere, at which the protons need to be located so that there's no resultant force acting on them.

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Key Concepts

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

Coulomb's Law
To elucidate the interaction between charges, we refer to Coulomb's Law, a fundamental principle in electrostatics. This quantifies the force between two point charges as directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The formula is expressed as: \[F = k \frac{{q_1 q_2}}{{r^2}}\]
where:
  • \(F\) is the electrostatic force between charges,
  • \(k\) is Coulomb's constant (\(8.9875 \times 10^9 N m^2/C^2\)),
  • \(q_1\) and \(q_2\) are the magnitudes of the charges,
  • and \(r\) is the distance between the centers of the two charges.

In the exercise at hand, the electrostatic force between the two protons is evaluated using this law. What might initially seem as a simple task, to find a condition for equilibrium, becomes more complex due to the presence of another charge – the sphere's uniform charge distribution. Coulomb's Law is integral not only in calculating the force between individual protons but also in understanding the electric field created by the sphere, which affects the forces at play.
Gauss's Law
Now, let's pivot our focus to another pivotal concept in electrostatics – Gauss's Law. It relates the electric fields emanating from a closed surface to the charge enclosed by that surface. The law is encapsulated by the equation:

\[\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q}{{\epsilon_0}}\]
where:
  • \(\Phi_E\) is the electric flux through a closed surface \(S\),
  • \(\vec{E}\) is the electric field,
  • \(d\vec{A}\) is a vector representing an infinitesimal area on surface \(S\), pointing outward,
  • and \(Q\) is the charge inside \(S\).

In simpler terms, Gauss's Law can be used to easily calculate the force on a charge within a symmetric charge distribution. In our exercise with the negatively charged sphere, Gauss's Law helps to deduce the force exerted on a proton within it. Due to the sphere's symmetry, the law simplifies the complex charge interactions into a more tractable form, offering us a glimpse at the otherwise challenging task of force calculation in a uniform electric field.
Force Equilibrium
Reaching a state of force equilibrium is a condition where all net forces acting on a particle are balanced, resulting in no acceleration for the particle. For instance, in our exercise, when we discuss the protons' required position for zero net force, we're looking for a point of force equilibrium.

The criterion for equilibrium can be written as:
\[\sum \vec{F} = \vec{0}\]
for all forces acting on the protons. This implies that \(F_{pp}\), the force between protons, should be equal in magnitude and opposite in direction to \(F_{sp}\), the force exerted by the sphere. By setting these forces equal, we essentially seek the point where the repulsive electrostatic force between the protons exactly cancels out with the attractive force from the negatively charged sphere. The equilibrium position is then solved by finding appropriate values of \(x\) that satisfy the equation. Achieving force equilibrium allows us to determine a stable configuration, which albeit simplified, provides a basic model of the structure seen in a hydrogen molecule.
Electric Charge Distribution
Delving into electric charge distribution, it's pivotal to recognize how charges arrange themselves and the impact this has on electric fields and forces. In electrostatics, a uniform charge distribution, as highlighted in our exercise, means that charge is spread evenly over a volume or a surface, and this simplifies the analysis considerably.

A key point to remember is that even within a uniform charge distribution, the forces on a charge are not merely the sum of the forces due to individual charges. Instead, thanks to the principle of superposition, the resultant force is the vector sum of the forces due to each and every particle in the distribution. Our exercise tackles a uniform negative charge distribution within a sphere and implies that wherever you place a proton inside this sphere, it will feel a force towards the center, this phenomenon being a direct consequence of the symmetrical charge distribution and is essential for understanding the force interactions between the protons and the sphere.

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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.

Field between two wires * Consider a high-voltage direct current power line that consists of two parallel conductors suspended 3 meters apart. The lines are oppositely charged. If the electric field strength halfway between them is \(15,000 \mathrm{~N} / \mathrm{C}\), how much excess positive charge resides on a \(1 \mathrm{~km}\) length of the positive conductor?

Potential energy of a cylinder * * Problem \(1.24\) gives one way of calculating the energy per unit length stored in a solid cylinder with radius \(a\) and uniform volume charge density \(\rho .\) Calculate the energy here by using Eq. (1.53) to find the total energy per unit length stored in the electric field. Don't forget to include the field inside the cylinder. You will find that the energy is infinite, so instead calculate the energy relative to the configuration where all the charge is initially distributed uniformly over a hollow cylinder with large radius \(R\). (The field outside radius \(R\) is the same in both configurations, so it can be ignored when calculating the relative energy.) In terms of the total charge \(\lambda\) per unit length in the final cylinder, show that the energy per unit length can be written as \(\left(\lambda^{2} / 4 \pi \epsilon_{0}\right)(1 / 4+\ln (R / a))\).

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.

Work for a rectangle ** Two protons and two electrons are located at the corners of a rectangle with side lengths \(a\) and \(b\). There are two essentially different arrangements. Consider the work required to assemble the system, starting with the particles very far apart. Is it possible for the work to be positive for either of the arrangements? If so, how must \(a\) and \(b\) be related? You will need to solve something numerically.

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