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Chapter 1: Problem 31

Decreasing energy? A hollow spherical shell with radius \(R\) has charge \(Q\) uniformly distributed over it. The task of Problem \(1.32\) is to show that the energy stored in this system is \(Q^{2} / 8 \pi \epsilon_{0} R\). (You can derive this here if you want, or you can just accept it for the purposes of this problem.) Now imagine taking all of the charge and concentrating it in two point charges \(Q / 2\) located at diametrically opposite positions on the shell. The energy of this new system is \((Q / 2)^{2} / 4 \pi \epsilon_{0}(2 R)=\) \(Q^{2} / 32 \pi \epsilon_{0} R\), which is less than the energy of the uniform spherical shell. Does this make sense? If not, where is the error in this reasoning?

Short Answer

Expert verified
The given reasoning is flawed because it neglected the work required to concentrate the charges from a uniform sphere distribution to a concentrated point distribution. This work would add to the potential energy, so the correct energy of the point charge configuration would be greater than the one calculated.

Step by step solution

01

Understanding Energy of Charge Distributions

The energy stored in a system with charges is due to the potential energy resulting from the charges interacting with each other. For a uniform spherical distribution of charge, the charges are distributed evenly, so the potential energy would be spread out over a larger area. Conversely, for point charges, all of the potential energy is concentrated in very small areas.
02

Comparing the Energies

The given expression for the energy stored in the spherical shell distribution is \(Q^{2} / 8 \pi \epsilon_{0} R\). For the point charge distribution, the energy is given as \(Q^{2} / 32 \pi \epsilon_{0} R\), which is 4 times less than the energy of the spherical shell distribution.
03

Is it Logical?

Upon scrutiny, it appears counter-intuitive that concentrating the same amount of charge at certain points would result in less energy than distributing the charges evenly. However, it's important to remember that in order to concentrate the charges, work needs to be done against electric forces. This work gets stored as potential energy in the point charge configuration. However, this detail was not taken into account when calculating the energy of the point charges, hence the reasoning was flawed.

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Key Concepts

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

Electric Potential Energy
Electric potential energy is a fundamental concept when analyzing charge distributions and their interactions. To understand this, think of electric potential energy as the energy that a charge, or a distribution of charges, possesses due to its position in an electric field. It's similar to how a ball at the top of a hill has gravitational potential energy.

In the context of our problem, the energy stored in the original spherical shell is a result of the electric potential energy spread across the entire distribution. Adding or removing charge from the shell requires work, and this work changes the potential energy of the system. Imagine bringing charges together from the shell to form concentrated point charges; one must work against the repulsive forces between similar charges, and this work is 'stored' as the increased potential energy of the newly formed point charge system.

However, when calculating the energy of the point charges, it's critical to account for all aspects of the work done in the process. Omitting this part can lead to a misunderstanding of the system's actual potential energy.
Uniform Charge Distribution
A uniform charge distribution means that the charge is spread out evenly across the entire surface or volume of the object. Our hollow spherical shell in this exercise is a perfect example: the charge is distributed without any concentration at one point. The significance of a uniform charge distribution lies in the way the charges interact with each other.

The forces are evenly distributed and the potential at any point on a conductor's surface is the same. This creates a consistent electric field emanating outward or inward, depending on the nature of the charge. In a uniform distribution, the electric potential energy of the system is calculated by considering the constant separation between the tiny increments of charge distributed across the surface.
Point Charge
A point charge is a hypothetical charge located at a single point in space. This model is a useful simplification used in physics to study the behavior of electric charges. When charges are concentrated into a point charge, as with the experiment discussed in the problem, the charge's influence becomes more defined.

Concentrating the charge Q over the spherical shell into two point charges located at diametrically opposite ends creates a new electric potential and field configuration. This has a crucial impact on the system's electric potential energy. However, as indicated in the step-by-step solution, simply comparing the energies of a uniform charge distribution to that of point charges without accounting for the work done in moving the charges leads to a flawed interpretation.
Electric Forces
Electric forces are pivotal to understanding charge interactions. They are the forces that charged objects exert on each other and are governed by Coulomb's Law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In context, when dealing with uniform charge distributions, these electric forces ensure that potential energy is evenly spread across the distribution.

But when transitioning to a system of point charges, the forces between the specific charges become more significant. To move charges in the manner described in the exercise鈥攃oncentrating a uniform charge to point charges鈥攐ne must analyze the electric forces and the work done against them closely. This is crucial for accurately determining potential energy changes and understanding electric interactions in systems.

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Most popular questions from this chapter

Deriving the energy density ** Consider the electric field of two protons a distance \(b\) apart. According to Eq. (1.53) (which we stated but did not prove), the potential energy of the system ought to be given by $$ \begin{aligned} U &=\frac{\epsilon_{0}}{2} \int \mathbf{E}^{2} d v=\frac{\epsilon_{0}}{2} \int\left(\mathbf{E}_{1}+\mathbf{E}_{2}\right)^{2} d v \\ &=\frac{\epsilon_{0}}{2} \int \mathbf{E}_{1}^{2} d v+\frac{\epsilon_{0}}{2} \int \mathbf{E}_{2}^{2} d v+\epsilon_{0} \int \mathbf{E}_{1} \cdot \mathbf{E}_{2} d v \end{aligned} $$ where \(\mathbf{E}_{1}\) is the field of one particle alone and \(\mathbf{E}_{2}\) that of the other. The first of the three integrals on the right might be called the "electrical self-energy" of one proton; an intrinsic property of the particle, it depends on the proton's size and structure. We have always disregarded it in reckoning the potential energy of a system of charges, on the assumption that it remains constant; the same goes for the second integral. The third integral involves the distance between the charges. Evaluate this integral. This is most easily done if you set it up in spherical polar coordinates with one of the protons at the origin and the other on the polar axis, and perform the integration over \(r\) before the integration over \(\theta\). Thus, by direct calculation, you can show that the third integral has the value \(e^{2} / 4 \pi \epsilon_{0} b\), which we already know to be the work required to bring the two protons in from an infinite distance to positions a distance \(b\) apart. So you will have proved the correctness of Eq. (1.53) for this case, and by invoking superposition you can argue that Eq. (1.53) must then give the energy required to assemble any system of charges.

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?

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.)

(a) A point charge \(q\) is located at an arbitrary position inside a sphere (just an imaginary sphere in space) with radius \(R\). Show that the average electric field over the surface of the sphere is zero. Hint: Use an argument involving Newton's third law, along with what you know about spherical shells. (b) If the point charge \(q\) is instead located outside the sphere, a distance \(r\) from the center, show that the average electric field over the surface of the sphere has magnitude \(q / 4 \pi \epsilon_{0} r\). (c) Return to the case where the point charge \(q\) is located inside the sphere of radius \(R\). Let the distance from the center be \(r\). Use the above results to show that the average electric field over the entire volume of the sphere of radius \(R\) has magnitude qr \(/ 4 \pi \epsilon_{0} R^{3}\) and points toward the center (if \(q\) is positive).

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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