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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.

Short Answer

Expert verified
Yes, the work can be positive for the vertical arrangement. For the work done to be positive, the distance \(a\) between like-charges must be less than the distance \(b\) between unlike-charges. This is because the work done is equal to the change in potential energy, and the potential energy is positive when like charges are brought close to each other.

Step by step solution

01

Calculate the work required for the Horizontal Arrangement

Consider the first arrangement where protons and electrons are alternately arranged on longer side (which is \(a\)). The work done in assembling this configuration is the amount of potential energy in this system. The potential energy \(PE_1\) between two charges is given by the formula \(k*Q*Q'/r\), where \(Q\) and \(Q'\) are the charges, \(r\) is the distance between them, and \(k\) is Coulomb's constant. Here, each pair of charges (proton-electron) contributes a potential energy of \(-k*|e|^2/a\) since the charges are opposite and the total potential energy is twice this amount or \(-2k*|e|^2/a\).
02

Calculate the work required for the Vertical Arrangement

Now consider the second arrangement where two protons are at one end and two electrons at the other end of the longer side (\(b\)) of the rectangle. The potential energy \(PE_2\) for this system includes contributions from pairs with unlike and like charges. The potential energy contribution from the unlike pairs is \(-k*|e|^2/b\) while that from the like pairs is \(k*|e|^2/a\). So, the total potential energy of the system is \(-2k*|e|^2/b + 2k*|e|^2/a\).
03

Compare the Work Done in Both Arrangements

Now, if the work done in the second arrangement is to be more than the first one, \(PE_2\) must be more than \(PE_1\). Equating the two and simplifying we get \(a>b\). Hence for the work or potential energy to be higher in the second arrangement, the distance between like charges must be shorter than that between unlike charges.

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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 fundamental in understanding electrostatic interactions. It describes the force between two charges in space. According to the law, the electrostatic force \(F\) between two point charges is proportional to the product of their magnitudes \(Q\) and inversely proportional to the square of the distance \(r\) between them. Mathematically, this is represented as:\[F = k \frac{{|Q \, Q'|}}{{r^2}}\]where \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, N \, m^2/C^2\). The force is attractive if the charges are of opposite types (positive and negative) and repulsive if they are of the same type.
Understanding this concept is crucial for calculating potential energy in systems with multiple charges, such as protons and electrons in a rectangle, as outlined in the original exercise.
Proton-Electron Interactions
In a system involving protons and electrons, interactions are central to determining the total energy. A proton, being positively charged, and an electron, being negatively charged, will attract each other. This attraction is described by Coulomb's law and results in a potential energy that is negative, indicating a stable, bound state.
In the exercise, when arranging a proton and electron alternately along the rectangle's side, the system's energy is largely influenced by these negative potential energies. Each proton-electron pair contributes a negative amount to the total potential energy of the system, causing a net stabilization.
Charge Configuration
Charge configuration in an electric field refers to the particular arrangement of charges. The exercise provides two configurations: horizontal and vertical, determined by how protons and electrons are arranged at opposite corners of a rectangle.
  • In the horizontal configuration, each proton-electron pair along the longer side \(a\) contributes to a lower potential energy due to increased attraction, minimizing repulsion by ensuring that like charges (protons and protons or electrons and electrons) are further apart.
  • In the vertical setup, like charges are closer on the longer side \(b\), potentially increasing energy due to repulsive forces.
Choosing the ideal configuration depends on balancing these attractions and repulsions to achieve lower or higher potential energy, which relates to the work required to assemble the system.
Work Energy Theorem
The work-energy theorem connects the work done by forces on a system to its energy change. In an electrostatic context, the work required to assemble charges in a configuration is equal to the change in their electric potential energy.
In the provided exercise, calculating the work involves determining the total potential energy for different charge configurations and comparing these values:
  • The horizontal arrangement yields a certain negative potential energy, showing lesser work required due to more favorable proton-electron interactions.
  • The vertical arrangement increases potential energy, thus more work or energy input may be required to bring charges into the configuration, especially if \(a > b\).
For some configurations, depending on the distances \(a\) and \(b\), the work done can be positive, implying energy must be added to assemble the system. Using these concepts helps analyze and predict outcomes in various charge distributions.

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Most popular questions from this chapter

Field from two sheets : Two infinite plane sheets of surface charge, with densities \(3 \sigma_{0}\) and \(-2 \sigma_{0}\), are located a distance \(\ell\) apart, parallel to one another. Discuss the electric field of this system. Now suppose the two planes, instead of being parallel, intersect at right angles. Show what the field is like in each of the four regions into which space is thereby divided.

Force between two strips ** (a) The two strips of charge shown in Fig. \(1.47\) have width \(b\), infinite height, and negligible thickness (in the direction perpendicular to the page). Their charge densities per unit area are \(\pm \sigma .\) Find the magnitude of the electric field due to one of the strips, a distance \(x\) away from it (in the plane of the page). (b) Show that the force (per unit height) between the two strips equals \(\sigma^{2} b(\ln 2) / \pi \epsilon_{0}\). Note that this result is finite, even though you will find that the field due to a strip diverges as you get close to it

Find a geometrical arrangement of one proton and two electrons such that the potential energy of the system is exactly zero. How many such arrangements are there with the three particles on the same straight line? You should find that the ratio of two of the distances involved is the golden ratio.

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?

Escaping field lines * Charges \(2 q\) and \(-q\) are located on the \(x\) axis at \(x=0\) and \(x=a\), respectively. (a) Find the point on the \(x\) axis where the electric field is zero, and make a rough sketch of some field lines. (b) You should find that some of the field lines that start on the \(2 q\) charge end up on the \(-q\) charge, while others head off to infinity. Consider the field lines that form the cutoff between, these two cases. At what angle (with respect to the \(x\) axis) do these lines leave the \(2 q\) charge? Hint: Draw a wisely chosen Gaussian surface that mainly follows these lines.

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