Problem 32 - Thomas Calculus in SI Units

Question

Two electrons \(r\) meters apart repel each other with a force of $$F=\frac{23 \times 10^{-29}}{r^{2}} \text { newtons. }$$ a. Suppose one electron is held fixed at the point (1,0) on the \(x\) -axis (units in meters). How much work does it take to move a second electron along the \(x\) -axis from the point (-1,0) to the origin? b. Suppose an electron is held fixed at each of the points (-1,0) and (1, 0). How much work does it take to move a third electron along the \(x\) -axis from (5,0) to (3,0)\(?\)

Step-by-Step Solution

Answer

a) \(1.15 \times 10^{-29}\) J, b) \(7.67 \times 10^{-30}\) J.

Step 1: Understand the Formula for Electric Repulsion Force

The expression for the force between two electrons is given as \( F = \frac{23 \times 10^{-29}}{r^{2}} \) newtons, where \( r \) is the distance in meters between the electrons. The force is inversely proportional to the square of the distance between them.

Step 2: Define Work Done for Moving an Electron

The work \( W \) done to move an electron through an electric field is given by the integral \( W = \int F \, dr \), where \( F \) is the force expressed as a function of \( r \), and \( dr \) is the displacement over which the force is applied.

Step 3: Calculate Work for Moving Between Points (-1,0) to (0,0)

For part (a), assume the electron is moving from \( x = -1 \) to \( x = 0 \). The work done \( W \) is calculated by integrating \( F = \frac{23 \times 10^{-29}}{r^{2}} \) from \( r = 2 \) to \( r = 1 \). Perform the integration:\[ W = \int_{2}^{1} \frac{23 \times 10^{-29}}{r^{2}} \, dr = -23 \times 10^{-29} \left[ -\frac{1}{r} \right]_{r=2}^{r=1} \].Substitute the limits to find:\[ W = -23 \times 10^{-29} \left(-1 + \frac{1}{2}\right) = \frac{23 \times 10^{-29}}{2} \approx 1.15 \times 10^{-29} \text{ J}\].

Step 4: Calculate Work for Moving Between Points (5,0) to (3,0) with Two Fixed Electrons

For part (b), two electrons are fixed at (-1,0) and (1,0), while the third electron moves from (5,0) to (3,0). The total work done is the sum of the work against each repulsion field. Each field exerts a force \( F = \frac{23 \times 10^{-29}}{r^{2}} \), where \( r \) changes based on position:1. Electron at (-1,0): Work is \( W_1 = \int_{6}^{4} \frac{23 \times 10^{-29}}{r^{2}} \ dr \).2. Electron at (1,0): Work is \( W_2 = \int_{4}^{2} \frac{23 \times 10^{-29}}{r^{2}} \ dr \).Calculate each:- \( W_1 = 23 \times 10^{-29} \left( \frac{1}{4} - \frac{1}{6} \right) \).- \( W_2 = 23 \times 10^{-29} \left( \frac{1}{2} - \frac{1}{4} \right) \).Solve these:- \( W_1 = 23 \times 10^{-29} \cdot \frac{1}{12} \approx 1.92 \times 10^{-30} \text{ J}\).- \( W_2 = 23 \times 10^{-29} \cdot \frac{1}{4} \approx 5.75 \times 10^{-30} \text{ J}\).Total work is \( W = W_1 + W_2 = 1.92 \times 10^{-30} + 5.75 \times 10^{-30} = 7.67 \times 10^{-30} \text{ J}\).

Step 5: Conclusion

The work to move the second electron from \((-1,0)\) to the origin is approximately \(1.15 \times 10^{-29}\) joules. The work to move the third electron from \((5,0)\) to \((3,0)\) when two electrons are fixed is about \(7.67 \times 10^{-30}\) joules.

Key Concepts

Work and Energy

Work and energy are fundamental concepts in physics, describing how forces cause movement and how energy is transferred. Work is essentially the measure of energy transferred when an object is moved by a force. It is calculated using the formula for work, \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of movement.
In the context of this exercise, we are moving electrons within an electric field created by other electrons. This requires understanding that the work done is not only due to the applied force but also influenced by how the force changes over the distance. Since the force varies with the distance \( r \), integration is used to find the work over that distance. In the situations given, the work involves moving charges (electrons) through a non-uniform electric field, hence, integrating over the path gives the total work done.

Integration in Calculus

Integration in calculus is a mathematical tool used to determine the total accumulation of quantities over continuous domains. In physics, integration allows us to calculate things like work done by a variable force across a distance.
The integral for work \( W = \int F \cdot dr \) sums the infinitesimal contributions of force over a range of \( r \). This is crucial when forces vary, such as when an electron is repelled by another electron at varying distances. The exercise's integration reflects how forces like electrostatic repulsion are characterized by nonconstant forces. Here, the steps involved taking the inverse square nature of the force \( F = \frac{23 \times 10^{-29}}{r^2} \) and integrating it over a specified range of distances, \( r \), which resulted in an accurate measure of work performed.

Electrostatics

Electrostatics is the study of electric charges at rest. It concerns forces, fields, and potentials created by stationary charges. Here, we're dealing with the electrostatic force between electrons.
The electrostatic force between two point charges, like electrons, is defined by Coulomb's law. It states that the force \( F \) is directly proportional to the product of the charges and inversely proportional to the square of the distance \( r \) separating them. This exercise uses principles of electrostatics to move electrons against repulsive forces. Importantly, as two electrons repel each other, the work needed to move one from point to point along a line can be calculated using the concept of potential energy changes across that field.

Physics Problem Solving

Physics problem solving involves the methodical application of principles to unravel the scenarios presented, often involving calculations and logical thinking. It's a skill developed through understanding the foundational theories and applying them to practical problems.
In this problem, it required identifying that movement of electrons involves calculating work against a variable electrostatic force. The step-by-step method started with understanding the formula for the electric force given and then correctly applying calculus (integration) to solve the work required for electron displacement. Key to physics problem solving here was to identify variable forces and precise limits to integrate between, ensuring calculations fitted the specific path and scenario of moving an electron along an axis in the presence of other fixed electrons. This showcases applying theoretical principles to compute exact physical values.