Problem 34 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

(a) 34.5 × 10⁻²⁹ Joules, (b) -7.67 × 10⁻²⁹ Joules.

Step 1: Understanding the Force Equation

The force between two electrons separated by a distance \( r \) is given by \( F = \frac{23 \times 10^{-29}}{r^2} \) newtons. This is an inverse square law similar to Coulomb's law.

Step 2: Calculate Work for Part A

To calculate the work done in moving the electron from \((-1,0)\) to \((0,0)\), denote the variable electron at \( x \) and the fixed electron at the point \((1,0)\). The distance \( r \) between the electrons is \( |x - 1| \). The work done, \( W_1 \), is the integral of force along the path from \( -1 \) to \( 0 \):\[ W_1 = \int_{-1}^{0} \frac{23 \times 10^{-29}}{(x-1)^2} \, dx. \]

Step 3: Integrate to Find Work for Part A

Calculate \( W_1 \):\[ W_1 = \int_{-1}^{0} \frac{23 \times 10^{-29}}{(x-1)^2} \, dx = 23 \times 10^{-29} \left[-\frac{1}{x-1} \right]_{-1}^{0}. \]Evaluate the integral:\[ W_1 = 23 \times 10^{-29} \left( -\frac{1}{0-1} + \frac{1}{(-1)-1} \right) = 23 \times 10^{-29} \left(1 - \frac{1}{-2} \right) = 23 \times 10^{-29} \cdot \left(1 - (-0.5) \right) = 23 \times 10^{-29} \cdot 1.5. \]This results in:\[ W_1 = 34.5 \times 10^{-29} \, \text{joules}. \]

Step 4: Setup the Problem for Part B

For this part, the electron is moved from \((5,0)\) to \((3,0)\) with fixed electrons at \((-1,0)\) and \((1,0)\). Calculate total work done, \( W_2 = W_{-1} + W_1 \), where \( W_{-1} \) is due to the electron at \((-1,0)\) and \( W_1 \) is due to \((1,0)\).

Step 5: Calculate Work Done by Electron at (-1,0)

For the fixed electron at \((-1,0)\):\[ W_{-1} = \int_{5}^{3} \frac{23 \times 10^{-29}}{(x+1)^2} \, dx. \]Calculate \( W_{-1} \):\[ W_{-1} = 23 \times 10^{-29} \left[-\frac{1}{x+1} \right]_{5}^{3} = 23 \times 10^{-29} \left( -\frac{1}{3+1} - (-\frac{1}{5+1}) \right). \]This simplifies to:\[ W_{-1} = 23 \times 10^{-29} \left(-\frac{1}{4} + \frac{1}{6}\right) = 23 \times 10^{-29} \cdot \left( -0.25 + 0.1667 \right) \approx 23 \times 10^{-29} \cdot (-0.0833). \]This gives:\[ W_{-1} = -1.9167 \times 10^{-29} \, \text{joules}. \]

Step 6: Calculate Work Done by Electron at (1,0)

For the fixed electron at \((1,0)\):\[ W_1 = \int_{5}^{3} \frac{23 \times 10^{-29}}{(x-1)^2} \, dx. \]Calculate \( W_1 \):\[ W_1 = 23 \times 10^{-29} \left[-\frac{1}{x-1} \right]_{5}^{3} = 23 \times 10^{-29} \left( -\frac{1}{3-1} + \frac{1}{5-1} \right). \]This simplifies:\[ W_1 = 23 \times 10^{-29} \left(-\frac{1}{2} + \frac{1}{4}\right) = 23 \times 10^{-29} \cdot \left(-0.5 + 0.25\right). \]This gives:\[ W_1 = -5.75 \times 10^{-29} \, \text{joules}. \]

Step 7: Calculate Total Work for Part B

Total work done, \( W_2 = W_{-1} + W_1 \):\[ W_2 = -1.9167 \times 10^{-29} + (-5.75 \times 10^{-29}) = -7.6667 \times 10^{-29} \, \text{joules}. \]

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in physics that describes the electric force between two charged particles. It states that the force (\( F \)) between two charges is directly proportional to the product of the magnitudes of the charges (\( q_1 \) and \( q_2 \)) and inversely proportional to the square of the distance (\( r \)) between them. The formula for Coulomb's Law is:\[F = k \frac{|q_1 q_2|}{r^2}\]where \( k \) is the electrostatic constant.

• Force is measured in newtons (N).
• The charges are in coulombs (C).
• Distance \( r \) is in meters (m).

In the given problem, although the equation provided is not exactly the classic form of Coulomb's Law, it follows a similar inverse square law. The force equation provided:\[F = \frac{23 \times 10^{-29}}{r^2}\]echoes the way forces behave inversely with the distance squared, typical in mutual interactions like gravitational and electric forces. Understanding this relationship is critical for solving problems related to electric fields and forces. It emphasizes how quickly the force diminishes as objects move apart.

Integration in Physics

Integration is a powerful mathematical tool used in physics to calculate quantities that accumulate over a path or area, like work or energy. When applying integration in physics, think of it like summing up an infinite number of tiny increments.
For instance, when calculating work done by a variable force, integration is used because the force may change along the path of application. The formula for work (\( W \)) done by a force (\( F(x) \)) over a path from \( a \) to \( b \) is:\[W = \int_{a}^{b} F(x) \, dx\]In the provided exercise, the force is represented as a function of (\( x \)), specifically \( F = \frac{23 \times 10^{-29}}{(x-1)^2} \), indicating it changes depending on position (distance from another electron).

• Integrating the force over a distance finds the total work done.
• The process involves finding the antiderivative and substituting limits.
• This results in the work required to move a charge within an electric field.

Mastering integration in physics allows you to tackle complex scenarios where forces aren't constant, providing deeper insights into how energy is transferred or transformed.

Electric Force Calculations

Electric force calculations are essential when dealing with problems involving charges and how they interact. The challenge is often due to variables like distance changing as charges or objects move. The original problem considers two scenarios which involve calculating the work to move electrons, a classic application of electric forces.
For Part A of the exercise, an electron is moved from (-1,0) to the origin, under the influence of a stationary electron. By integrating the force from \(-1 \) to \( 0 \) and evaluating it, we find the work required.
Similarly, Part B extends this by adding complexity: calculating the total work when moving a third electron influenced by two fixed charges. Here, the net work combines contributions from both electrons, using integral limits from \( 5 \) to \( 3 \).

• Start with identifying the formula for force between the moving and fixed electrons.
• Set limits of integration based on movement path.
• Sum contributions from multiple sources if needed.

These steps not only teach how to perform electric force calculations but also to understand the physical implications—knowing that these forces govern interactions between charged particles, shaping everything from atomic structures to large-scale fields.