Problem 38 - Thomas Calculus

Question

Forcing electrons together Two electrons \(r\) meters apart repel each other with a force of $$ F=\frac{23 \times 10^{-29}}{r^{2}} $$ 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. 23 x 10^{-29} J, b. 1.3417 x 10^{-28} J.

Step 1: Understanding Work Done by a Force

Work done by a force when moving an object from point A to point B is given by the formula: \[ W = \int_{A}^{B} F \, dr \] where \( F \) is the force applied and \( dr \) is the differential displacement in the path of the object. Since force and displacement are along the same axis, this becomes an integration problem.

Step 2: Calculating Work for Part (a)

The work done to move the electron from \((-1, 0)\) to the origin is given by:\[ W = \int_{-1}^{0} \frac{23 \times 10^{-29}}{r^2} \, dr \]Integrating this expression, we have: \[ W = 23 \times 10^{-29} \left[ -\frac{1}{r} \right]_{-1}^{0} \]Evaluate the bounds:\[ W = 23 \times 10^{-29} \left( 0 - (1) \right) = 23 \times 10^{-29} \]

Step 3: Calculating Work for Part (b)

For part (b), the force acting on the electron at position \( x \) due to two electrons fixed at \((-1, 0)\) and \((1, 0)\) is given by: \[ F(x) = \frac{23 \times 10^{-29}}{(x+1)^2} + \frac{23 \times 10^{-29}}{(x-1)^2} \]The work done is then:\[ W = \int_{5}^{3} \left( \frac{23 \times 10^{-29}}{(x+1)^2} + \frac{23 \times 10^{-29}}{(x-1)^2} \right) \, dx \]Integrate this term by term:\[ W = 23 \times 10^{-29} \left[ -\frac{1}{x+1} \right]_{5}^{3} + 23 \times 10^{-29} \left[ -\frac{1}{x-1} \right]_{5}^{3} \]Evaluating the integrals,\[ W = 23 \times 10^{-29} \left( \frac{1}{4} - \frac{1}{6} \right) + 23 \times 10^{-29} \left( \frac{1}{2} - \frac{1}{4} \right) \]Simplify the fractions:\[ W = 23 \times 10^{-29} \left( \frac{1}{12} + \frac{1}{4} \right) = 23 \times 10^{-29} \times \frac{4+3}{12} = 23 \times 10^{-29} \times \frac{7}{12} \]\[ W = \frac{161}{12} \times 10^{-29} \approx 1.3417 \times 10^{-28} \]

Step 4: Final Calculation and Conclusion

Sum up the work from each step to find the total work required in each scenario. (a) Total work to move the electron from \((-1,0)\) to the origin: \[ W_a = 23 \times 10^{-29} \] (b) Total work to move the third electron from \((5,0)\) to \((3,0)\): \[ W_b \approx 1.3417 \times 10^{-28} \]

Key Concepts

Electron Dynamics

Electron dynamics, in physics, is the study of how electrons move under the influence of forces. In our exercise, the electrons repel each other due to an electromagnetic force. This force is essential in understanding why electrons behave the way they do in different conditions. When two electrons are brought close, they experience a repulsive force. This force is inversely proportional to the square of the distance separating them.

• Due to their charge, electrons generate and respond to electromagnetic forces.
• The formula for the force affecting two electrons at a distance is given by \( F = \frac{23 \times 10^{-29}}{r^2} \).
• Electrons, as charged particles, are constantly in motion, and their dynamics is essential in fields like electronics and quantum physics.

By studying how these forces interact, physicists can predict the behavior of electrons in electric circuits, magnetic fields, and other fields.

Work and Energy in Physics

Work and energy are foundational concepts in physics that describe the energy transfer that occurs when an object is moved by a force. In this exercise, when you push an electron from one point to another against the electromagnetic force, you do work on it.

• The definition of work in physics is the product of the force applied to an object and the displacement of that object in the direction of the force. Formally it's represented as \( W = \int_{A}^{B} F \, dr \).
• This integral accounts for the sum of all work done as you move the electron from point A to B, showing how force and displacement along the path result in energy transfer.

This relation helps us understand various phenomena such as the amount of energy required to move electrical charges in a circuit or mechanical systems.

Integral Calculus Applications

Integral calculus is a mathematical tool that helps calculate the total accumulation of quantities, such as area under a curve or total work done. In the context of our exercise, it helps in determining the work needed to move an electron across a force field generated by other electrons.

• By setting up the integral from a starting point to an end point of the movement, one can find the total work done against the force.
• Here, \( W = \int_{-1}^{0} \frac{23 \times 10^{-29}}{r^2} \, dr \) demonstrates how integration is used to find the work in moving an electron from one location to another.
• Calculus allows physicists and engineers to model real-world scenarios where forces and movements vary continuously across spaces.

Integral calculus is indispensable for predicting forces and energy requirements in dynamic systems.