Problem 49 - Calculus: Early Transcendental Functions

Question

The Coulomb force for a unit charge at the origin and charge \(q\) at point \(P_{1}=\left(x_{1}, y_{1}, z_{1}\right)\) is \(\mathbf{F}=\frac{k q}{r^{2}} \hat{\mathbf{r}},\) where \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) and \(\hat{\mathrm{r}}=\frac{\langle x, y, z\rangle}{r} .\) Show that the work done by \(\mathbf{F}\) to move the charge \(q\) from \(P_{1}\) to \(P_{2}=\left(x_{2}, y_{2}, z_{2}\right)\) is equal to \(\frac{k q}{r_{1}}-\frac{k q}{r_{2}}\) where \(r_{1}=\sqrt{x_{1}^{2}+y_{1}^{2}+z_{1}^{2}}\) and \(r_{2}=\sqrt{x_{2}^{2}+y_{2}^{2}+z_{2}^{2}}.\)

Step-by-Step Solution

Answer

The work done by the Coulomb force to move the charge q from \(P_{1}\) to \(P_{2}\) is \(W = kq(\frac{1}{r1} - \frac{1}{r2})\).

Step 1: Understand basic principles

Understand the concept of Coulomb's law and the work-energy theorem. Coulomb's law dictates the force between two charge particles as \(F = \frac{kq}{r^2}\) where k is Coulomb's constant, q is the charge, and r is the distance between the charges. The direction is given by unit vector \(\hat{r}\). Hence the force vector is \(\mathbf{F}=\frac{k q}{r^{2}} \hat{\mathbf{r}}\). The work done- energy theorem states that work done is equal to the change in kinetic energy. We also know that work is done by a force when a charge moves a distance under the influence of it and is given by the dot product \(W = \mathbf{F} . \mathbf{d}r\). This is the key knowledge for this exercise.

Step 2: Write down the equations

Start with the formula for work done \(W = \mathbf{F} . \mathbf{d}r\). Substitute the given expression for \(\mathbf{F}\) in terms of Coulomb's expression into the work done formula.

Step 3: Perform the dot product

Perform the dot product. Given that \( \mathbf{F} = \frac{k q}{r^{2}} \hat{\mathbf{r}} \), and \( \mathbf{d}r = \hat{\mathbf{r}} dr \), we find that: \(W = (\mathbf{F} . \mathbf{d}r) = (\frac{k q}{r^{2}} \hat{\mathbf{r}}) . (\hat{\mathbf{r}} dr)\). Given \(\hat{\mathbf{r}} . \hat{\mathbf{r}} = 1\), this simplifies to \(W = - \frac{qk dr}{r^2} \).

Step 4: Integrate

The definite integral of the work done from position \(P1\) to \(P2\) is then: \(W = \int_{r1}^{r2} - \frac{qk dr}{r^2}\). Solving this yields \( W = kq(\frac{1}{r1} - \frac{1}{r2})\).

Key Concepts

Work-Energy Theorem

The work-energy theorem is a powerful concept in physics. It states that the work done by a force on an object is equal to the change in the object's kinetic energy. In simpler terms, if a force moves an object, it transfers energy to that object, changing its state of motion. This theorem is expressed mathematically as: \[ W = \Delta KE \] Where \( W \) is the work done by the force and \( \Delta KE \) represents the change in kinetic energy. For electrical charges and fields, like in our exercise using Coulomb's Law, the work done by the electric field changes the potential energy of the charges rather than kinetic energy if the charge is initially stationary. Moving a charge in the presence of an electric force involves work that affects this energy, providing a clear application of the work-energy theorem. Understanding this theorem is crucial for analyzing how energy is transferred in electric fields.

Vector Calculus

Vector calculus is essential for understanding forces and movements in 3-dimensional space. When dealing with phenomena like electric fields, which are vector fields, we use vector calculus to interpret changes and interactions. Vectors have both magnitude and direction, making them perfect tools to represent forces, velocities, and accelerations. When calculating work done by a force, the formula \( W = \mathbf{F} \cdot \mathbf{d}r \) uses the concept of a dot product. This calculates the component of the force in the direction of displacement, which is crucial for knowing how much work has been done. A unit vector \( \hat{\mathbf{r}} \) denotes direction, and by multiplying it with the force magnitude, we get the force vector working in that direction. In our exercise, understanding that \( \mathbf{F} \cdot \mathbf{d}r = \frac{k q}{r^{2}} \hat{\mathbf{r}} \cdot (\hat{\mathbf{r}} dr) \) is vital. The property \( \hat{\mathbf{r}} \cdot \hat{\mathbf{r}} = 1 \) simplifies calculations, reducing the complexities from directional aspects and focusing on magnitudes during integration. Vector calculus thus simplifies and solves what would be cumbersome in plain algebraic terms.

Integral Calculus

Integral calculus provides the mathematical tools to analyze changing quantities. When working with forces over a distance like in our Coulomb's Law example, integration is used to accumulate the effect of small incremental work along the path. That’s why we integrate the force over the path to find total work done. The process involves solving: \[ W = \int_{r1}^{r2} - \frac{qk}{r^2} dr \] This integral computes the total work by considering all small elements of work done, \(- \frac{qk}{r^2} dr \), from the initial to the final position. Through integration, we achieve the expression \( W = kq\left( \frac{1}{r1} - \frac{1}{r2} \right) \). Integral calculus here tells us that work depends on positions \( r_1 \) and \( r_2 \) rather than the path taken, as is common with conservative forces like electric forces. It's a prime example of how integrating allows us to apply Coulomb's force law over distances between charges, leading to fundamental insights about energy changes in electrostatics.