Problem 88 - Calculus Early Transcendentals

Question

The nucleus of an atom is positively charged because it consists of positively charged protons and uncharged neutrons. To bring a free proton toward a nucleus, a repulsive force \(F(r)=k q Q / r^{2}\) must be overcome, where \(q=1.6 \times 10^{-19} \mathrm{C}\) (coulombs) is the charge on the proton, \(k=9 \times 10^{9} \mathrm{N}-\mathrm{m}^{2} / \mathrm{C}^{2}, Q\) is the charge on the nucleus, and \(r\) is the distance between the center of the nucleus and the proton. Find the work required to bring a free proton (assumed to be a point mass) from a large distance \((r \rightarrow \infty)\) to the edge of a nucleus that has a charge \(Q=50 q\) and a radius of \(6 \times 10^{-11} \mathrm{m}\)

Step-by-Step Solution

Answer

Answer: The work required to bring a free proton from a large distance to the edge of the nucleus is approximately \(6.4 \times 10^{-15}\) Joules.

Step 1: List down the given values

We are given: - The charge on the proton, \(q=1.6 \times 10^{-19} C\) - Coulomb's constant, \(k=9 \times 10^{9} Nm^2/C^2\) - The charge on the nucleus, \(Q=50q\) - The radius of the nucleus, \(r=6 \times 10^{-11} m\)

Step 2: Write the expression for the force

The repulsive force between the proton and the nucleus is given by Coulomb's Law: \(F(r) = \frac{kqQ}{r^2}\)

Step 3: Calculate the work done

Now, let's calculate the work done to bring the proton from infinity to the edge of the nucleus by integrating the force expression: \(W = \int_{\infty}^{r} F(r) \, dr\) Substitute the repulsive force we found: \(W = \int_{\infty}^{r} \frac{kqQ}{r^2} \, dr\) Substitute the given values: \(W = \int_{\infty}^{6 \times 10^{-11}} \frac{(9 \times 10^9)(1.6 \times 10^{-19})(50)(1.6 \times 10^{-19})}{r^2} \, dr\)

Step 4: Evaluate the integral

We now need to integrate the function: \(W = -(9 \times 10^9)(1.6 \times 10^{-19})(50)(1.6 \times 10^{-19})\int_{\infty}^{6 \times 10^{-11}} r^{-2} \, dr\) \(W = -(9 \times 10^9)(1.6 \times 10^{-19})(50)(1.6 \times 10^{-19})\Big[-\frac{1}{r}\Big]_{\infty}^{6 \times 10^{-11}}\) Now, evaluate the expression by plugging in the limits: \(W = -(9 \times 10^9)(1.6 \times 10^{-19})(50)(1.6 \times 10^{-19})\Big[-\frac{1}{6 \times 10^{-11}} - (-\frac{1}{\infty})\Big]\) Since \(1/\infty = 0\), the expression becomes: \(W = -(9 \times 10^9)(1.6 \times 10^{-19})(50)(1.6 \times 10^{-19})(-\frac{1}{6 \times 10^{-11}})\)

Step 5: Calculate the final value

Now, compute the work: \(W = 9 \times 10^{9} \times (1.6 \times 10^{-19})^2 \times 50 /\Big( 6 \times 10^{-11}\Big)\) \(W \approx 6.4 \times 10^{-15}\) Joules Hence, the work required to bring a free proton from a large distance to the edge of the nucleus is approximately \(6.4 \times 10^{-15}\) Joules.

Key Concepts

Understanding Coulomb's Law

Coulomb's Law, named after Charles-Augustin de Coulomb, is a foundational principle in electromagnetism. It quantifies the electrostatic force between two point charges and is expressed by the equation:
\( F(r) = \frac{kq_1q_2}{r^2} \).
Here, \( F(r) \) represents the force between the charges, \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \)), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the separation distance between the charges. The force is attractive if the charges have opposite signs, whereas it is repulsive for like charges, as seen in our exercise involving a proton and a nucleus.

Coulomb's Law is a vector equation and includes the direction. When calculating forces in one dimension, as with the proton approaching a nucleus, we can use scalar forms and add a negative sign if the force is repulsive. The strength of the force diminishes with the square of the distance, demonstrating an inverse-square law.

Calculating Work Done by Electric Force

The work done by electric force when moving a charge in an electric field is a measure of energy transfer. According to physics, work is calculated by integrating the force over the path of movement:
\( W = \int_{\text{path}} \mathbf{F} \
\cdot d\mathbf{s} \),
where \( W \) denotes work, \( \mathbf{F} \) is force, and \( d\mathbf{s} \) is the differential displacement vector along the path. When the force is conservative, like the electrostatic force, the path does not matter and only the endpoints are important.

In our example involving a proton and a nucleus, work is done against the electrostatic force to bring the proton from infinity (where the potential energy is considered zero) to the edge of the nucleus. Mathematically, this is represented as the integral of the force over the radial distance from \( r = \infty \) to \( r = 6 \times 10^{-11} \, \mathrm{m} \), which was evaluated to find the work required for this process.

Applying Integration in Physics

Integration stands as a powerful tool in physics for solving a variety of problems involving energy, motion, and other quantities that vary with respect to another variable. Essentially, integration enables us to add up tiny pieces to determine a whole.
In the context of our exercise, we used integration to find the work done to move a proton, by integrating the force over the distance from infinity to the nuclear edge. This integral takes the form:
\( W = \int_{\infty}^{r} \frac{kqQ}{r^2} dr \).Integration is often used in physics when the quantity to be calculated varies continuously. The act of integrating allows us to accumulate the tiny slices of work over each infinitesimal segment of the path that the proton travels, giving us the total work done when the proton reaches the nucleus.