Question

A long, horizontal, conducting wire has the charge density \(\lambda=6.055 \cdot 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton (mass \(=1.673 \cdot 10^{-27} \mathrm{~kg}\) ) is placed a distance \(d\) above the wire and released. The magnitude of the initial acceleration of the proton is \(1.494 \cdot 10^{7} \mathrm{~m} / \mathrm{s}^{2} .\) What is the distance \(d ?\)

Short answer

Answer: The distance at which the proton is placed above the conducting wire is approximately \(1.38 \times 10^{-7} \mathrm{m}\).

Step-by-step solution

Find the Electric Field Produced by the Wire

Using Coulomb's Law, the electric field produced by an infinite wire of charge density \(\lambda\) at a distance \(d\) above it is given by: \(E = \frac{1}{2\pi\epsilon_0}\frac{\lambda}{d}\), where \(\epsilon_0\) is the vacuum permittivity (\(8.854 \times 10^{-12} \mathrm{~F}/ \mathrm{m}\)).

Find the Electrostatic Force Acting on the Proton

The electrostatic force acting on the proton due to the electric field is given by: \(F_e = qE\), where \(q\) is the charge of the proton (\(1.602 \times 10^{-19} \mathrm{~C}\)).

Apply Newton's Second Law and Calculate the Acceleration

According to Newton's second law, the net force acting on the proton is equal to the product of its mass and acceleration: \(F_e = ma\), where \(m\) is the mass of the proton and \(a\) is its acceleration.

Substitute the Electrostatic Force and Given Acceleration and Solve for Distance

From steps 2 and 3, we have \(qE = ma\). Substituting the electric field formula from step 1, we get: \(q \left( \frac{1}{2\pi\epsilon_0}\frac{\lambda}{d} \right) = ma\) Rearrange the equation to solve for the distance \(d\): \(d = \frac{1}{2\pi\epsilon_0}\frac{\lambda}{m}\frac{q}{a}\) Now, plug in the given values: \(d = \frac{1}{2\pi(8.854\times10^{-12}\mathrm{~F}/\mathrm{m})}\frac{6.055\times10^{-12}\mathrm{~C}/\mathrm{m}}{1.673\times10^{-27}\mathrm{~kg}}\frac{1.602\times10^{-19}\mathrm{~C}}{1.494\times10^7\mathrm{~m}/\mathrm{s^2}}\) After calculating, we get: \(d \approx 1.38 \times 10^{-7} \mathrm{m}\)

Final Answer

The distance \(d\) at which the proton is placed above the conducting wire is approximately \(1.38 \times 10^{-7} \mathrm{m}\).

Key concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electromagnetism, stating that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It's mathematically represented as:
\[ F = k_e \frac{{|q_1q_2|}}{{r^2}} \]
where \( F \) is the electrostatic force between the charges, \( k_e \) is Coulomb's constant (\( 8.99 \times 10^9 N\cdot m^2/C^2 \)), \( q_1 \) and \( q_2 \) are the amounts of the charges, and \( r \) is the distance between the centers of the two charges. For an infinitely long wire, the formula is adapted, considering a uniform linear charge density, to calculate the electric field, which in turn helps us find the force exerted on other charges within that field.

Electric Charge Density

Electric charge density is a measure of how much electric charge is accumulated in a particular field. There are three types of charge density: volume charge density (\( \rho \)), surface charge density (\( \sigma \)), and linear charge density (\( \lambda \)). In our problem, we're dealing with linear charge density given by:
\[ \lambda = \frac{{Q}}{{L}} \]
where \( \lambda \) is the charge per unit length, \( Q \) is the total charge, and \( L \) is the length of the charged segment. Understanding the charge density is crucial because it helps us determine the electric field created by the charged object. For an infinitely long wire, the electric field at a distance \( d \) is calculated using the known charge density, which was the focus of the first step in our problem.

Newton's Second Law

Newton's second law of motion is one of the most foundational principles in classical mechanics. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. Mathematically, it's represented as:
\[ a = \frac{{F}}{{m}} \]
where \( a \) is acceleration, \( F \) is the net force exerted on the object, and \( m \) is its mass. In the context of our exercise, Newton's second law explains how the electric field exerts an electrostatic force on the proton, causing it to accelerate. The law interconnects electrical and mechanical concepts, allowing us to solve for the distance the proton is from the wire by equating the electrostatic force with the product of the proton's mass and the given acceleration.