Question

An infinitely long line of charge has linear charge density 5.00 \(\times 10^{-12}\) C\(/\)m. A proton (mass 1.67 \(\times 10^{-27}\) kg, charge \(+\)1.60 \(\times 10^{-19}\) C) is 18.0 cm from the line and moving directly toward the line at 3.50 \(\times 10^3\) m\(/\)s. (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge?

Short answer

a) KE ≈ 1.02 × 10⁻²⁰ J, b) Proton gets extremely close, nearly touching the line.

Step-by-step solution

Calculate Initial Kinetic Energy

First, we need to find the initial kinetic energy (KE) of the proton using the formula for kinetic energy: \( KE = \frac{1}{2}mv^2 \), where \( m = 1.67 \times 10^{-27} \text{ kg} \) is the mass of the proton and \( v = 3.50 \times 10^3 \text{ m/s} \) is its velocity. Calculating, we have:\[ KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times (3.50 \times 10^3)^2 \]\[ KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times 12.25 \times 10^6 \]\[ KE = \frac{1}{2} \times 1.67 \times 12.25 \times 10^{-21} \]\[ KE \approx 1.02 \times 10^{-20} \text{ J} \]

Electric Potential Energy Difference Calculation

To find how close the proton gets, we'll use energy conservation. First, calculate the electric potential energy (PE) when the proton is 18.0 cm from the line of charge. We'll use the formula for the electric field of an infinite line charge: \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \), where \( \lambda = 5.00 \times 10^{-12} \text{ C/m} \) and \( \varepsilon_0 \approx 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \).The electric potential energy is given by: \( PE = q \times V \), where \( V = -\int E \cdot dr \). The work done by the electric field as the proton moves from \( r_1 = 0.18 \text{ m} \) to \( r_2 \) is\[ \Delta PE = -q \cdot \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_2}{r_1}\right) \]\( q = 1.60 \times 10^{-19} \text{ C} \) is the proton's charge.

Set Initial and Final Energy Equal

By the conservation of energy, the initial kinetic energy plus potential energy at the initial position must equal the potential energy at the closest approach:\[ KE_{initial} + PE_{initial} = PE_{final} \]Since \( PE_{initial} \) can be taken as zero when the proton is far away initially, so\[ KE_{initial} = -q \cdot \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_{final}}{r_{initial}}\right) \] Substituting for all known values, solve for \( r_{final} \).

Solve for Closest Distance r

Insert known values into the equation:\[ 1.02 \times 10^{-20} \text{ J} = -1.60 \times 10^{-19} \times \frac{5.00 \times 10^{-12}}{2\pi\times 8.85\times 10^{-12}} \ln\left(\frac{r_{final}}{0.18}\right) \] Calculate the constants:\[ \frac{5.00 \times 10^{-12}}{2\pi\times 8.85\times 10^{-12}} \approx 0.0899 \]So the equation becomes:\[ 1.02 \times 10^{-20} = -1.60 \times 0.0899 \ln\left(\frac{r_{final}}{0.18}\right) \]Solving gives:\[ \ln\left(\frac{r_{final}}{0.18}\right) \approx -71.11 \]\[ \frac{r_{final}}{0.18} \approx e^{-71.11} \]\( r_{final} \) is very close to 0 cm, implying the proton stops a negligible distance from the line before turning around.

Key concepts

Linear Charge Density

Linear charge density is a key concept when dealing with infinite or long lines of charge. It represents the amount of electric charge distributed per unit length along a line. In this context, linear charge density is denoted by the symbol \( \lambda \) and has the unit of Coulombs per meter (C/m). For the problem at hand, the line of charge has a linear charge density of \( 5.00 \times 10^{-12} \) C/m.

Understanding linear charge density helps in calculating the electric field created by the line of charge. The electric field strength is crucial when figuring out the interaction between the charge line and any nearby charged particles, like the proton in the problem. An infinite line of charge creates an electric field around it, and this electric field determines the potential energy and force that acts on other charges.

By recognizing the relationship between linear charge density and the resulting electric field, we can determine the potential energy changes and predict how other charges will behave in potential scenarios. This is fundamental for solving problems related to electrostatics involving lines of charge.

Kinetic Energy

Kinetic energy (KE) is the energy that an object possesses due to its motion. In the exercise, we calculate the initial kinetic energy of a proton using the formula:

• \( KE = \frac{1}{2}mv^2 \)
• Where \( m \) is the mass of the proton \( 1.67 \times 10^{-27} \text{ kg} \)
• \( v \) is the velocity of \( 3.50 \times 10^3 \text{ m/s} \)

Plugging in these values provides an initial kinetic energy of approximately \( 1.02 \times 10^{-20} \text{ J} \). Understanding kinetic energy is vital to analyzing how much of the proton's energy is converted into potential energy as it moves towards the line of charge.

In many physics problems, kinetic energy sets the scene for exploring energy transfer or transformation between different forms like potential energy. For this exercise, kinetic energy represents the initial energy the proton has as it heads towards the line of charge, before being affected by the electric field.

Conservation of Energy

The conservation of energy principle is essential to understanding this problem. It states that the total energy of an isolated system remains constant, highlighting how energy transitions between kinetic energy and potential energy. In this exercise:

• Proton's initial kinetic energy is transformed into electric potential energy as it approaches the line of charge.
• This transformation happens due to the electric field created by the line.

The equation captures this conservation:\[ KE_{initial} + PE_{initial} = PE_{final} \]The initial potential energy \( PE_{initial} \) is often assumed to be zero if infinitely far from the line. Thus, the equation simplifies such that the initial kinetic energy equals the change in potential energy as the proton gets closer.

The closest approach, or \( r_{final} \), is calculated by considering the balance in kinetic and potential energy changes. Overall, energy conservation explains why the proton slows down, comes to a halt momentarily, and eventually reverses direction when too close to the line of charge. Understanding this principle is crucial for predicting motion and energy changes in electromagnetic fields.