Problem 38 - College Physics

Question

\(\bullet\) A proton is traveling horizontally to the right at \(4.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 \(\mathrm{cm} .\) (b) How much time does it take the proton to stop after entering the field? (c) What mini- mum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Step-by-Step Solution

Answer

(a) E = 3.30 x 10^5 N/C to the left; (b) t ≈ 1.42 x 10^-8 s; (c) E = 1.80 x 10^3 N/C to the right.

Step 1: Calculate the acceleration needed to stop the proton

We need to stop the proton, which is initially traveling at a velocity \( v_i = 4.50 \times 10^6 \, \text{m/s} \). Given the distance \( d = 3.20 \, \text{cm} = 0.032 \, \text{m} \), we use the kinematic equation \( v_f^2 = v_i^2 + 2a d \) with final velocity \( v_f = 0 \). Solving for acceleration \( a \), we get: \( 0 = (4.50 \times 10^6)^2 + 2a(0.032) \). Rearranging yields \( a = -\frac{(4.50 \times 10^6)^2}{2 \times 0.032} \approx -3.16 \times 10^{14} \, \text{m/s}^2 \).

Step 2: Calculate the required electric field for the proton

Using the relation \( F = ma \) and \( F = qE \), where \( q = 1.6 \times 10^{-19} \, \text{C} \) is the charge of the proton, we equate to find \( E = \frac{ma}{q} \). Substitute \( m = 1.67 \times 10^{-27} \, \text{kg} \) (mass of a proton) and \( a = -3.16 \times 10^{14} \, \text{m/s}^2 \) to find \( E \approx \frac{1.67 \times 10^{-27} \times -3.16 \times 10^{14}}{1.6 \times 10^{-19}} \approx -3.30 \times 10^5 \, \text{N/C} \). This means the electric field points to the left, opposite the proton's initial velocity.

Step 3: Calculate the time to stop the proton

Using the kinematic equation \( v_f = v_i + at \), and solving for time \( t \), we get \( 0 = 4.50 \times 10^6 + (-3.16 \times 10^{14})t \). Solve for \( t \) to find \( t = \frac{4.50 \times 10^6}{3.16 \times 10^{14}} \approx 1.42 \times 10^{-8} \, \text{s} \).

Step 4: Calculate the electric field for an electron

Since the electron has an opposite charge to the proton, its acceleration will be the same magnitude but in the opposite field direction. Use \( E = \frac{ma}{q} \), where \( m = 9.11 \times 10^{-31} \, \text{kg} \) (mass of the electron) and \( q = -1.6 \times 10^{-19} \, \text{C} \). Substitute \( a \approx 3.16 \times 10^{14} \, \text{m/s}^2 \). Thus, \( E \approx \frac{9.11 \times 10^{-31} \times 3.16 \times 10^{14}}{1.6 \times 10^{-19}} \approx 1.80 \times 10^3 \, \text{N/C} \). The field direction should be to the right, opposite the forward direction of the electron's velocity.

Key Concepts

Proton Motion

A proton moves through an electric field, and understanding how it behaves can help in solving many physics problems. When a proton is placed in an electric field, it experiences a force due to its charge. This force will influence the motion of the proton, either accelerating or decelerating it depending on the direction of the field in relation to the proton's movement. In this situation, we want to bring a proton to rest from an initial speed of \(4.50 \times 10^{6} \text{ m/s}\) over a distance of \(3.20 \text{ cm}\). We use kinematic equations and electric force concepts to determine the strength and direction of the electric field needed to achieve this deceleration. Since the proton is positively charged, the electric field that stops it must oppose its current direction of motion.

Kinematics

Kinematics is the branch of mechanics that focuses on the motion of objects without considering the forces that cause this motion. When solving problems involving a proton or any charged particle moving in an electric field, we often rely on kinematic equations to describe its motion. One useful formula is the equation for constant acceleration:

• \(v_f^2 = v_i^2 + 2ad\)

Use this equation to find the acceleration needed to stop the proton. Here, \(v_f\) is the final velocity (zero, since the proton comes to rest), \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(d\) is the distance over which the proton travels.To find the time \(t\) required for the proton to stop, use:

• \(v_f = v_i + at\)

This second equation will help determine the duration it takes for the proton to cease moving once it's acted upon by the electric force resulting from the electric field.

Electric Force

The electric force acting on a charge in an electric field is one of the fundamental forces in physics and can be calculated with the formula:

• \(F = qE\)

where \(F\) is the force, \(q\) is the charge of the particle (positive for protons), and \(E\) is the electric field strength. For a proton coming to rest, the direction of the electric force must be opposite to the proton's direction of travel. Using Newton's second law, \(F = ma\), where \(m\) is the mass of the proton and \(a\) is the acceleration calculated through kinematics, equate the two expressions for \(F\) to solve for \(E\):

• \(E = \frac{ma}{q}\)

The negative sign in the calculated acceleration reflects the electric field's direction being contrary to the proton's velocity.

Electron Motion

Electrons, due to their negative charge, behave oppositely to protons in an electric field. If an electron initially moves in the same direction as a proton, the electric field required to stop an electron will have an inverted direction. Calculating the field needed to bring an electron to rest uses the same principles but pays attention to the opposing charge sign.The mass of an electron is much smaller than that of a proton, which means a much weaker electric field can achieve the same acceleration effect on an electron. In this scenario, we use:

• \(E = \frac{m_e a}{q_e}\)

Here, \(m_e\) is the mass of the electron and \(q_e\) is the electron's charge. By substituting the acceleration from the kinematics result—keeping in mind it should be directed oppositely to that for a proton—you can calculate the necessary electric field magnitude and ensure it points in the desired direction to halt the electron's motion.