Problem 25 - University Physics with Modern Physics

Question

A proton is traveling horizontally to the right at 4.50 \(\times 10^6\) m\(/\)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 cm. (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Step-by-Step Solution

Answer

(a) 3.3 × 10³ N/C to the left. (b) 1.42 × 10⁻⁸ s. (c) 1.8 × 10⁻¹ N/C to the left.

Step 1: Known Values and Conversion

First, list the given values and convert units where necessary:- Initial velocity of proton, \( v_i = 4.50 \times 10^6 \) m/s- Final velocity of proton, \( v_f = 0 \) m/s- Distance, \( d = 3.20 \) cm = 0.032 m.We'll convert the distance from centimeters to meters for consistency in SI units.

Step 2: Find Acceleration Needed

Use the kinematic equation to find the acceleration \(a\):\[v_f^2 = v_i^2 + 2ad\]Solving for \(a\):\[a = \frac{v_f^2 - v_i^2}{2d} = \frac{0 - (4.50 \times 10^6)^2}{2 \times 0.032}\]

Step 3: Calculate Acceleration

Calculate the acceleration value:\[a = \frac{- (4.50 \times 10^6)^2}{2 \times 0.032} = -3.16 \times 10^{14}\, \text{m/s}^2\]The negative sign indicates the direction of acceleration is opposite to the direction of motion.

Step 4: Find the Electric Field Magnitude

Use the relation between acceleration and electric field:\[a = \frac{qE}{m}\]For a proton, \(q = 1.6 \times 10^{-19}\, \text{C}\), \(m = 1.67 \times 10^{-27}\, \text{kg}\). Solving for \(E\):\[E = \frac{ma}{q}\]

Step 5: Calculate Electric Field Magnitude for Proton

Substitute the values:\[E = \frac{1.67 \times 10^{-27} \times 3.16 \times 10^{14}}{1.6 \times 10^{-19}} = 3.3 \times 10^3\, \text{N/C}\]The direction of the electric field is opposite to the proton's initial velocity (to the left) since it needs to decelerate the proton.

Step 6: Calculate Time to Stop the Proton

Use the kinematic equation for time:\[v_f = v_i + at\]\[t = \frac{v_f - v_i}{a} = \frac{0 - 4.50 \times 10^6}{-3.16 \times 10^{14}}\]Calculate the time.

Step 7: Calculate Time Result

Compute the time:\[t = \frac{-4.50 \times 10^6}{-3.16 \times 10^{14}} \approx 1.42 \times 10^{-8}\, \text{s}\]

Step 8: Find Minimum Field for Electron

Use similar calculations as with the proton:For an electron, \( q = 1.6 \times 10^{-19}\, \text{C} \), \( m = 9.11 \times 10^{-31}\, \text{kg} \). \[E = \frac{ma}{q}\]Again substitute the values previously found for \(a\).

Step 9: Calculate Electric Field Magnitude for Electron

Substitute values for the electron:\[E = \frac{9.11 \times 10^{-31} \times 3.16 \times 10^{14}}{1.6 \times 10^{-19}} = 1.8 \times 10^{-1}\, \text{N/C}\]The direction will also be opposite to the electron's initial velocity, which is to the left.

Key Concepts

Proton Motion

Protons are subatomic particles found in the nucleus of an atom. In physics problems, especially those involving electric fields, understanding the motion of protons is crucial.
When a proton travels through an electric field, its motion can be influenced by the electric force acting upon it. This is because the proton carries a positive charge. In this exercise, the proton starts with an initial horizontal velocity of \(4.50 \times 10^6\) m/s.
The goal is to apply an electric field to bring the proton to a stop over a specified distance (3.20 cm). Key points in analyzing proton motion in electric fields include:

• Protons will accelerate or decelerate based on the direction and magnitude of the electric field.
• The direction of the electric field needed to stop a proton must be opposite to its initial motion.
• Understanding proton dynamics allows us to predict the electric field strength required to cause specific changes in motion.

Kinematic Equations

Kinematic equations are mathematical formulas used to describe motion. They help us calculate variables like displacement, velocity, acceleration, and time.
For uniformly accelerated motion, such as a proton stopping under the influence of an electric field, we can use:

• \[ v_f^2 = v_i^2 + 2ad \]This formula calculates the acceleration needed to stop a particle over a certain distance. Here, \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(d\) is the distance traveled.
• \[ v_f = v_i + at \]This equation calculates the time taken to bring the particle to rest, where \(t\) is the time.

Utilizing these equations allows us to understand how adjustments in electric field strength can affect particle motion, specifically in slowing down or stopping particles such as protons.

Acceleration

Acceleration is a key concept when analyzing the effect of forces, including electric fields, on particle motion. It describes the rate of change of velocity over time. When a proton enters an electric field, the acceleration can be calculated using the principle:

• The force exerted by the electric field is given by \( F = qE \), where \( q \) is the charge and \( E \) is the electric field strength.
• Newton's second law, \( F = ma \), links this force to acceleration, \( a \), and mass, \( m \). Thus, \( a = \frac{qE}{m} \).

In this problem, the calculated acceleration is \(-3.16 \times 10^{14}\, \text{m/s}^2\), indicating a rapid deceleration. The negative sign denotes that the acceleration is directed in the opposite direction to the motion, which is necessary for the proton to stop.
Understanding acceleration helps us not only evaluate how quickly particles will come to rest but also assists in determining the required electric field to achieve this.

Electron Motion

Electrons, similar to protons, are subatomic particles but with a negative charge. This charge difference means their interactions with electric fields are opposite to that of protons.
When stopping an electron with an electric field under similar conditions as a proton, we must consider its lighter mass and charge:

• Electrons require a much weaker electric field to stop over the same distance due to their smaller mass (\(9.11 \times 10^{-31}\, \text{kg}\)).
• The calculated field strength needed for an electron is \(1.8 \times 10^{-1}\, \text{N/C}\), significantly lower than that needed for a proton.
• The direction of the required electric field is still opposite to the electron's motion.

By considering the distinct properties of the electron, such as its charge and mass, we better understand how varying field strengths and directions influence its motion.