Problem 25 - University Physics with Modern Physics

Question

A proton is placed in a uniform electric field of \(2.75 \times\) \(10^{3} \mathrm{N} / \mathrm{C}\) . Calculate: (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00\(\mu \mathrm{s}\) in the field, assuming it starts from rest.

Step-by-Step Solution

Answer

(a) Force: \(4.40 \times 10^{-16} \text{ N}\); (b) Acceleration: \(2.63 \times 10^{11} \text{ m/s}^2\); (c) Speed: \(2.63 \times 10^{5} \text{ m/s}\).

Step 1: Calculate the Electric Force on the Proton

To find the electric force experienced by the proton, use the formula \( F = qE \), where \( F \) is the force, \( q \) is the charge of the proton, and \( E \) is the electric field. The charge of a proton is \( q = 1.60 \times 10^{-19} \, \text{C} \). Thus, the force is:\[F = (1.60 \times 10^{-19} \, \text{C})(2.75 \times 10^{3} \, \text{N/C}) = 4.40 \times 10^{-16} \, \text{N}\]

Step 2: Calculate the Proton's Acceleration

The acceleration \( a \) can be found using Newton's second law \( F = ma \), where \( m \) is the mass of the proton \((m = 1.67 \times 10^{-27} \, \text{kg})\). Rearrange to find:\[a = \frac{F}{m} = \frac{4.40 \times 10^{-16} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}} = 2.63 \times 10^{11} \, \text{m/s}^2\]

Step 3: Determine the Proton's Speed After 1.00 \(\mu\text{s}\)

Use the formula for final velocity for constant acceleration, \( v = u + at \), where \( u \) is the initial velocity (0 m/s since it starts from rest), \( a \) is the acceleration, and \( t \) is the time (\(1.00 \, \mu\text{s} = 1.00 \times 10^{-6} \, \text{s}\)). So:\[v = 0 + (2.63 \times 10^{11} \, \text{m/s}^2)(1.00 \times 10^{-6} \, \text{s}) = 2.63 \times 10^{5} \, \text{m/s}\]

Key Concepts

Electric Force

The electric force is a fundamental interaction that occurs between charged particles. It describes how these particles exert influence over each other within an electric field. The strength of this force depends directly on the charge of the particles and the electric field strength.
The formula for calculating the electric force (\( F \)) on a charge (\( q \)) in an electric field (\( E \)) is given by:

• \( F = qE \)

For a proton, which has a charge of \( q = 1.60 \times 10^{-19} \, \text{C} \), the electric force in an electric field with a magnitude of \( 2.75 \times 10^{3} \, \text{N/C} \) can be calculated as:

• \( F = (1.60 \times 10^{-19} \, \text{C})(2.75 \times 10^{3} \, \text{N/C}) = 4.40 \times 10^{-16} \, \text{N} \)

This calculation shows how a relatively small charge, like that of a proton, can experience significant forces due to a strong electric field.

Proton Acceleration

When a proton is subjected to an electric force, it accelerates. The acceleration of the proton can be determined using Newton's second law, which states that the force on an object is equal to its mass times its acceleration.
The formula for finding acceleration (\( a \)) is:

• \( F = ma \)
• \( a = \frac{F}{m} \)

The mass of a proton is \( m = 1.67 \times 10^{-27} \, \text{kg} \). Using the previously calculated force of \( 4.40 \times 10^{-16} \, \text{N} \), the acceleration can be calculated as:

• \( a = \frac{4.40 \times 10^{-16} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}} = 2.63 \times 10^{11} \, \text{m/s}^2 \)

This result indicates that protons can achieve very high accelerations when exposed to strong electric fields.

Uniform Electric Field Effect

A uniform electric field implies that the field strength is consistent across a specified region. This means that any charged particle within this field will experience a constant force and therefore undergo constant acceleration.
In this context, a proton placed in a uniform electric field will have predictable motion. It begins to accelerate immediately due to the electric force acting on it, and this force does not change because the field is uniform. The knowledge of such consistent acceleration allows us to determine parameters like speed and kinetic energy over time, as demonstrated in the specific application of this uniform electric field to a proton.

• Uniform fields simplify calculations as the force \( F \) and acceleration \( a \) remain constant.
• This consistency allows for straightforward application of kinematic equations to find velocity and displacement at any given time.

Newton's Second Law

Newton's Second Law of Motion is a fundamental principle that describes how the velocity of an object changes when it is subjected to an external force. It is expressed through the equation:

• \( F = ma \)

This law articulates that the force exerted on an object is equal to the mass of the object multiplied by its acceleration.
In the case of a proton in an electric field, Newton’s Second Law helps us understand how the electric force (\( F \)) influences the proton's acceleration (\( a \)). Given that the mass (\( m \)) of the proton is a known value, we can rearrange the formula to solve for the acceleration:

• \( a = \frac{F}{m} \)

This approach provides an insightful way to determine the change in motion of a proton when it is acted upon by an electric force. The principle is central to solving many physics problems involving forces and motion.