Problem 23 - University Physics with Modern Physics

Question

A proton is placed in a uniform electric field of 2.75 \(\times 10^3 \space N/C\). 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 \(\mu\)s in the field, assuming it starts from rest.

Step-by-Step Solution

Answer

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

Step 1: Calculate the Electric Force

First, we need to calculate the magnitude of the electric force experienced by the proton. The electric force \( F \) is given by the formula \( F = qE \), where \( q \) is the charge of the proton and \( E \) is the electric field strength. The charge of a proton \( q \) is \( 1.60 \times 10^{-19} \ C\), and the electric field \( E \) is \( 2.75 \times 10^3 \ N/C \). Thus, \( F = (1.60 \times 10^{-19} \ C)(2.75 \times 10^3 \ N/C) = 4.40 \times 10^{-16} \ N \).

Step 2: Calculate the Proton's Acceleration

Next, we use Newton's second law to find the proton's acceleration. According to \( F = ma \), where \( m \) is the mass of the proton \( 1.67 \times 10^{-27} \ kg \). Solving for acceleration \( a \), we have: \( a = \frac{F}{m} = \frac{4.40 \times 10^{-16} \ N}{1.67 \times 10^{-27} \ kg} = 2.63 \times 10^{11} \ m/s^2 \).

Step 3: Calculate the Proton's Speed

Finally, to find the proton's speed after 1.00 \( \mu s \) (microsecond), we use the formula for motion, \( v = u + at \), where \( u \) (initial speed) is zero, \( a \) is the acceleration from Step 2, and \( t \) is the time \( 1.00 \times 10^{-6} \ s \). Thus, \( v = 0 + (2.63 \times 10^{11} \ m/s^2)(1.00 \times 10^{-6} \ s) = 2.63 \times 10^{5} \ m/s \).

Key Concepts

Uniform Electric Field

A uniform electric field is a region where the electric force on a charged particle is constant in both magnitude and direction. Imagine a flat expanse where every point experiences the same push or pull. This is different from a non-uniform field, where the force varies at different points. Fields like these are often created between two parallel plates connected to a voltage source.
The strength of a uniform electric field, denoted as \( E \), is measured in newtons per coulomb (\( N/C \)). Each charged particle within this field, such as a proton, experiences an electric force \( F \), calculated by the formula \( F = qE \). Here, \( q \) is the charge of the particle.

• For a proton, \( q = 1.60 \times 10^{-19} \, C \).
• Given field strength \( E \) is \( 2.75 \times 10^3 \, N/C \).

Using the formula \( F = qE \), the electric force on the proton comes out to be \( 4.40 \times 10^{-16} \, N \). This constant force characterizes how an electric field acts uniformly on a charge.

Proton Acceleration

Acceleration occurs when an object changes its velocity over time due to an external force. For a proton in an electric field, the force calculated earlier leads to its acceleration. We determine this using Newton's second law of motion, \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration.
The mass of a proton is \( 1.67 \times 10^{-27} \, kg \). Using

• \( F = 4.40 \times 10^{-16} \, N \) from the uniform electric field,
• we solve for \( a \): \( a = \frac{F}{m} \)

This calculation results in an acceleration of \( 2.63 \times 10^{11} \, m/s^2 \). This large acceleration shows how strongly the proton is pushed by the electric force in such a field.

Calculating Speed from Acceleration

After determining the proton's acceleration, the next step is calculating how fast it moves after some time in the electric field. We use the basic physics equation for linear motion: \( v = u + at \). Here, \( v \) is the final speed, \( u \) is the initial speed, \( a \) is acceleration, and \( t \) is the time duration.
Starting from rest means \( u = 0 \), and the given time \( t \) is \( 1.00 \times 10^{-6} \, s \). With the acceleration \( a = 2.63 \times 10^{11} \, m/s^2 \), the calculation becomes

• \( v = 0 + (2.63 \times 10^{11} \, m/s^2)(1.00 \times 10^{-6} \, s) \)

This results in a final speed \( v = 2.63 \times 10^5 \, m/s \). This formula highlights how motion parameters intertwine when influenced by an external force across a known period.