Problem 38 - University Physics with Modern Physics

Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 \(\mathrm{cm}\) distant from the first, in a time interval of \(1.50 \times 10^{-6} \mathrm{s}\) . (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

Step-by-Step Solution

Answer

(a) \( E \approx 1487.5 \ N/C \); (b) \( v \approx 21,300 \ m/s \).

Step 1: Understand the problem

We are given a uniform electric field between two charged plates, a proton is released from rest from one plate and travels to the other. The distance between the plates is 1.60 cm, and the time is given as \( 1.50 \times 10^{-6} \ s \). We need to find the magnitude of the electric field and the speed of the proton when it hits the opposite plate.

Step 2: Write down known values

Distance \( d = 1.60 \ cm = 0.016 \ m \), time \( t = 1.50 \times 10^{-6} \ s \), charge of the proton \( q = 1.60 \times 10^{-19} \ C \), mass of the proton \( m = 1.67 \times 10^{-27} \ kg \), initial velocity \( v_0 = 0 \ m/s \).

Step 3: Use kinematic equation to find acceleration

Use the kinematic equation for constant acceleration: \[ d = v_0 t + \frac{1}{2} a t^2 \]Substituting known values:\[ 0.016 = 0 \cdot t + \frac{1}{2} a (1.50 \times 10^{-6})^2 \]Solving for \( a \):\[ a = \frac{2 \times 0.016}{(1.50 \times 10^{-6})^2} \approx 1.42 \times 10^{10} \ m/s^2 \]

Step 4: Calculate the electric field using acceleration

Now, use Newton's second law and the relation for electric force:\[ F = ma = qE \]\[ E = \frac{ma}{q} \]Substituting known values:\[ E = \frac{1.67 \times 10^{-27} \ times 1.42 \times 10^{10}}{1.60 \times 10^{-19}} \]Calculating gives:\[ E \approx 1487.5 \ N/C \]

Step 5: Calculate the final speed of the proton

Use the kinematic equation:\[ v = v_0 + at \]Since \( v_0 = 0 \),\[ v = a \cdot t = 1.42 \times 10^{10} \times 1.50 \times 10^{-6} \]Calculating gives:\[ v \approx 21.3 \times 10^3 \ m/s \]

Step 6: Summary of results

The magnitude of the electric field is approximately \( 1487.5 \ N/C \), and the speed of the proton when it strikes the negatively charged plate is approximately \( 21,300 \ m/s \).

Key Concepts

Kinematics

Kinematics is the branch of physics that describes the motion of objects without considering the causes of this motion. It uses equations that relate distance, velocity, acceleration, and time. In our problem, a proton moves between two plates under the influence of an electric field. Since it starts from rest, its initial velocity is zero.

• We use the kinematic equation: \[ d = v_0 t + \frac{1}{2} a t^2 \]

This formula helps us find the acceleration the proton experiences during its travel. Here, \( v_0 \) is the initial velocity, \( a \) is the acceleration, and \( t \) is time. By substituting the known values of distance and time, we can solve for the acceleration of the proton.
Understanding these connections helps us see how motion evolves in the electric field.

Newton's Second Law

Newton's Second Law of Motion explains how the velocity of an object changes when it is subjected to an external force. It is commonly expressed as \( F = ma \), where \( F \) is the force applied to the object, \( m \) is the mass, and \( a \) is the acceleration.
In this scenario, the external force on the proton is the electric force, derived from the electric field in the region.

• The relationship \( F = qE \) ties the electric force to the electric field \( E \) and the proton's charge \( q \).
• By substituting known values we directly find the electric field's magnitude.

This reveals how external electric fields affect charged particles like protons, making Newton's Second Law a critical tool in analyzing motion in electric fields.

Proton Motion

Protons are positively charged subatomic particles that move based on the forces acting upon them. In this exercise, the proton is influenced solely by the electric field as it travels between the plates.
The proton's journey starts from rest, implying that its initial speed is zero. Then, as the electric force acts on it, the proton gains speed. This acceleration is linear due to the constant force from the uniform electric field.

• The final velocity \( v \) of the proton can be computed by the equation: \[ v = v_0 + at \]

Using the acceleration found from kinematics, we can determine the speed of the proton upon hitting the opposite plate. Such calculations showcase how charged particles react in electric fields, expanding our understanding of particle motion.

Uniform Electric Field

A uniform electric field is one in which the field strength is constant throughout. This means that any charge placed in this field experiences a constant force, leading to uniform acceleration.
In our exercise, the electric field between the plates is uniform, creating a predictable environment for the proton's motion.

• The formula for the electric field strength \( E \) is given by: \[ E = \frac{F}{q} \]

Knowing the acceleration from kinematics and applying Newton's Second Law allows us to find the magnitude of this constant electric field. Uniform fields simplify calculations, helping us predict how charges will behave over time. Understanding these principles deepens our grasp of electric phenomena and their effects on matter.