Problem 32 - 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 cm distant from the first, in a time interval of 3.20 \(\times \space 10^{-6}\) 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

Electric field: 327 N/C, Speed of proton: \(1.00 \times 10^5 \text{ m/s}\).

Step 1: Identify the given values

We are given that the distance between the plates is \( d = 1.60 \text{ cm} = 0.016 \text{ m} \) and the time interval is \( t = 3.20 \times 10^{-6} \text{ s} \). The charge of a proton is \( q = 1.60 \times 10^{-19} \text{ C} \) and the mass of a proton is \( m = 1.67 \times 10^{-27} \text{ kg} \).

Step 2: Use kinematics to find acceleration

Since the proton starts from rest and travels a distance \( d \), we use the kinematic equation: \( d = \frac{1}{2} a t^2 \) to find acceleration \( a \). Rearranging gives \( a = \frac{2d}{t^2} \). Substituting the values: \[ a = \frac{2 \times 0.016}{(3.20 \times 10^{-6})^2} = 3.13 \times 10^{10} \text{ m/s}^2. \]

Step 3: Calculate the electric field

The force on the proton due to the electric field is given by \( F = ma = qE \). Solving for \( E \), we have \( E = \frac{ma}{q} \). Substituting the known values: \[ E = \frac{1.67 \times 10^{-27} \times 3.13 \times 10^{10}}{1.60 \times 10^{-19}} = 327 \text{ N/C}. \]

Step 4: Calculate the final velocity

Using the equation \( v = at \) since the proton starts from rest. Substitute the acceleration and time:\[ v = 3.13 \times 10^{10} \times 3.20 \times 10^{-6} = 1.00 \times 10^{5} \text{ m/s}. \]

Step 5: Summary

The magnitude of the electric field is 327 N/C, and the speed of the proton when it strikes the negatively charged plate is \(1.00 \times 10^{5} \text{ m/s} \).

Key Concepts

Proton Motion

When a proton is placed in an electric field, it experiences a force because of its positive charge. This force causes the proton to accelerate and change its motion. In this exercise, the proton is released from rest near the surface of a positively charged plate. It moves towards an oppositely charged, negatively charged plate due to the influence of the electric field generated between the two plates. This scenario is a classic demonstration of linear motion under uniform electric influence.
The proton's motion is typically linear in this context because the electric field between the parallel plates is uniform. This means that anywhere between the plates, the proton feels the same electric force, ensuring it moves in a straight line. Understanding the proton's initial conditions — like starting from rest — is crucial in predicting its subsequent motion and determining its speed when it reaches the other plate.

Kinematic Equations

Kinematic equations are mathematical formulas that describe and predict the motion of objects. They're especially useful for objects moving with constant acceleration, like a proton moving in a uniform electric field. In this exercise, the third kinematic equation is used:

• Distance: \( d = \frac{1}{2} a t^2 \)

This equation relates the distance traveled by an object, its acceleration, and the time it took to travel that distance.
By rearranging the equation, we can solve for the proton's acceleration using the given distance between the plates and the time it takes to travel that distance. This makes kinematic equations powerful tools for analyzing motion when simple observations—like distance and time—are available.

Electric Force

Electric force is described by Coulomb's law in physics, which tells us that the electric force \( F \) on a charge \( q \) in an electric field \( E \) is given by:

• Formula: \( F = qE \)

This force is responsible for the acceleration of the proton between the electric plates. When a charged particle like a proton enters an electric field, it experiences a force proportional to both its charge and the strength of the electric field.
In this exercise, the electric force acts as the only significant force on the proton, as gravitational forces are negligible in comparison. This means the proton's acceleration can also be expressed using Newton’s second law, \( F = ma \), where \( m \) is the proton's mass. By equating these expressions (\( ma = qE \)), we can solve for the electric field strength \( E \) once we know the proton's acceleration.

Proton Acceleration

The acceleration of the proton is a critical element in understanding its motion in the electric field. It is determined using the kinematic equation from the earlier section, allowing us to find the value of \( a \) knowing the distance \( d \) and time \( t \). The calculated acceleration tells us how quickly the proton is speeding up as it moves due to the electric force.
Using the equation \( a = \frac{2d}{t^2} \), we calculate acceleration directly from the known displacement \( d \) and time \( t \). This helps further deduce the electric field's influence using the relationship \( F = ma = qE \). Once we know \( a \), we can obtain the electric field, strengthening our understanding of how much force the electric field imparts on charged particles such as protons. This entire calculation process links back seamlessly to real-world observations and measurements in electric fields.