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Chapter 21: Problem 23

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.

Short Answer

Expert verified
(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 by step solution

01

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, \( F = (1.60 \times 10^{-19} \ C)(2.75 \times 10^3 \ N/C) = 4.40 \times 10^{-16} \ N \).
02

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: \( a = \frac{F}{m} = \frac{4.40 \times 10^{-16} \ N}{1.67 \times 10^{-27} \ kg} = 2.63 \times 10^{11} \ m/s^2 \).
03

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, \( v = 0 + (2.63 \times 10^{11} \ m/s^2)(1.00 \times 10^{-6} \ s) = 2.63 \times 10^{5} \ m/s \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Suppose you had two small boxes, each containing 1.0 g of protons. (a) If one were placed on the moon by an astronaut and the other were left on the earth, and if they were connected by a very light (and very long!) string, what would be the tension in the string? Express your answer in newtons and in pounds. Do you need to take into account the gravitational forces of the earth and moon on the protons? Why? (b) What gravitational force would each box of protons exert on the other box?

Two point charges are located on the \(y\)-axis as follows: charge \(q_1 = -1.50 \)nC at \(y = -\)0.600 m, and charge \(q_2 = +\)3.20 nC at the origin \((y = 0)\). What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_3 = +\)5.00 nC located at \(y = -\)0.400 m ?

A point charge \(q_1 = -\)4.00 nC is at the point \(x =\) 0.600 m, \(y =\) 0.800 m, and a second point charge \(q_2 = +\)6.00 nC is at the point \(x =\) 0.600 m, \(y =\) 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

Two small spheres, each carrying a net positive charge, are separated by \(0.400 m\). You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (\(charge \space q_1\)) at the origin and the other sphere (\(charge \space q_2\)) at \(x = +\)0.400 m. Available to you are a third sphere with net charge \(q_3 = 4.00 \times 10^{-6}\) C and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the \(x\)-axis at \(x =\) 0.200 m; you measure the net force on it to be 4.50 N in the \(+ x\)-direction. Then you move the third sphere to \(x = +\)0.600 m and measure the net force on it now to be 3.50 N in the \(+ x\)-direction. (a) Calculate \(q_1\) and \(q_2\). (b) What is the net force (magnitude and direction) on \(q_3\) if it is placed on the \(x\)-axis at \(x = -\)0.200 m? (c) At what value of \(x\) (other than \(x = \pm \infty\)) could \(q_3\) be placed so that the net force on it is zero?

In a follow-up experiment, a charge of \(+40\) pC was placed at the center of an artificial flower at the end of a 30-cm long stem. Bees were observed to approach no closer than 15 cm from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? (a) \(2.4 \space N/C; (b) 16 \space N/C; (c) 2.7 \times 10^{-10} \space N/C; (d) 4.8 \times 10^{-10} \space N/C.\)
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