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In an experiment in space, one proton is held fixed and another proton is released from rest a distance of \(2.50 \mathrm{~mm}\) away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration-time and velocity-time graphs of the released proton's motion.

Short Answer

Expert verified
The initial acceleration of the proton is equal to the electrostatic force divided by the mass of the proton. The acceleration-time graph is an upward slope (as the acceleration increases with time). The velocity-time graph is curved and rises more steeply with time due to the increasing acceleration.

Step by step solution

01

Calculate the force between the protons

The force between two charged particles can be calculated using Coulomb's law, which is \(F = k \cdot \frac{q_1 \cdot q_2}{{d^2}}\), where \(k = 8.99 × 10^{9} N \cdot (m^2/C^2)\), \(q_1 = q_2 = 1.60 × 10^{-19} C\) (charge of a proton), and \(d = 2.50 × 10^{-3} m\) is the distance between the protons.
02

Calculate the acceleration

Acceleration \(a\) can be found by dividing the force \(F\) by the mass of a proton \(m = 1.67 \times 10^{-27} kg\). It is obtained from the second law of motion (Newton's second law), which states that the force acting on an object is equal to its mass multiplied by its acceleration: \( F = m \cdot a \). Then, the acceleration is \(a = \frac{F}{m}\).
03

Sketch the acceleration-time graph

Since the distance between the two protons is decreasing, the force (and thus the acceleration) is increasing with time. This should be represented as an ascending curve on the acceleration-time graph.
04

Sketch the velocity-time graph

As the proton accelerates, its velocity increases. Furthermore, since acceleration is increasing, this increase in velocity is not linear but rather exponential. Thus, the velocity-time graph should be a curve ascending more steeply with time.

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Key Concepts

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

Electric Force
The concept of electric force lies at the heart of understanding numerous phenomena in physics, especially when studying charged particles. The electric force between two charges is described mathematically by Coulomb's law, which is formulated as \(F = k \cdot \frac{q_1 \cdot q_2}{{d^2}}}\), wherein \(k\) is Coulomb's constant (\(8.99 \times 10^{9} \text{N} \cdot (\text{m}^2/\text{C}^2)\)), \(q_1\) and \(q_2\) represent the magnitudes of the charges involved, and \(d\) is the distance separating them. For protons, which are positively charged particles with a charge of \(1.60 \times 10^{-19} \text{C}\), the electric force is repulsive and will decrease as the distance between them increases.

When working with electric forces in practice, it is important not only to compute the magnitude of the force using Coulomb's law but also to consider its direction, which is along the line connecting the two charges. This understanding is essential when predicting the motion of charged particles in fields.
Proton Acceleration
Acceleration is a measure of how quickly an object's velocity changes. When a proton is released from rest, as in the exercise, it accelerates due to the electric force acting upon it. Here, we can calculate the initial acceleration of a proton using Newton's second law, \( F = m \cdot a \) and Coulomb's law combined.

This involves computing the force between the protons with Coulomb's law and then rearranging Newton's second law to \( a = \frac{F}{m} \) to find acceleration. The mass of a proton (\(m = 1.67 \times 10^{-27} \text{kg}\)) is a constant, meaning the acceleration of the proton can be found by dividing the electric force that we have previously found by this mass. The resulting figure represents the rate at which the proton's velocity will change per second as a result of the electrical interaction with another proton.
Acceleration-Time Graph
When representing motion graphically, an acceleration-time graph offers insights into how an object's acceleration changes over time. In the case of a proton released in the vicinity of another fixed proton, we can imagine how its acceleration evolves. Initially, as the distance between the protons is maximal, the acceleration is minimal. But as the proton moves closer, the force—and consequently the acceleration—increases.

Therefore, the acceleration-time graph for this scenario would start with a lower value at \(t = 0\) and would curve upwards, indicating an increase in acceleration. The curve rises steeper as the distance between the protons reduces, reflecting the inverse square nature of Coulomb's law, where force, and hence acceleration, increases dramatically as distance decreases.
Velocity-Time Graph

Understanding Velocity Change

A velocity-time graph helps us understand how an object's speed and direction of travel change over time. For a proton accelerating due to electric force, its velocity starts at zero and increases as time progresses. The curve of the graph depends on the acceleration: a constant acceleration produces a straight line, while varying acceleration produces a curve.

Given that the proton's acceleration increases with time due to a decrease in distance from the fixed proton, the slope of the velocity-time graph also increases. In our case, the graph should represent a curve starting at the origin and becoming steeper, indicating that not only is the proton speeding up, but it is doing so at an increasing rate.
Newton's Second Law
One of the foundational principles in physics, Newton's second law, relates the net force acting on an object to its mass and the acceleration it experiences. Stated as \( F = m \cdot a \), this law provides the framework for predicting the motion of objects. In the context of this exercise, we use Newton's second law to determine the proton's acceleration.

It implies that for any object with a constant mass, such as a proton, the net force and acceleration are directly proportional. As a result, if we know the force on the proton due to another charge, we can determine how quickly it will accelerate. This law is fundamental for interpreting interaction outcomes between particles and the resulting motion, making it a cornerstone for exercises involving forces and movements of charged particles.

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Most popular questions from this chapter

Two point charges are located on the \(y\) -axis as follows: charge \(q_{1}=-1.50 \mathrm{nC}\) at \(y=-0.600 \mathrm{~m},\) and charge \(q_{2}=+3.20 \mathrm{nC}\) at the origin \((y=0) .\) What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_{3}=+5.00 \mathrm{nC}\) located at \(y=-0.400 \mathrm{~m} ?\)

A small object with mass \(m,\) charge \(q,\) and initial speed \(v_{0}=5.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\) is projected into a uniform electric field between two parallel metal plates of length \(26.0 \mathrm{~cm}\) (Fig. \(\mathrm{P} 21.78\) ). The electric field between the plates is directed downward and has magnitude \(E=800 \mathrm{~N} / \mathrm{C}\). Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance \(d=1.25 \mathrm{~cm}\) from its original direction of motion and reaches a collecting plate that is \(56.0 \mathrm{~cm}\) from the edge of the parallel plates. Ignore gravity and air resistance. Calculate the object's charge-to-mass ratio, \(q / m\).

A particle has charge \(-5.00 \mathrm{nC}\). (a) Find the magnitude and direction of the electric field due to this particle at a point \(0.250 \mathrm{~m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of \(12.0 \mathrm{~N} / \mathrm{C} ?\)

In a follow-up experiment, a charge of \(+40 \mathrm{pC}\) was placed at the center of an artificial flower at the end of a \(30-\mathrm{cm}\) -long stem. Bees were observed to approach no closer than \(15 \mathrm{~cm}\) 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) \(2.4 \mathrm{~N} / \mathrm{C} ;\) (b) \(16 \mathrm{~N} / \mathrm{C} ;\) (c) \(2.7 \times 10^{-10} \mathrm{~N} / \mathrm{C}\) (d) \(4.8 \times 10^{-10} \mathrm{~N} / \mathrm{C}\).

Two small spheres, each carrying a net positive charge, are separated by \(0.400 \mathrm{~m}\). 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 \(q_{1}\) ) at the origin and the other sphere (charge \(q_{2}\) ) at \(x=+0.400 \mathrm{~m}\). Available to you are a third sphere with net charge \(q_{3}=4.00 \times 10^{-6} \mathrm{C}\) 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 \(x=0.200 \mathrm{~m} ;\) you measure the net force on it to be \(4.50 \mathrm{~N}\) in the \(+x\) -direction. Then you move the third sphere to \(x=+0.600 \mathrm{~m}\) and measure the net force on it now to be \(3.50 \mathrm{~N}\) 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 \(x=-0.200 \mathrm{~m} ?\) (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?

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