/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 In an experiment in space, one p... [FREE SOLUTION] | 91Ó°ÊÓ

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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
(a) Calculate initial acceleration using Coulomb's law and Newton's second law; (b) Sketch show decreasing acceleration and increasing velocity over time.

Step by step solution

01

Identify the forces involved

The force acting between the two protons is the electrostatic force, calculated using Coulomb’s law: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \] where \(k\) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \), \( q_1 \) and \( q_2 \) are the charges of the protons \( (1.6 \times 10^{-19} \, \text{C}) \), and \( r = 2.50 \times 10^{-3} \, \text{m} \) is the distance between them.
02

Calculate the electrostatic force

Substitute the known values into Coulomb's law: \[ F = \frac{(8.99 \times 10^9) \times (1.6 \times 10^{-19})^2}{(2.50 \times 10^{-3})^2} \] Perform the calculation to find the force in newtons.
03

Calculate the initial acceleration

Use Newton's second law to calculate the acceleration: \[ a = \frac{F}{m} \] where \( m \) is the mass of a proton \( (1.67 \times 10^{-27} \, \text{kg}) \). Substitute the force calculated in Step 2.
04

Interpret initial motion qualitatively

For the sketch of graphs: - The acceleration-time graph will show that the acceleration starts at the initial maximum and decreases over time, as the force decreases with increasing distance between protons. - The velocity-time graph will show that the velocity starts from zero, increasing continuously over time, as the proton accelerates away from the fixed proton.

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Key Concepts

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

Coulomb's Law
Understanding how forces act between charged particles is essential in physics. Coulomb's Law serves as the foundation for electrostatics and provides the means to calculate forces between two point charges. The equation is formulated as follows:\[F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\]where:- \( F \) represents the electrostatic force between the charges.- \( k \) is Coulomb's constant, with a value of approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).- \( q_1 \) and \( q_2 \) are the magnitudes of the charges involved.- \( r \) is the distance separating the charges.The inverse square relationship in Coulomb's Law means that as the distance \( r \) increases, the force \( F \) decreases. This is crucial when considering how charged particles behave over varying distances.
Acceleration of Protons
Protons, being fundamental charged particles, have a known mass and charge that allow us to predict their motion under specific forces. In this scenario, once a proton is released, it is only subjected to the electrostatic force by the fixed proton. To determine its initial acceleration, Newton's Second Law comes into play:\[a = \frac{F}{m}\]Here \( F \) is the force calculated using Coulomb’s Law, and \( m \) is the mass of a proton, \( 1.67 \times 10^{-27} \, \text{kg} \). By substituting the electrostatic force value, we calculate the initial acceleration. This initial acceleration will decrease over time as the proton moves away, reducing the electrostatic attraction.
Newton's Second Law
Newton's Second Law is one of the cornerstones of classical mechanics, providing a quantitative description of the motion of objects. It is expressed as:\[F = m \cdot a\]where:- \( F \) is the net force on the object.- \( m \) is the mass of the object.- \( a \) is the acceleration of the object.In the context of the proton problem, this law tells us that the acceleration of an object (or proton, in this case) is directly proportional to the net force acting on it and inversely proportional to the object’s mass. As the proton is released, the electrostatic force provides the net force, and using the mass of the proton, we calculate its acceleration. This force decreases as the proton moves further away from the fixed proton, which affects its acceleration over time. Thus, Newton's Second Law not only aids us in computational calculations but also deepens our understanding of how forces and motion are interrelated.

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Most popular questions from this chapter

A \(+2.00-n C\) point charge is at the origin, and a second \(-5.00-n C\) point charge is on the \(x\)-axis at \(x=0.800 \mathrm{~m} .\) (a) Find the electric field (magnitude and direction) at each of the following points on the \(x\)-axis: (i) \(x=0.200 \mathrm{~m}\); (ii) \(x=-0.200 \mathrm{~m}\). (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{~N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\), and \(\alpha=30.0^{\circ}\)

Three point charges are arranged on a line. Charge \(q_{3}=+5.00 \mathrm{nC}\) and is at the origin. Charge \(q_{2}=-3.00 \mathrm{nC}\) and is at \(x=+4.00 \mathrm{~cm}\). Charge \(q_{1}\) is at \(x=+2.00 \mathrm{~cm}\). What is \(q_{1}\) (magnitude and sign) if the net force on \(q_{3}\) is zero?

A negative charge of \(-0.550 \mu \mathrm{C}\) exerts an upward \(0.200-\mathrm{N}\) force on an unknown charge \(0.300 \mathrm{~m}\) directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the \(-0.550-\mu \mathrm{C}\) charge?

(a) What must the charge (sign and magnitude) of a \(1.3 \mathrm{~g}\) particle be for it to remain stationary when placed in a downward-directed electric field of magnitude \(650 \mathrm{~N} / \mathrm{C}\) ? (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

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