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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?

Short Answer

Expert verified
The tension in the string is primarily due to electrostatic forces, not gravitational, and is approximately 121.7 Newtons or about 27.4 pounds. The gravitational force is negligible.

Step by step solution

01

Understand the Problem

You are given two boxes, each with 1.0 g of protons, and they are connected by a string between Earth and Moon. You need to find the tension in the string due to the electrostatic force between the protons and compare it with gravitational forces.
02

Calculate the Total Charge

Protons have a charge of approximately \(1.6 \times 10^{-19}\) C. Calculate the number of protons in 1.0 g using the formula \(n = \frac{1.0 \text{ g}}{1.67 \times 10^{-24} \text{ g/p}}\). Then, multiply \(n\) by the charge of a proton to get the total charge \(Q = n \times 1.6 \times 10^{-19}\) C for one box.
03

Find the Electrostatic Force

Use Coulomb's Law to find the electrostatic force: \[ F = \frac{k \cdot |Q_1| \cdot |Q_2|}{r^2} \] where \(k\) is Coulomb's constant \(8.99 \times 10^9 \text{ N m}^2/ ext{C}^2\), \(Q_1 = Q_2\) is the charge of each box, and \(r\) is the distance between Earth and Moon, approximately \(3.84 \times 10^8\) m.
04

Check Effect of Gravitational Forces

The gravitational force between two masses can be calculated using \[ F_g = \frac{G m_1 m_2}{r^2} \] where \(m_1 = m_2 = 1.0 \text{ g} = 0.001 \text{ kg}\) and \(G\) is the gravitational constant \(6.674 \times 10^{-11} \text{ N m}^2/ ext{kg}^2\). Compare this force \(F_g\) to the electrostatic force obtained in Step 3.
05

Determine Tension in the String

The tension in the string is primarily due to the electrostatic force because it is significantly larger than the gravitational force. Convert the force from newtons to pounds using the conversion factor \(1 ext{ N} = 0.224809 ext{ lb}\). Calculating this gives you tension in both units.
06

Short Calculation of Gravitational Force

Using the gravitational formula from Step 4, calculate the force that each box exerts on the other. Given the small masses and large distance, this force will be negligible compared to the electrostatic force.

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Key Concepts

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

Coulomb's Law
Coulomb's Law is at the heart of understanding how charged particles interact. Named after Charles-Augustin de Coulomb, this fundamental principle in electromagnetism calculates the force between two point charges. The formula used is: \[ F = \frac{k \cdot |Q_1| \cdot |Q_2|}{r^2} \] where:
  • \( F \) is the electrostatic force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).
  • \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance between the centers of the two charges.
Coulomb's Law demonstrates that the force is inversely proportional to the square of the distance between the charges. This means that as you increase the distance between two charged objects, the force decreases rapidly. For our exercise, you calculate the tension in the string between the two boxes of protons by applying Coulomb's Law. Since the protons are identical in each box, \( Q_1 = Q_2 \). The charge of a single proton is about \( 1.6 \times 10^{-19} \text{ C} \), so for 1.0 g of protons, the total charge is significantly larger.
Gravitational Force
Gravitational force is one of the fundamental forces of nature. It describes the attraction between two masses. Isaac Newton described this force with his universal law of gravitation. The formula to compute gravitational force between two masses is: \[ F_g = \frac{G m_1 m_2}{r^2} \] Where:
  • \( F_g \) is the gravitational force.
  • \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \).
  • \( m_1 \) and \( m_2 \) are the masses of the objects.
  • \( r \) is the distance between the centers of the masses.
In the context of our exercise, the question asks whether the gravitational forces need to be considered. By calculating the gravitational force between the two boxes of protons, we find that due to their small mass (1.0 g each) and large separation (the distance between Earth and Moon), the gravitational force is negligible. This is in stark contrast to the much larger electrostatic force calculated using Coulomb's law.
Proton Charge
A proton is a subatomic particle found in the nucleus of every atom. It carries a positive electrical charge of about \( 1.6 \times 10^{-19} \text{ C} \). This charge is considered the fundamental unit of positive charge, and is equal in magnitude but opposite in sign to the charge of an electron. The concept of proton charge is crucial in electromagnetism and chemistry, as it defines how protons interact with electrons and other protons.In the given exercise, understanding the proton charge helps determine the total charge of each box. Since each box contains 1.0 g of protons, we first calculate the number of protons by considering the mass of a single proton, approximately \( 1.67 \times 10^{-24} \text{ g} \). Once the number of protons \( n \) is known, the total charge \( Q \) of each box is given by \( Q = n \times 1.6 \times 10^{-19} \text{ C} \). This allows us to apply Coulomb’s Law to find the electrostatic force and, consequently, the tension in the string between the two boxes.

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Most popular questions from this chapter

If a proton and an electron are released when they are 2.0 x 10\(^{-10}\) m apart (a typical atomic distance), find the initial acceleration of each particle.

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.

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.50 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.

Two very large horizontal sheets are 4.25 cm apart and carry equal but opposite uniform surface charge densities of magnitude \(\sigma\). You want to use these sheets to hold stationary in the region between them an oil droplet of mass 486 \(\mu\)g that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should \(\sigma\) be?

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