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Chapter 21: Problem 73

Imagine two \(1.0 \mathrm{~g}\) bags of protons, one at the earth's north pole and the other at the south pole. (a) How many protons are in each bag? (b) Calculate the gravitational attraction and the electric repulsion that each bag exerts on the other. (c) Are the forces in part (b) large enough for you to feel if you were holding one of the bags?

Short Answer

Expert verified
There are \(3.6 脳 10^{23}\) protons in each bag. The gravitational attraction and electric repulsion each bag exerts on the other are \(9.5 脳 10^{-11} N\) and \(1.1 脳 10^38 N\) respectively, which are too small and too large (respectively) to be perceptible by human senses.

Step by step solution

01

Calculate the number of protons

The number of protons in each bag can be calculated using Avogadro's number \(6.022 x 10^{23} \) and the mass of a proton \(1.67 x 10^{-24} g\). The number of protons \(N\) is given by the formula: \(N = \frac{mass}{proton mass}\), substituting the known values and calculating the result.
02

Calculate gravitational attraction

The gravitational attraction between two masses can be calculated using Newton's Law of Gravitation. According to it, the gravitational force \(F\) is given by \(F = \frac{Gm_1m_2}{r^2}\), where \(G\) is the Gravitational Constant, \(m_1\) and \(m_2\) are the masses and \(r\) is the distance between them. The Gravitational Constant is \(6.674 x 10^{-11} m^3kg^{-1}s^{-2}\), the total mass of protons calculated in step 1 is the mass and the distance between the poles is approximately \(12,742 km\). Substituting these values, we can find the force.
03

Calculate electrical repulsion

The electric repulsion between two charges can be calculated using Coulomb's Law. According to it, the electric force \(F\) is given by \(F = \frac{kq_1q_2}{r^2}\), where \(k\) is the Coulomb's constant, \(q_1\) and \(q_2\) are the charges and \(r\) is the distance between them. The Coulomb's constant is \(8.99 x 10^9 Nm^2C^{-2}\), the charge is the product of proton's charge and number of protons and the distance is the same as in step 2. Using these values we can calculate the electric force.
04

Comparing Forces

Now, compare the electric force and gravitational force calculated in previous steps to determine if they are perceptible to human senses. An average human can perceive forces of about \(1N\).

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Key Concepts

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

Number of Protons
Protons are positively charged particles in an atom's nucleus. In the exercise, determining how many protons fit into a 1-gram bag involves understanding their mass. Each proton has a minuscule mass of approximately \(1.67 \times 10^{-24} \, \text{g}\).
By calculating the total number of protons in 1 gram, we use the formula:
  • \(N = \frac{\text{mass of protons in grams}}{\text{mass of one proton}} = \frac{1.0 \, \text{g}}{1.67 \times 10^{-24} \, \text{g}}\)
Substituting these values gives the number of protons, which turns out to be a gigantic number due to their tiny size. Understanding this quantity helps in calculating subsequent gravitational and electric forces.
Newton's Law of Gravitation
Isaac Newton's Law of Gravitation is crucial for understanding how masses attract each other. It posits that every particle attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically, it is expressed as:
  • \(F = \frac{G m_1 m_2}{r^2}\)
where:
  • \(F\) is the gravitational force between the bodies.
  • \(G\) is the gravitational constant, \(6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2}\).
  • \(m_1\) and \(m_2\) are the masses of the two bodies.
  • \(r\) is the distance between the centers of the two masses.
The exercise asks to compute this force between two bags of protons positioned at the Earth's poles, requiring conversion of units and distances. Understanding Newton's approach helps in visualizing cosmic and everyday forces.
Coulomb's Law
Coulomb's Law is the electric force equivalent to Newton's gravitational law, but for charged particles. It states that the electric force between two charges is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between their centers.
The formula is:
  • \(F = \frac{k q_1 q_2}{r^2}\)
where:
  • \(F\) is the electric force between the charges.
  • \(k\) is Coulomb's constant, \(8.99 \times 10^9 \, \text{N} \text{m}^2 \text{C}^{-2}\).
  • \(q_1\) and \(q_2\) are the magnitudes of the charges.
  • \(r\) is the separation distance.
The electric force calculation, like in the problem, involves huge charges due to the large number of protons. This shows the overwhelming repulsive force generated when like charges are amassed.
Gravitational Constant
The gravitational constant \(G\) is a fundamental constant in physics, crucial for calculating gravitational forces.
It is approximately \(6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2}\). This tiny value reflects how weak gravity is compared to other fundamental forces.
In calculations like those in the exercise, this constant helps quantify the gravitational interaction between objects such as proton bags. Despite the large number of protons, the gravitational force is almost negligible due to the tiny value of \(G\). This illustrates why gravity is perceived mainly in interactions involving massive objects, like planets.
Coulomb's Constant
Coulomb's constant \(k\) is key in calculating electromagnetic forces between charged particles.
Its value is \(8.99 \times 10^9 \, \text{N} \text{m}^2 \text{C}^{-2}\), much larger compared to the gravitational constant \(G\). This huge value makes electric forces much more significant than gravitational forces for small charges and distances.
The exercise highlights this contrast, showing that although gravity holds planets in orbit, in atomic interactions, electric forces dominate, being vastly stronger for separated protons in the bags. Understanding \(k\) helps in grasping the influence of charges as opposed to masses.

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Most popular questions from this chapter

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{C}\) is placed at the point \(x=+0.150 \mathrm{~m}, y=0,\) and an identical point charge is placed at \(x=-0.150 \mathrm{~m}, y=0 .\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{~m}, y=0\) (c) \(x=0.150 \mathrm{~m}, y=-0.400 \mathrm{~m}\) (d) \(x=0, y=0.200 \mathrm{~m}\)

Two \(1.20 \mathrm{~m}\) nonconducting rods meet at a right angle. One rod carries \(+2.50 \mu \mathrm{C}\) of charge distributed uniformly along its length, and the other carries \(-2.50 \mu \mathrm{C}\) distributed uniformly along it (Fig. \(\mathbf{P} 2 \mathbf{1 . 8 5}\) ). (a) Find the magnitude and direction of the electric field these rods produce at point \(P,\) which is \(60.0 \mathrm{~cm}\) from each rod. (b) If an electron is released at \(P\), what are the magnitude and direction of the net force that these rods exert on it?

A proton is traveling horizontally to the right at \(4.50 \times 10^{6} \mathrm{~m} / \mathrm{s}\). (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of \(3.20 \mathrm{~cm}\). (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Negative charge \(-Q\) is distributed uniformly around a quarter-circle of radius \(a\) that lies in the first quadrant, with the center of curvature at the origin. Find the \(x\) - and \(y\) -components of the net electric field at the origin.

Point charge \(A\) is on the \(x\) -axis at \(x=-3.00 \mathrm{~cm}\). At \(x=1.00 \mathrm{~cm}\) on the \(x\) -axis its electric field is \(2700 \mathrm{~N} / \mathrm{C}\). Point charge \(B\) is also on the \(x\) -axis, at \(x=5.00 \mathrm{~cm}\). The absolute magnitude of charge \(B\) is twice that of \(A .\) Find the magnitude and direction of the total electric field at the origin if (a) both \(A\) and \(B\) are positive; (b) both are negative; (c) \(A\) is positive and \(B\) is negative; (d) \(A\) is negative and \(B\) is positive.
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