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Chapter 18 Electric Charge and Electric Field (126/148) -- College Physics

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Chapter 18 Electric Charge and Electric Field

Chapter 18 Electric Charge and Electric Field 18.3 Coulomb’s Law Summary - State Coulomb’s law in terms of how the electrostatic force changes with the distance between two objects. - Calculate the electrostatic force between two charged point forces, such as electrons or protons. - Compare the electrostatic force to the gravitational attraction for a proton and an electron; for a human and the Earth. Through the work of scientists in the late 18th century, the main features of the electrostatic force—the existence of two types of charge, the observation that like charges repel, unlike charges attract, and the decrease of force with distance—were eventually refined, and expressed as a mathematical formula. The mathematical formula for the electrostatic force is called Coulomb’s law after the French physicist Charles Coulomb (1736–1806), who performed experiments and first proposed a formula to calculate it. Coulomb’s Law Coulomb’s law calculates the magnitude of the force FF between two point charges, [latex]{q_1}[/latex] and [latex]{q_2}[/latex], separated by a distance [latex]{r}[/latex]. In SI units, the constant [latex]{k}[/latex] is equal to The electrostatic force is a vector quantity and is expressed in units of newtons. The force is understood to be along the line joining the two charges. (See Figure 2.) Although the formula for Coulomb’s law is simple, it was no mean task to prove it. The experiments Coulomb did, with the primitive equipment then available, were difficult. Modern experiments have verified Coulomb’s law to great precision. For example, it has been shown that the force is inversely proportional to distance between two objects squared [latex]{(F \propto 1/r^2)}[/latex] to an accuracy of 1 part in [latex]{10^{16}}[/latex]. How Strong is the Coulomb Force Relative to the Gravitational Force? Compare the electrostatic force between an electron and proton separated by [latex]{0.530 \times 10^{-10} \;\text{m}}[/latex] with the gravitational force between them. This distance is their average separation in a hydrogen atom. Strategy To compare the two forces, we first compute the electrostatic force using Coulomb’s law, [latex]{F = k}[/latex] [latex]{\frac{|q_1 q_2|}{r2}}[/latex]. We then calculate the gravitational force using Newton’s universal law of gravitation. Finally, we take a ratio to see how the forces compare in magnitude. Solution Entering the given and known information about the charges and separation of the electron and proton into the expression of Coulomb’s law yields Thus the Coulomb force is The charges are opposite in sign, so this is an attractive force. This is a very large force for an electron—it would cause an acceleration of [latex]{8.99 \times 10^{22} \;\text{m/s}^2 }[/latex] (verification is left as an end-of-section problem).The gravitational force is given by Newton’s law of gravitation as: where [latex]{G = 6.67 \times 10^{-11} \;\text{N} \cdot \text{m}^2 / \text{kg}^2}[/latex]. Here [latex]{m}[/late
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