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Chapter 1: Problem 16

Use the fact that the electrostatic force acting between two electric charges, \(q_{1}\) and \(q_{2}\), is given by \(q_{1} q_{2} / 4 \pi \epsilon_{0} r^{2}\), where \(r\) is the distance between the charges, to determine the dimensions of \(\epsilon_{0}\), the permittivity of a vacuum. In the table of constants (Appendix E), \(\epsilon_{0}\) is expressed in units of farads per metre. Express one farad in terms of the SI base units.

Short Answer

Expert verified
The dimensions of the permittivity of a vacuum (\(\epsilon_{0}\)) are [A^2 s^4 / kg m^3] and one farad can be expressed as [A^2 s^4 / kg m^2] in terms of the SI base units.

Step by step solution

01

The Dimensions of the Permittivity of a Vacuum (\(\epsilon_{0}\))

The permittivity of a vacuum (\(\epsilon_{0}\)) is a constant in Coulomb's Law, which is \(F = q_{1} q_{2} / 4 \pi \epsilon_{0} r^{2}\) where \(F\) is the force, \(q_{1}\) and \(q_{2}\) are the charges, and \(r\) is the distance between the charges. By rearranging the equation to isolate \(\epsilon_{0}\), we can determine its dimensions. The rearranged equation is \(\epsilon_{0} = q_{1} q_{2} / 4 \pi F r^{2}\). The dimensions of each variable are: force (\(F\)) is in Newtons [kg m / s^2], charge is in Coulombs [A s], and distance (\(r\)) is in metres [m]. Therefore, the dimensions of \(\epsilon_{0}\) are [A^2 s^4 / kg m^3].
02

Express One Farad in Terms of the SI Base Units

A farad is the SI unit of electrical capacitance, which measures how much electric charge is stored at a given electric potential. It can be expressed as [A^2 s^4 / kg m^2] in terms of the SI base units, by using the definition of a coulomb (charge) which is [A s]. Therefore, we have [A^2 s^4 / kg m^2] as the SI base unit representation of one farad.

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Key Concepts

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

Electrostatic Force
Electrostatic force is the interaction between static electrically charged particles. In simple terms, it's the push or pull that occurs between objects with charge. This force can either attract or repel charges, depending on their nature: like charges repel each other, whereas opposite charges attract.

The strength of the electrostatic force is significant because it directly affects how charges move or stay in place. This concept is fundamental in understanding how atoms form bonds, how objects become electrically charged through friction, and how electrical circuits work. Moreover, this force plays a critical role in various applications, from everyday static cling to the intricate workings of electronic devices.
Permittivity of a Vacuum
The permittivity of a vacuum, commonly denoted by the symbol \(\epsilon_0\), is a physical constant that describes how an electric field affects and is affected by a vacuum. It's an essential parameter that sets the strength of the electrostatic interaction in a vacuum. The permittivity of a vacuum is unique because it has the same value throughout the universe and does not change with other physical conditions like temperature or pressure.

When we talk about the dimensions of \(\epsilon_0\), we are typically interested in its units. Its dimensions are [A\(^2\) s\(^4\) / kg m\(^3\)]. This means that the permittivity value tells us how much electric field (\(E\)) can 'pass through' or be 'permittable' through a vacuum when there is an electric potential (\(V\)) applied across a certain distance (\(d\)). Understanding this concept is crucial for students studying physics or engineering because it helps predict how electric fields interact in different mediums.
Coulomb's Law
Coulomb's Law is a mathematical equation that calculates the electrostatic force (\(F\)) between two point charges. It was named after Charles-Augustin de Coulomb, who first published the law in the 18th century. The law states that the electrostatic force is directly proportional to the product of the magnitudes of the two charges (\(q_1\) and \(q_2\)) and inversely proportional to the square of the distance (\(r\)) between them.

Mathematical Expression

Coulomb's Law is expressed mathematically as: \[ F = \frac{q_1 q_2}{4 \pi \epsilon_0 r^2}\] This equation tells us that the force decreases swiftly as the distance between charges increases. Likewise, if either of the charges is increased, the force between them will also increase. Coulomb's Law is a central concept in electrostatics, providing the backbone for understanding electric forces and fields. This equation is also instrumental in the derivation of many other important electrodynamics formulas and applications, including the study of electric field lines and the behavior of electric charges in various materials.

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Most popular questions from this chapter

Use a binomial expansion to evaluate \(1 / \sqrt{4.2}\) to five places of decimals, and compare it with a more accurate answer obtained using a calculator.

It is shown in the text that the exponential function \(\exp (x)\) is identical in value to the power \(e^{x}\) for all \(x\). It therefore follows from the relation \(e^{2 x}=e^{x} \times e^{x}\) that $$ \exp (2 x)=[\exp (x)]^{2} $$ Write both sides of this equation in terms of the relevant series and, by considering \(\exp (x)\) as \(1+p(x)\), verify (*) term-by-term up to and including the cubic term in \(x\).

The following is a student's proposed formula for the energy flux \(S\) (the magnitude of the so-called Poynting vector) associated with an electromagnetic wave in a vacuum, the electric field strength of the wave being \(E\) and the associated magnetic flux density being \(B\) : $$ S=\frac{1}{2}\left[\left(\frac{\epsilon_{0}}{\mu_{0}}\right)^{1 / 2} E^{2}+\left(\frac{\mu_{0}}{\epsilon_{0}}\right)^{1 / 2} B^{2}\right] $$ The dimensions of \(\epsilon_{0}\), the permittivity of free space, are \(M^{-1} L^{-3} T^{4} I^{2}\), and those of its permeability \(\mu_{0}\) are \(M L T^{-2} I^{-2}\). Given, further, that the force acting on a rod of length \(\ell\) that carries a current \(I\) at right angles to a field of magnetic flux density \(B\) is \(B I \ell\), determine whether the student's formula could be correct and, if not, locate the error as closely as possible.

Given that \(a>b>0\), prove algebraically that $$ \frac{a}{b}>\frac{a+c}{b+c} $$ whenever \(c>0\) and for \(c<0\) when \(|c|>b\), but that for \(c<0\) with \(|c|\) sign should be replaced by a \(<\) sign. [Note: It may help to illustrate these results graphically on an annotated sketch of \((a+c) /(b+c)\) as a function of \(c\), for fixed \(a\) and \(b\). For definiteness the values \(a=4, b=3\) could be used.]

Find the number for which the cube of its square root is equal to twice the square of its cube root.
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