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Force, and potential squared * (a) In Gaussian units, show that the square of a potential difference \(\left(\phi_{2}-\phi_{1}\right)^{2}\) has the same dimensions as force. (In SI units, \(\epsilon_{0}\left(\phi_{2}-\phi_{1}\right)^{2}\) has the same units as force.) This tells us that the electrostatic forces between bodies will largely be determined, as to order of magnitude, by the potential differences involved. Dimensions will enter only in ratios, and there may be some constants like \(4 \pi\). What is the order of magnitude of force youexpect with 1 statvolt potential difference between something and something else? Practically achievable potential differences are rather severely limited, for reasons having to do with the structure of matter. The highest man-made difference of electric potential is about \(10^{7}\) volts, achieved by a Van de Graaff electrostatic generator operating under high pressure. (Billion-volt accelerators do not involve potential differences that large.) How many pounds force are you likely to find associated with a "square megavolt"? These considerations may suggest why electrostatic motors have not found much application.

Step by step solution

01

Compare the dimensions of potential difference and force

The physical dimension of force in Gaussian units is M \(L/T^2\) (Mass \times Length/Time Squared). Meanwhile the dimensions of potential difference (or electric potential) is M \(L^3/QT^2\) (Mass \times Length Cubed/Charge x Time Squared). Squaring the potential difference, we get M \(L^6/Q^2T^4\). Upon dividing the squared potential difference by \(L/Q\), we get M \(L/T^2\), which is the same dimensions as force. Thus, \(\left(\phi_{2}-\phi_{1}\right)^{2}\) and force share the same dimensions.

02

Calculate order of magnitude for 1 statvolt

The force between two objects due to electrostatic interactions can be given by the formula \(F=k\left(\phi_{2}-\phi_{1}\right)^{2}\), where \(F\) is the force, \(k\) is Coulomb's constant (\(2 \times 10^7 N m^2/C^2\) in SI units), and \(\left(\phi_{2}-\phi_{1}\right)\) is the potential difference. For 1 statvolt of potential difference (\(\phi_{2} - \phi_{1}\) = 1), we find that the force \(F=2 \times 10^7 N\). This is the magnitude of force we expect for a difference of 1 statvolt between the charges on two objects.

03

Calculate force for 'square megavolt'

The highest man-made difference of electric potential or 'square megavolt' is about \(10^{7}\) volts. Plugging this into our formula for \(F\), we get \(F=k\left(10^{7}\right)^{2} = 2 \times 10^{7} N \times 10^{14} = 2 \times 10^{21} N\). In pounds, this equals \(4.45 \times 10^{20}\) pounds-force. This is a huge force which explains why electrostatic motors are not commonly used.

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Key Concepts

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

Potential Difference

Understanding the potential difference is crucial when studying electrostatics. Potential difference, often referred to as voltage, measures the work done per unit charge as a charge moves between two points in an electric field. Picturing it is easy if one thinks of water in a pipe. When water flows from a high place to a lower one, it does so because of a difference in height—the same goes for electric charges.

To calculate this in the context of electrostatic force, one must look at the energy difference caused by the charges. If one considers two points, \(\phi_{2}\) and \(\phi_{1}\), the potential difference is given by \(\phi_{2} - \phi_{1}\). It's this difference that drives the flow of charge and creates an electrostatic force which can do work. This is why, as seen in the exercise, the square of the potential difference, \(\left(\phi_{2} - \phi_{1}\right)^2\), has dimensions that are equivalent to force—because this squared value relates to the energy that can be exerted on a charged particle, and force is a measure of energy transfer per distance moved.

Electric Potential Units

The Units of Potential

In the realm of electric potential, we deal with units that quantify the energy per charge. In the International System (SI) of units, the basic unit of electric potential is the volt (V), named after Alessandro Volta. One volt is equivalent to one joule per coulomb \(\mathrm{J/C}\).

When looking at calculations regarding electric potential, you might find units like 'statvolt' used in the Gaussian system, which is another way to express electric potential. However, it's essential to know how to convert between different systems. For instance, in SI units, electrostatic force involves using the permittivity of free space \(\epsilon_{0}\), which doesn't appear in the Gaussian system. Converting between these systems usually requires knowing some fundamental constants that relate the two systems.

Gaussian Units

In science, different systems of units are used for various fields. Gaussian units are one such system, which are part of the broader 'cgs' (centimeter-gram-second) system. In cgs units, the centimeter, gram, and second are the fundamental units for length, mass, and time, respectively.

What distinguishes Gaussian units in electrostatics is the way charges and electric fields are quantified. For instance, the electrostatic force in Gaussian units is not prefaced with the permittivity of free space \(\epsilon_0\), as it is with SI units. Instead, Coulomb's law is expressed by a simpler form, which can be more intuitive. However, this also means that when dealing with Gaussian units, one must be able to bridge the understanding to SI units especially in an educational setting where SI units are predominantly used. This involves recognizing equivalencies and conversion factors that allow one to translate measurements and equations between the two. The exercise given demonstrates this need for equivalencies, as it shows that the square of a potential difference corresponds dimensionally with force in both Gaussian and SI units.