Question

This problem investigates what fraction of the available chayge must he tranferred from one conductor to another to produre a typical contact potential. (a) As a rough appnximation, treat the conductors as \(10 \mathrm{~cm} \times\) \(10 \mathrm{c} \mathrm{m}\) square plates \(2 \mathrm{~cm}\) apart - a parallel-plate capactor \(-\) so that \(q=C V\), where \(C=\varepsilon_{\mathrm{p}}\left(0.01 \mathrm{~m}^{2} / 0.02 \mathrm{~m}\right)\). How much charge must be iransferred from one plate to the other to produce a potential difference of \(2 \mathrm{~V}\) ? (b) Approximately what fruction would this be of the total number of conduction electrons in a \(100 \mathrm{~g}\) piece of copper, which has one conduction electron per atem?

Short answer

The fraction of charge transferred from the total number of conduction electrons is found after calculating the charge transferred to produce 2V potential difference and the total number of conduction electrons in a 100g piece of copper.

Step-by-step solution

Calculate the transferred charge

Using the formula \(q = CV\), where C is the capacitance and V is the potential difference. We know that the capacitance \(C=\varepsilon_{p}(0.01 m^{2} / 0.02 m)\) and the potential difference V is 2 volts. Substituting the known values into the formula, the charge can be calculated.

Estimate the total number of conduction electrons in copper

The total number of conduction electrons in a 100g piece of copper can be estimated with the given fact that copper has one conduction electron per atom. We can use the atomic mass of copper and Avogadro's number to calculate the total number of conduction electrons.

Calculate the fraction

After getting the transferred charge (in terms of electrons) and the total number of conduction electrons, we can find the required fraction by dividing the former by the latter.

Key concepts

Parallel-Plate Capacitor

A parallel-plate capacitor is one of the simplest and most commonly used capacitor configurations. It consists of two conductive plates separated by an insulating material known as the dielectric. When a voltage is applied across the plates, an electric field develops between them, inducing positive charge on one plate and an equivalent negative charge on the other. This results in an electrical potential difference between the plates.

The ability of a capacitor to store charge is quantified by its capacitance, represented by the symbol 'C'. The capacitance of a parallel-plate capacitor can be calculated using the formula: \[C = \varepsilon \times \frac{A}{d}\], where \(\varepsilon\) is the permittivity of the dielectric, 'A' is the area of one plate, and 'd' is the separation between the plates. In the context of our exercise, the plates are approximated to have an area of \(0.01 m^2\) each and are \(0.02 m\) apart. The permittivity \(\varepsilon_p\) represents the permittivity of the free space or any other medium between the plates.

When discussing a contact potential between two conductors, a parallel-plate capacitor is an idealized model that helps visualize how charges distribute themselves and the resulting potential difference due to this distribution.

Capacitance and Charge Relationship

The relationship between capacitance and charge in a capacitor is key to understanding its function. Capacitance is defined as the amount of electric charge stored per unit of potential difference across its plates. The formula that represents this relationship is \[q = C V\], where 'q' is the charge, 'C' is the capacitance, and 'V' is the potential difference (also known as voltage).

In the original exercise, a potential difference of 2 volts implies that the voltage 'V' in our equation is 2. If we apply this formula, knowing the capacitance 'C' found from the previous section, we can calculate the exact amount of charge 'q' that must be transferred to achieve this potential difference.

Understanding this relationship is crucial when estimating how much charge is required to produce a specific voltage, which directly impacts the operation of electronic devices. In practical terms, this could determine how long a device can operate before needing to recharge, as the available charge dictates the energy stored.

Conduction Electrons in Copper

Copper is a highly conductive material, primarily owing to its conduction electrons, which are the electrons in the outermost shell of copper atoms that are free to move within the material. These electrons allow for the flow of electric current when a potential difference is applied.

In a solid piece of copper, nearly every atom contributes one free electron that becomes a conduction electron. Considering a 100g piece of copper, we can use the atomic mass of copper (approximately 63.55 grams per mole) and Avogadro's number (\(6.022 \times 10^{23}\) atoms per mole) to estimate the total number of conduction electrons. This exercise illustrates the basic principle that the conductivity of a metal is directly related to the number of free electrons available to transport charge.

In the step-by-step solution, a fraction is calculated by comparing the charge transferred to a parallel-plate capacitor (representing a contact potential scenario) to the total number of conduction electrons in the piece of copper. This comparison offers insight into the scale of electric phenomena, from the transfer of a relatively small number of electrons to the vast number available in a conductor. Such an understanding is fundamental in fields such as electronics and materials science where the manipulation of charges is central to the operation of devices and the development of new materials.