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Chapter 31: Problem 1

When \(1 \cdot 0 \times 10^{12}\) electrons are transferred from one conductor to another, a potential difference of \(10 \mathrm{~V}\) appears between the conductors. Calculate the capacitance of the two-conductor system.

Short Answer

Expert verified
The capacitance is \(1.6 \times 10^{-8} \text{ F}\).

Step by step solution

01

Understanding the Charge Transferred

To find the capacitance, we first need to compute the total charge transferred. Given that one electron has a charge of approximately \(-1.6 \times 10^{-19} \text{ C}\), and \(1.0 \times 10^{12}\) electrons are involved, we calculate the charge \(Q\) using the formula \(Q = n \cdot e\), where \(n\) is the number of electrons and \(e\) is the charge of an electron.
02

Calculating Total Charge

Using the formula \(Q = n \cdot e\), substitute \(n = 1.0 \times 10^{12}\) and \(e = 1.6 \times 10^{-19}\). This gives us: \[Q = 1.0 \times 10^{12} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-7} \text{ C} \]
03

Using the Capacitance Formula

Capacitance \(C\) is determined by the formula \(C = \frac{Q}{V}\), where \(V\) is the potential difference. Here, \(V = 10 \text{ V}\).
04

Calculating Capacitance

Substitute \(Q = 1.6 \times 10^{-7} \text{ C}\) and \(V = 10 \text{ V}\) into the capacitance formula: \[ C = \frac{1.6 \times 10^{-7}}{10} = 1.6 \times 10^{-8} \text{ F} \] The capacitance of the two-conductor system is \(1.6 \times 10^{-8} \text{ F}\).

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Key Concepts

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

Electrons
Electrons are fundamental particles with a negative electric charge. They play a crucial role in electricity and circuits. In this exercise, the movement of electrons is key to understanding the concept of capacitance. Each electron carries a charge of approximately \(-1.6 \times 10^{-19}\) Coulombs (C). When a certain number of electrons move from one conductor to another, they create an electric charge. This movement is what we refer to when talking about electron transfer.
  • Electrons are the primary charge carriers in many electrical phenomena.
  • The number of electrons transferred affects the total charge generated.
  • Understanding the individual charge per electron helps in calculating overall effects in circuits.
When considering any electrical system, it is important to remember the negative charge of electrons can contribute to creating a potential difference between two points. This aspect is crucial in the study and application of capacitance in conductors.
Charge Transfer
Charge transfer is the movement of electric charge from one place or object to another. It represents a fundamental concept in physics and electrical engineering. In the given problem, the transfer of electrons results in the accumulation of charge on the conductors. This charge accumulation is vital for capacitance. Capacitance is the ability of a system to store an electric charge, and it is directly related to how much charge is transferred and stored.
  • Charge is often measured in Coulombs.
  • Knowing the quantity of electrons helps to determine the total charge.
  • A stable charge difference between two conductors leads to a stable potential difference.
The formula for total charge transferred \(Q = n \cdot e\) is essential here; where \(n\) is the number of electrons and \(e\) is the charge of each electron. This formula helps to calculate the exact charge that contributes to the potential difference.
Potential Difference Calculation
The potential difference, also known as voltage, between two points can be thought of as the "pressure" that drives the charge around a circuit. In this exercise, we deal with a potential difference of 10 volts between the two conductors. This difference is caused by the transfer of electrons.
The potential difference \(V\) plays a key role in calculating capacitance \(C\), which is done using the formula \(C = \frac{Q}{V}\). This tells us how effectively a system can store charge per unit voltage.
  • The greater the charge transferred, the higher the potential difference for a given capacitance.
  • This calculation helps assess how efficiently a conductor can store and maintain electrical energy.
  • Potential difference is a crucial parameter for understanding energy storage in capacitors.
By using the capacitance formula, we find that the capacitance of the given system is \(1.6 \times 10^{-8}\) Farads. This illustrates how well the system can store charge for the given electron transfer and potential difference.

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Most popular questions from this chapter

Two conducting spheres of radii \(R_{1}\) and \(R_{2}\) are kept widely separated from each other. What are their individual capacitances ? If the spheres are connected by a metal wire, what will be the capacitance of the combination ? Think in terms of series-parallel connections.

A capacitor having a capacitance of \(100 \mu \mathrm{F}\) is charged to a potential difference of \(50 \mathrm{~V}\). (a) What is the magnitude of the charge on each plate ? (b) The charging battery is disconnected and a dielectric of dielectric constant \(2 \cdot 5\) is inserted. Calculate the new potential difference between the plates. (c) What charge would have produced this potential difference in absence of the dielectric slab. (d) Find the charge induced at a surface of the dielectric slab.

Figure (31-E26) shows two identical parallel plate capacitors connected to a battery through a switch \(S\). Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3 . Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

The separation between the plates of a parallel-plate capacitor is \(0.500 \mathrm{~cm}\) and its plate area is \(100 \mathrm{~cm}^{2}\). A \(0.400 \mathrm{~cm}\) thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.

A capacitor of capacitance \(5 \cdot 00 \mu \mathrm{F}\) is charged to \(24 \cdot 0 \mathrm{~V}\) and another capacitor of capacitance \(6 \cdot 0 \mu \mathrm{F}\) is charged to \(12 \cdot 0 \mathrm{~V}\). (a) Find the energy stored in each capacitor. (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the new charges on the capacitors. (c) Find the loss of electrostatic energy during the process. (d) Where does this energy go ?
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