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Chapter 19: Problem 53

A capacitor has a capacitance of \(2.5 \times 10^{-8} \mathrm{~F}\). In the charging process, electrons are removed from one plate and placed on the other plate. When the potential difference between the plates is \(450 \mathrm{~V}\), how many electrons have been transferred?

Short Answer

Expert verified
Approximately \( 7.03 \times 10^{13} \) electrons have been transferred.

Step by step solution

01

Understand the Relation Between Charge and Capacitance

The relationship between the charge \( Q \) on the plates of a capacitor, the capacitance \( C \), and the potential difference \( V \) is given by the formula \( Q = C \times V \). We need to calculate \( Q \), which represents the charge movement due to electron transfer.
02

Calculate the Charge (Q)

Substitute the given capacitance \( C = 2.5 \times 10^{-8} \mathrm{~F} \) and the potential difference \( V = 450 \mathrm{~V} \) into the equation \( Q = C \times V \):\[ Q = (2.5 \times 10^{-8} \mathrm{~F}) \times (450 \mathrm{~V}) = 1.125 \times 10^{-5} \mathrm{~C} \].This value of \( Q \) represents the total charge transferred due to electrons.
03

Convert Charge to Number of Electrons

The charge of one electron is approximately \( 1.6 \times 10^{-19} \mathrm{~C} \). To find the number of electrons \( n \) moved, use the formula \( n = \frac{Q}{e} \), where \( e \) is the elementary charge:\[ n = \frac{1.125 \times 10^{-5} \mathrm{~C}}{1.6 \times 10^{-19} \mathrm{~C/electron}} \approx 7.03125 \times 10^{13} \].

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Key Concepts

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

Charge
In electronics, charge represents the quantity of electricity held or transported by a object or just a single particle. In the context of a capacitor, charge is the result of either transferring electrons to or from its plates. A capacitor stores this charge for various uses in electrical circuits. It does so by utilizing two conductive plates and accumulating charge on each plate during the charging process.The total amount of charge stored in a capacitor, denoted by the symbol \( Q \), is determined by its capacitance \( C \) and the potential difference \( V \) applied across its plates. The relationship is represented by the formula: \( Q = C \times V \).
  • Capacitance \( C \) refers to the capacity of the capacitor to store charge. It's determined by factors such as plate size, plate separation, and dielectric material between the plates.
  • The potential difference \( V \) refers to the voltage applied, which influences how much charge the capacitor will store.
No charge flow happens without potential difference, and the greater the potential difference, the more charge can be moved. The charge stored in a capacitor can be utilized in discharge processes to perform work, like powering electronic components momentarily.
Potential Difference
Potential difference, often referred to as voltage, is key in driving charge movement and storage in capacitors. It expresses the amount of work needed to move a charge between two points in an electric field.
The voltage applied across the plates of a capacitor incites electron flow, which moves electrons from one plate to another, eventually storing energy.
  • Higher potential differences result in a greater amount of charge stored in a capacitor. This occurs because the electric field between the plates enhances as voltage rises, thus attracting more electrons to move across the field.
  • Potential difference is measured in volts (\( V \)), and it dictates how energetically charges are pushed across the plates.
Potential difference plays a vital role in calculating the charge in capacitors, as seen in the provided formula \( Q = C \times V \). Understanding its role helps predict and control how capacitors function within electrical circuits.
Electron Transfer
Electron transfer is the essential mechanism by which capacitors store energy. When a voltage is applied, electrons are removed from one plate and transferred to the opposite plate. This movement creates a disparity in electron concentration, which stores energy as an electric field between the plates.
The amount of electron transfer is quantified by the charge moved, denoted by \( Q \) in the equation \( Q = C \times V \).
  • The electron charge is a fundamental unit, with each electron carrying a charge of approximately \( 1.6 \times 10^{-19} \; \text{C} \).
  • The number of electrons transferred is calculated by dividing the total charge \( Q \) by the charge of a single electron \( e \), using the formula \( n = \frac{Q}{e} \).
This aspect of a capacitor's charging process highlights the microscopic interactions essential for electrical operation, translating theoretical concepts into real-world applications by measuring the basic units of electron flow.

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Most popular questions from this chapter

Suppose that the electric potential outside a living cell is higher than that inside the cell by \(0.070 \mathrm{~V}\). How much work is done by the electric force when a sodium ion (charge \(=+e\) ) moves from the outside to the inside?

An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed \(v_{\mathrm{e}}\), while the proton acquires a speed \(v_{\mathrm{p}}\). (a) As each particle accelerates from rest, it gains kinetic energy. Does it gain or lose electric potential energy? (b) Does the electron gain more, less, or the same amount of kinetic energy as the proton does? (c) Is \(v_{\mathrm{e}}\) greater than, less than, or equal to \(v_{\mathrm{p}}\) ? Justify your answers. Find the ratio \(v_{\mathrm{e}} / v_{\mathrm{p}}\). Verify that your answer is consistent with your answers to the Concept Questions.

A positive point charge is surrounded by an equipotential surface \(A\), which has a radius of \(r_{A}\). A positive test charge moves from surface \(A\) to another equipotential surface \(B\), which has a radius of \(r_{B}\). In the process, the electric force does negative work. (a) Does the electric force acting on the test charge have the same or opposite direction as the displacement of the test charge? (b) Is \(r_{B}\) greater than or less than \(r_{A}\) ? Explain your answers. Problem The positive point charge is \(q=+7.2 \times 10^{-8} \mathrm{C}\), and the test charge is \(q_{0}=+4.5 \times 10^{-11} \mathrm{C}\). The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9} \mathrm{~J}\). The radius of surface \(A\) is \(r_{A}=1.8 \mathrm{~m}\). Find \(r_{B}\). Check to see that your answer is consistent with your answers to the Concept Questions.

Multiple-Concept Example 4 provides useful background for this problem. Point \(\mathrm{A}\) is at a potential of \(+250 \mathrm{~V}\), and point \(B\) is at a potential of \(-150 \mathrm{~V}\). An \(\alpha\) -particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An \(\alpha\) -particle starts from rest at \(A\) and accelerates toward \(B\). When the \(\alpha\) -particle arrives at \(B\), what kinetic energy (in electron volts) does it have?

The potential at location \(A\) is \(452 \mathrm{~V}\). A positively charged particle is released there from rest and arrives at location \(B\) with a speed \(v_{B}\). The potential at location \(C\) is \(791 \mathrm{~V},\) and when released from rest from this spot, the particle arrives at \(B\) with twice the speed it previously had, or \(2 v_{B} .\) Find the potential at \(B\).
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