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An electron gun in a TV set shoots out a beam of electrons. The beam current is \(1.0 \times 10^{-5} \mathrm{~A}\). How many electrons strike the TV screen each second? How much charge strikes the screen in a minute?

Short Answer

Expert verified
6.25 脳 10鹿鲁 electrons strike per second; 6.0 脳 10鈦烩伌 C strikes per minute.

Step by step solution

01

Understand the relationship between current and charge

The current, given in amperes, relates to the flow of charge over time. Specifically, current \(I\) is the charge \(Q\) flowing per unit time \(t\), described by the equation \(I = \frac{Q}{t}\). We are given a current \(I = 1.0 \times 10^{-5} \text{ A}\).
02

Calculate the number of electrons striking per second

First, find the charge striking the screen in one second. For that, we use the relationship \(Q = I \times t\). In one second, \(t = 1 \text{ s}\), so \(Q = 1.0 \times 10^{-5} \text{ C}\). Next, since one electron has a charge of \(1.6 \times 10^{-19} \text{ C}\), the number of electrons \(n\) is \(n = \frac{Q}{e} = \frac{1.0 \times 10^{-5}}{1.6 \times 10^{-19}}\). This simplifies to \(n = 6.25 \times 10^{13}\).
03

Calculate the charge striking the screen in a minute

We need to determine the charge over 60 seconds. Use \(Q = I \times t\) once more, with \(t = 60 \text{ s}\). Thus, \(Q = 1.0 \times 10^{-5} \text{ C/s} \times 60 \text{ s} = 6.0 \times 10^{-4} \text{ C}\).

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Key Concepts

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

Charge Calculation
Calculating charge in any scenario involves understanding the fundamental relationship between charge, current, and time. The key equation for charge calculation is \( Q = I \times t \), where \( Q \) represents the charge in coulombs, \( I \) is the current in amperes, and \( t \) is the time in seconds. This formula stems from the definition of current as the rate of charge flow.

When given a current of \(1.0 \times 10^{-5} \) amperes, you can determine the amount of charge flowing in a certain time duration. For instance, over a span of one second, the charge is:
  • \( Q = 1.0 \times 10^{-5} \times 1 \)
  • \( Q = 1.0 \times 10^{-5} \text{ C} \)
In this example, the aim is to find how much charge reaches the screen in a longer period, such as a minute. Using the same principle:
  • \( Q = 1.0 \times 10^{-5} \times 60 \)
  • \( Q = 6.0 \times 10^{-4} \text{ C} \)
This calculation shows the scalability of charge determination over different time frames.
Electron Flow
Electron flow is the cornerstone of understanding current in electronic devices. An electric current is essentially the flow of electrons, tiny particles carrying a negative charge. Each electron carries a charge of \( -1.6 \times 10^{-19} \text{ C} \). This small charge amount allows for billions of electrons to flow in typical devices.

To find out how many electrons strike a surface per second, you first need to calculate the total charge for that duration. Using the previous example, we calculated a charge of \(1.0 \times 10^{-5} \text{ C} \) in one second. The number of electrons \(n\) can be found using:
  • \( n = \frac{Q}{e} \)
  • \( n = \frac{1.0 \times 10^{-5}}{1.6 \times 10^{-19}} \)
  • This results in about \(6.25 \times 10^{13} \) electrons.
Each electron individually contributes to the total movement seen in currents, illustrating how electric current is really a collective behavior of numerous tiny events.
Current and Time Relationship
The intimate relationship between current and time can be distilled into a simple idea: current tells you how quickly charge flows. When you know the current in a circuit, you can predict how much charge will pass through a point during a set period.

Consider the current \(I = 1.0 \times 10^{-5} \text{ A} \). If time \(t\) is 1 second, the charge \(Q\) is \(1.0 \times 10^{-5} \text{ C} \). When you extend the time to 60 seconds, multiplying the time by 60, the charge also multiplies to \(6.0 \times 10^{-4} \text{ C} \).

Using these types of calculations, you can estimate how long it takes for a certain amount of charge to accumulate at a point if you know the current. This practical application is highly important in designing and analyzing circuits, ensuring they function within desired parameters.

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Most popular questions from this chapter

The resistance of a coil of copper wire is \(3.35 \Omega\) at \(0^{\circ} \mathrm{C}\). What is its resistance at \(50^{\circ} \mathrm{C}\) ? For a copper alloy \(\alpha=4.3 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\). \(R=R_{0}+=R\left(T-T_{0}\right)=3.35 \Omega+\left(4.3 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\right)(3.35 \Omega)\left(50{ }^{\circ} \mathrm{C}\right)=4.1 \Omega\)

A resistor is to have a constant resistance of \(30.0 \Omega\), independent of temperature. For this, an aluminum resistor with resistance \(R_{0}\) at \(0{ }^{\circ} \mathrm{C}\) is used in series with a carbon resistor with resistance \(R_{02}\) at \(0{ }^{\circ} \mathrm{C}\). Evaluate \(R_{01}\) and \(R_{02}\), given that \(\alpha_{1}=3.9 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) for aluminum and \(\alpha_{2}=-0.50 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) for carbon.

What happens to the resistance of a copper wire if its diameter is doubled, all else kept constant?

When the insulation resistance between a motor winding and the motor frame is tested, the value obtained is \(1.0\) megohm \(\left(10^{6} \Omega\right)\). How much current passes through the insulation of the motor if the test voltage is \(1000 \mathrm{~V}\) ?

A battery charger supplies a current of \(10 \mathrm{~A}\) to charge a storage battery that has an open-circuit voltage of \(5.6 \mathrm{~V}\). If the voltmeter connected across the charger reads \(6.8 \mathrm{~V}\), what is the internal resistance of the battery at this time?

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