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Chapter 19: Problem 2

Lightning strikes. During lightning strikes from a cloud to the ground, currents as high as 25,000 A can occur and last for about \(40 \mu\) s. How much charge is transferred from the cloud to the earth during such a strike?

Short Answer

Expert verified
1 coulomb of charge is transferred.

Step by step solution

01

Identify Given Information

We are given the current, which is 25,000 A, and the duration of the lightning strike, which is 40 \( \mu \)s (microseconds).
02

Convert Time Units

First, convert the duration from microseconds to seconds since the SI unit for time is seconds. 40 microseconds \(= 40 \times 10^{-6} \) s.
03

Understand the Formula for Charge

The electric charge \( Q \) transferred is given by the formula \( Q = I \times t \), where \( I \) is the current and \( t \) is the time duration.
04

Substitute Values into the Formula

Now substitute the given values into the formula: \( Q = 25,000 \times 40 \times 10^{-6} \).
05

Calculate the Charge

Perform the multiplication: \( Q = 25,000 \times 40 \times 10^{-6} = 1 \) C.

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter carried by certain particles. Two types of electric charges exist: positive and negative. Particles with the same type of charge repel each other, while those with opposite charges attract. The unit of electric charge in the International System of Units (SI) is the coulomb (C). In simple terms, electric charge quantifies the amount of electricity that flows during an electrical process, such as a lightning strike. Understanding electric charge is essential when calculating the amount of electricity transferred in an event, whether large scale like a strike or small, like a battery powering a remote.
Current
Current represents the flow of electric charge and is a crucial concept in electricity. Measured in amperes (A), it indicates how many charges pass through a point in a circuit per second. For instance, in our case of a lightning strike, the current is given as 25,000 A, meaning a vast number of charges move between the cloud and the earth every second. To visualize this better, think of electric current like water flowing through a pipe. A higher current means more water (or charge) flowing through per second. It helps us understand the intensity and quantity of electricity involved in processes.
Time Conversion
Time conversion is essential in calculations involving electricity because different units like seconds, minutes, and microseconds might be used. The standard SI unit for time is seconds, which means any given time in calculations should be converted to seconds for consistency. Consider converting microseconds to seconds: microseconds are one-millionth of a second, denoted as \( \mu s \). For example, \( 40 \mu s \) equals \( 40 \times 10^{-6} \) seconds. This conversion ensures that any operation or formula application involving time is accurate and universally applicable, thus avoiding calculation errors.
Formula Application
Applying the correct formula is vital for solving electricity-related problems. The formula linking electric charge \( Q \), current \( I \), and time \( t \) is \( Q = I \times t \). This straightforward relationship shows that charge depends directly on current and time. When calculating, substitute known values into the formula to find the unknown. In the lightning strike scenario, we calculate charge by substituting the current (25,000 A) and the time (\( 40 \times 10^{-6} \) s) into the formula: \( Q = 25,000 \times 40 \times 10^{-6} \), resulting in a charge of 1 C. This practical application shows the direct impact of each factor on the total electric charge.

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Most popular questions from this chapter

The power rating of a resistor is the maximum power it can safely dissipate without being damaged by overheating. (a) If the power rating of a certain \(15 \mathrm{k} \Omega\) resistor is \(5.0 \mathrm{~W},\) what is the maximum current it can carry without damage? What is the greatest allowable potential difference across the terminals of this resistor? (b) If a \(9.0 \mathrm{k} \Omega\) resistor is to be connected across a \(120 \mathrm{~V}\) potential difference, what power rating is required for that resistor?

The resistance of the body varies from approximately \(500 \mathrm{k} \Omega\) (when it is very dry) to about \(1 \mathrm{k} \Omega\) (when ?? it is wet). The maximum safe current is about \(5.0 \mathrm{~mA}\). At \(10 \mathrm{~mA}\) or above, muscle contractions can occur that may be fatal. What is the largest potential difference that a person can safely touch if his body is wet? Is this result within the range of common household voltages?

When you connect an unknown resistor across the terminals of a 1.50 V AAA battery having negligible internal resistance, you measure a current of 18.0 mA flowing through it. (a) What is the resistance of this resistor? (b) If you now place the resistor across the terminals of a \(12.6 \mathrm{~V}\) car battery having no internal resistance, how much current will flow? (c) You now put the resistor across the terminals of an unknown battery of negligible internal resistance and measure a current of 0.453 A flowing through it. What is the potential difference across the terminals of the battery?

In an ionic solution, a current consists of \(\mathrm{Ca}^{2}+\) ions (of charge \(+2 e\) ) and \(\mathrm{Cl}^{-}\) ions (of charge \(-e\) ) traveling in opposite directions. If \(5.11 \times 10^{18} \mathrm{Cl}^{-}\) ions go from \(A\) to \(B\) every \(0.50 \mathrm{~min},\) while \(3.24 \times 10^{18} \mathrm{Ca}^{2+}\) ions move from \(B\) to \(A,\) what is the current (in mA) through this solution, and in which direction (from \(A\) to \(B\) or from \(B\) to \(A\) ) is it going?

A nonideal \(10.0 \mathrm{~V}\) battery is connected across a resistor \(R .\) The internal resistance of the battery is \(2.0 \Omega\). Calculate the potential difference across the resistor for the following values: (a) \(R=100 \Omega\), (b) \(R=10 \Omega,\) (c) \(R=2 \Omega\). (d) Find the current through the battery for all three cases.
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