Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
1,000 Coulombs of charge is transferred.

Step by step solution

01

Understand the Problem

The problem involves calculating the amount of electric charge transferred during a lightning strike. The lightning current is given as 25,000 A and it lasts for 40 ms (milliseconds). We will use the formula for electric charge, which is the product of current and time.
02

Convert Time to Seconds

The time given is in milliseconds (ms). To use in the formula, convert it to seconds. Since 1 second = 1000 milliseconds, 40 ms = 40/1000 = 0.04 seconds.
03

Use the Formula for Charge

The formula to calculate the electric charge transferred is:\[ Q = I imes t \]where \( Q \) is the charge (in Coulombs), \( I \) is the current (in Amperes), and \( t \) is the time (in seconds). Substitute the known values: \( I = 25,000 \, A \) and \( t = 0.04 \, seconds \).
04

Calculate the Charge

Substitute the values into the formula:\[ Q = 25,000 \, A imes 0.04 \, s = 1,000 \, C \]So, the charge transferred during the lightning strike is 1,000 Coulombs.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Lightning Strike Current
Lightning is an impressive natural phenomenon and involves enormous electric currents. When lightning strikes, it results from a rapid discharge of electric charge accumulated in the clouds. The current in a lightning strike can reach extremely high values, typically ranging from thousands to tens of thousands of Amperes, like the 25,000 Amperes mentioned in the exercise.

This high current is the result of the massive buildup of electric charge in the cloud and its sudden release during a strike. This discharge causes the current to spike and, despite lasting only a brief moment, it is powerful and can transfer a significant amount of electric charge. Knowing the current helps in understanding how much electric power is being transferred from the cloud to the ground, which is a crucial factor in calculating the electric charge transferred.
Exploring the Electric Charge Formula
Electric charge is a fundamental property of matter and is measured in Coulombs (C). To calculate the amount of electric charge transferred in any process, you use the formula:
  • \[ Q = I \times t \]
where:
  • \( Q \) is the electric charge in Coulombs,
  • \( I \) is the current in Amperes,
  • \( t \) is the time in seconds.
This formula is based on the relationship that charge is equal to the product of current (flow of electric charge) and the time the current flows.

Essentially, if you have a steady flow of current that lasts for a particular duration, this formula gives you the total electric charge transferred during that time period. For instance, if a 25,000 Ampere current flows for 0.04 seconds, the charge transferred is \( 25,000 \times 0.04 = 1,000 \) Coulombs.
Converting Time to Seconds
Understanding time conversion is crucial when working with formulas as they often require consistent units. In the context of the exercise, the duration of the lightning strike is given in milliseconds (ms), a unit often used to measure very brief periods of time. However, to use this duration in the electric charge formula, we need to convert it to seconds.

The conversion is straightforward.
  • Since 1 second equals 1,000 milliseconds, the time conversion is simply dividing the milliseconds by 1,000.
  • For instance, 40 milliseconds becomes \( 40/1000 = 0.04 \) seconds.


By ensuring we convert any given time into seconds, we maintain consistency and accuracy in calculations involving electric charge, making it easier to apply formulas correctly and arrive at the right solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In another experiment, a piece of the web is suspended so that it can move freely. When either a positively charged object or a negatively charged object is brought near the web, the thread is observed to move toward the charged object. What is the best interpretation of this observation? The web is (a) a negatively charged conductor; (b) a positively charged conductor; (c) either a positively or negatively charged conductor; (d) an electrically neutral conductor.

An overhead transmission cable for electrical power is 2000 m long and consists of two parallel copper wires, each encased in insulating material. A short circuit has developed somewhere along the length of the cable where the insulation has worn thin and the two wires are in contact. As a power-company employee, you must locate the short so that repair crews can be sent to that location. Both ends of the cable have been disconnected from the power grid. At one end of the cable (point \(A\)), you connect the ends of the two wires to a 9.00-V battery that has negligible internal resistance and measure that 2.86 A of current flows through the battery. At the other end of the cable (point \(B\)), you attach those two wires to the battery and measure that 1.65 A of current flows through the battery. How far is the short from point A?

When a resistor with resistance \(R\) is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that \(R\) remains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100-W incandescent bulb uses only 23 W of power. The compact bulb lasts 10,000 hours, on the average, and costs \(11.00, whereas the incandescent bulb costs only \)0.75, but lasts just 750 hours. The study assumed that electricity costs $0.080 per kilowatt-hour and that the bulbs are on for 4.0 h per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100-W" fluorescent bulb? (Remember, it actually uses only 23 W of power and operates across 120 V.)

A "540-W" electric heater is designed to operate from 120-V lines. (a) What is its operating resistance? (b) What current does it draw? (c) If the line voltage drops to 110 V, what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free