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Lightning occurs when there is a flow of electric charge (principally electrons) between the ground and a thundercloud. The maximum rate of charge flow in a lightning bolt is about 20,000 C\(/\)s; this lasts for 100 \(\mu\)s or less. How much charge flows between the ground and the cloud in this time? How many electrons flow during this time?

Short Answer

Expert verified
2 Coulombs of charge flows, and approximately 1.25 x 10^19 electrons flow.

Step by step solution

01

Understand the given values

We are given that the current (rate of charge flow) in the lightning bolt is 20,000 C/s and the duration of this current is 100 microseconds (\[100 \mu s = 100 \times 10^{-6} s\] ). We need to find the total charge that flows and the number of electrons.
02

Calculate the total amount of charge

To find the total charge, use the formula \[Q = I \times t\] where \(I = 20,000 \) C/s and \(t = 100 \times 10^{-6} \) s. Substitute the values: \[Q = 20,000 \times 100 \times 10^{-6} = 2 \text{ Coulombs}\].
03

Determine the number of electrons

The charge of a single electron is approximately \(1.6 \times 10^{-19}\) C. To find the number of electrons, use the formula: \[n = \frac{Q}{e}\] where \(Q = 2 \) C and \(e = 1.6 \times 10^{-19} \) C. Calculate: \[n = \frac{2}{1.6 \times 10^{-19}} = 1.25 \times 10^{19} \].

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Key Concepts

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

Lightning
Lightning is a natural electrical phenomenon distinguished by a bright flash and a loud thunderclap. It occurs when there is a large discharge of electric charge—mainly electrons—between regions with different electrical potentials. This typically happens between thunderclouds and the ground. During a storm, the cloud develops a negative charge at its base and a positive charge at the top. When the charge difference becomes large enough, lightning occurs, neutralizing the charge difference.
Understanding lightning is crucial because it involves the rapid movement of a large number of electrons. This movement is so large that it creates an electrical current, resulting in dramatic weather effects. Lightning is a clear example of how electrical charge behaves in nature under extreme conditions.
- Occurs between charged regions (cloud and ground) - Involves rapid electron flow - Results in a bright light and sound (thunder) Knowing the basic mechanics of lightning helps us understand electric charge flow better.
Charge flow
Charge flow, or electrical current, refers to the movement of electric charges in a particular direction. In the context of the lightning exercise, the maximum rate of charge flow can reach up to 20,000 Coulombs per second (C/s). This is exceptionally high compared to everyday electrical currents. In simple terms, charge flow is comparable to water flowing through a pipe—the more water (or charge) that flows through per unit time, the stronger the flow (or current).
To calculate the total charge that flows during a lightning strike, we multiply the rate of charge flow by the time period over which it flows. This gives us a sense of how much electrical "water" has flowed from the cloud to the ground during the strike. Understanding charge flow is key to grasping how electricity behaves, particularly in large-scale phenomena like lightning.
  • Measured in Coulombs per second (C/s)
  • Determines the strength of the current
  • A central concept in understanding electrical circuits and phenomena
Electrons
Electrons are the tiny, negatively charged particles that form the flow of electric charge. In the context of lightning, electrons move from the negatively charged base of a thundercloud towards the ground, neutralizing the charge difference. These particles are the fundamental building blocks of electric currents, whether in a tiny wire of your household circuit or a massive lightning bolt.
A single electron carries a very small amount of charge, approximately \(1.6 \times 10^{-19}\) Coulombs. When we speak about the flow of electrons, we often measure it in terms of the collective charge they carry. For instance, during a lightning bolt, an enormous number of electrons move in a short time and are measured in Coulombs.
  • Negatively charged subatomic particles
  • Form the basis for electrical conductors
  • Responsible for carrying electric charge in circuits
While electrons are exceedingly small individually, together they create the powerful currents observed in natural phenomena like lightning.
Coulombs
Coulombs are the unit of measurement for electric charge, symbolized by the letter \(C\). They quantify how much electric charge is present in a system. In essence, Coulombs measure the quantity of electrons since each Coulomb contains approximately \(6.242 \times 10^{18}\) electrons.
The term originates from the French scientist Charles-Augustin de Coulomb, who researched electrostatic and magnetic forces. In our example of a lightning bolt, we have calculated that 2 Coulombs of charge flow between the cloud and the ground in an exceptionally short time period. This highlights the enormous power of lightning, as this huge amount of charge moves in mere microseconds.
- Measures amount of charge- One Coulomb equals \(6.242 \times 10^{18}\) electrons- Central to calculating and understanding electric phenomenaWhether analyzing household electrical devices or natural events like lightning, Coulombs are a critical part of the analysis.
Microseconds
Microseconds (\(\mu s\)) are units of time used to measure events happening on very short timescales. One microsecond is equivalent to one-millionth of a second (\(1 \times 10^{-6}\) seconds). In the context of lightning, microseconds are significant in capturing how quickly these immense discharges occur.
When solving problems related to time, especially with phenomena as fast as lightning strikes, knowing how to convert different time units is crucial. The original problem specifies a duration of 100 microseconds. We convert this to seconds to make calculations feasible using the formula for electric charge.
  • 1 microsecond = \(1 \times 10^{-6}\) seconds
  • Essential for calculations in fast phenomena
  • Helps in precise measurement of electrical effects
Using microseconds allows us to appreciate the speed of lightning: a massive transfer of electric charge happens in a blink of an eye.

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Most popular questions from this chapter

A \(-\)3.00-nC point charge is on the \(x\)-axis at \(x =\) 1.20 m. A second point charge, \(Q,\) is on the \(x\)-axis at -0.600 m. What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 N\(/\)C in the \(+\)x-direction, (b) 45.0 N\(/\)C in the \(-\)x-direction?

In a region where there is a uniform electric field that is upward and has magnitude 3.60 \(\times 10^4 \space N/\)C, a small object is projected upward with an initial speed of 1.92 m\(/\)s. The object travels upward a distance of 6.98 cm in 0.200 s. What is the object's charge-to-mass ratio \(q/m\)? Assume \(g = 9.80 \space m/s^2\), and ignore air resistance.

Two small aluminum spheres, each having mass 0.0250 kg, are separated by 80.0 cm. (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 g\(/\)mol, and its atomic number is 13.) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00 \(\times\) 10\(^4\) N (roughly 1 ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

A charge of \(-\)6.50 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25 cm. (a) Find the magnitude and direction of the electric field this disk produces at a point \(P\) on the axis of the disk a distance of 2.00 cm from its center. (b) Suppose that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. Find the magnitude and direction of the electric field at point \(P.\) (c) If the charge is all brought to the center of the disk, find the magnitude and direction of the electric field at point \(P.\) (d) Why is the field in part (a) stronger than the field in part (b)? Why is the field in part (c) the strongest of the three fields?

Point charges \(q_1 = -\)4.5 nC and \(q_2 = +\)4.5 nC are separated by 3.1 mm, forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of 36.9\(^\circ\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9}\) N \(\cdot\) m ?

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