/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Find the current when \(2.00 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the current when \(2.00 \mathrm{nC}\) jumps between your comb and hair over a \(0.500-\mu \mathrm{s}\) time interval.

Short Answer

Expert verified
The current is \(4.00 \times 10^{-3} \, A\) or 4.00 mA.

Step by step solution

01

Understand Current

Electric current, measured in amperes (A), is defined as the flow of electric charge across a surface per unit of time. The standard formula to calculate current (I) is: \( I = \frac{Q}{t} \), where \( Q \) is the charge in coulombs and \( t \) is the time in seconds the charge takes to flow.
02

Convert Units

Before calculating the current, convert the given charge and time to the proper SI units. For charge, convert nanocoulombs (nC) to coulombs (C) and for time, convert microseconds (\( \textmu s \)) to seconds (s). One nanocoulomb is \(10^{-9}\) coulombs and one microsecond is \(10^{-6}\) seconds. So, \(2.00 \, \text{nC} = 2.00 \times 10^{-9} \, C\) and \(0.500-\mu s = 0.500 \times 10^{-6} \, s\).
03

Calculate the Current

Using the converted units, substitute the values into the current formula: \( I = \frac{Q}{t} = \frac{2.00 \times 10^{-9} \, C}{0.500 \times 10^{-6} \, s} \). Perform the division to find the current.
04

Finalize the Answer

After performing the division, the result is the current in amperes: \( I = \frac{2.00 \times 10^{-9}}{0.500 \times 10^{-6}} = 4.00 \times 10^{-3} \, A \).

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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 that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charges: positive and negative, which are conventionally carried by protons and electrons, respectively. Like charges repel each other, while unlike charges attract.

In physics, the quantity of electric charge is measured in coulombs (C), named after Charles-Augustin de Coulomb, a French physicist who quantified the force between two charges. The charge of a single electron is approximately \( -1.602 \times 10^{-19} \, C \), which is considered as the elementary charge, often denoted by \( e \).

Understanding electric charge is essential when dealing with electric current, as current is simply the flow of charge through a conductor. The amount of charge that flows through a conductor in a given time frame is a crucial factor in current calculation, as depicted in the textbook exercise where a certain amount of charge jumps between a comb and hair.
SI Unit Conversion
SI (Système International d'Unités) is the modern form of the metric system and is the most widely used system of measurement. It standardizes units for various physical quantities, ensuring clarity and consistency across scientific and technical disciplines. SI unit conversion is a crucial skill in physics to ensure that equations dealing with physical quantities are dimensionally consistent.

For example, when calculating electric current and dealing with charges or time intervals, one might encounter units like nanocoulombs (nC) or microseconds (\(\mu s\)), not the standard SI units of coulombs and seconds. Moreover, to perform calculations correctly, it's necessary to convert these units into the standard SI units.
  • 1 nanocoulomb (nC) = \(10^{-9}\) coulombs (C)
  • 1 microsecond (\(\mu s\)) = \(10^{-6}\) seconds (s)

Converting correctly between these units is not only important for getting the right answer but also for deepening the understanding of the relationship between the size of different units and their physical implications.
Coulomb's Law
Coulomb's Law is a fundamental principle in electromagnetism, named after Charles-Augustin de Coulomb. This law quantifies the amount of electric force between two stationary, electrically charged particles.

The formula for Coulomb's Law is expressed as:\[ F = k \frac{|q_1 q_2|}{r^2} \], where:\
  • \(F\) is the force between the charges,
  • \(k\) is Coulomb's constant (\(8.9875 \times 10^9 N \cdot m^2/C^2\)),
  • \(q_1\) and \(q_2\) are the amounts of charge,
  • \(r\) is the distance between the centers of the two charges.

Coulomb's Law plays a vital role in explaining how charges interact with one another and is central to understanding the concept of the electric field. When relating this to current, the movement of charges (due to forces like those predicted by Coulomb's Law) is what we measure as electric current in circuits and various electrical phenomena. Indeed, the force between charges also underscores the exchange of electric charge in the mentioned exercise.

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Most popular questions from this chapter

Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on \(120 \mathrm{~V}\) and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on \(120 \mathrm{~V}\) and has 39 remaining identical bulbs, what is then the operating voltage of each?

In cars, one battery terminal is connected to the metal body. How does this allow a single wire to supply current to electrical devices rather than two wires?

A person with body resistance between his hands of \(10.0 \mathrm{k} \Omega\) accidentally grasps the terminals of a 20.0\(\mathrm{kV}\) power supply. (Do NOT do this!) (a) Draw a circuit diagram to represent the situation. (b) If the internal resistance of the power supply is \(2000 \Omega\), what is the current through his body? (c) What is the power dissipated in his body? (d) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in this situation to be \(1.00 \mathrm{~mA}\) or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-resistance devices? Explain your reasoning.

With a 1200-W toaster, how much electrical energy is needed to make a slice of toast (cooking time \(=1\) minute)? At \(9.0\) cents \(/ \mathrm{kW} \cdot \mathrm{h}\), how much does this cost?

The power dissipated in a resistor is given by \(P=V^{2} / R\), which means power decreases if resistance increases. Yet this power is also given by \(P=I^{2} R\), which means power increases if resistance increases. Explain why there is no contradiction here.

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