The electric field inside a 30-cm-long copper wire is 5.0 mV/m. What is the potential difference between the ends of the wire?
Ch 27: Current and Resistance
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 27, Problem 26b
A 1.5 V battery provides 0.50 A of current. How much work does the charge escalator do to lift 1.0 C of charge?
Verified step by step guidance1
Understand the relationship between voltage, work, and charge. The work done by the battery on a charge is given by the formula: , where is the work, is the charge, and is the voltage.
Identify the given values in the problem: the voltage of the battery is V, and the charge to be lifted is C.
Substitute the given values into the formula: .
Simplify the expression to calculate the work done. This will give the amount of energy (in joules) required to lift the charge.
Conclude that the work done by the charge escalator is directly proportional to the product of the charge and the voltage, as shown in the formula.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Voltage
Voltage, measured in volts (V), is the electric potential difference between two points in a circuit. It represents the energy per unit charge available to move charges through a circuit. In this question, the 1.5 V battery indicates that each coulomb of charge has 1.5 joules of energy available to do work.
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Current
Current, measured in amperes (A), is the flow of electric charge in a circuit. It quantifies how much charge passes through a point in the circuit per unit time. In this scenario, a current of 0.50 A means that 0.50 coulombs of charge flow through the circuit every second, which is essential for calculating the work done on the charge.
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Work Done by Electric Charge
The work done by an electric charge when moving through a potential difference is calculated using the formula W = Q × V, where W is work in joules, Q is charge in coulombs, and V is voltage in volts. In this case, lifting 1.0 C of charge through a 1.5 V potential difference results in 1.5 joules of work done, illustrating the relationship between charge, voltage, and work.
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Related Practice
Textbook Question
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