Chapter 19: Q6CQ (page 694)
Discuss how potential difference and electric field strength are related. Give an example.
Short Answer
The electric field strength is a measure of potential gradient at any point in the field \(\Delta V = E \times d\)
/*! 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}
Learning Materials
Features
Discover
Chapter 19: Q6CQ (page 694)
Discuss how potential difference and electric field strength are related. Give an example.
The electric field strength is a measure of potential gradient at any point in the field \(\Delta V = E \times d\)
All the tools & learning materials you need for study success - in one app.
Get started for free
What is an equipotential line? What is an equipotential surface?
Find the maximum potential difference between two parallel conducting plates separated by \(0.500{\rm{ }}cm\) of air, given the maximum sustainable electric field strength in air to be \(3.0 \times {10^6}{\rm{ }}V/m\).
Two parallel conducting plates are separated by \(10.0{\rm{ }}cm\), and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential \(8.00{\rm{ }}cm\) from the zero-volt plate (and \(2.00{\rm{ }}cm\) from the other) is \(450{\rm{ }}V\)? (b) What is the voltage between the plates?
Sketch the equipotential lines for the two equal positive charges shown in Figure. Indicate the direction of increasing potential.

Consider a battery used to supply energy to a cellular phone. Construct a problem in which you determine the energy that must be supplied by the battery, and then calculate the amount of charge it must be able to move in order to supply this energy. Among the things to be considered are the energy needs and battery voltage. You may need to look ahead to interpret manufacturer's battery ratings in ampere-hours as energy in joules.
What do you think about this solution?
We value your feedback to improve our textbook solutions.