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Chapter 17: Electric Fields
1.

Electric Fields and Potential

We saw in The Height Metaphor that we can think of electric potential as if it were height or elevation, and that positive charges will feel a force "downhill" as if they were boulders. By downhill, we don't just mean any direction that's "towards lower potential": we mean the direction that takes us downhill the fastest, and that is the line which is perpendicular to the equipotential lines, like those shown in the figure here. You can think of these as rivers running down the potential hillside (although don't take the metaphor too far: these aren't currents, and they aren't actually moving). These downhill lines are called electric field lines, and at every point on the hillside we define an electric field vector $\vec E$ to be a vector which points downhill, perpendicular to the equipotential line at that point.

A topographic map doesn't just tell us the downhill direction, it also tells us how steep the descent is. Lines which are closer together suggest a steeper descent or ascent, while lines which are farther apart describe a gentler slope. In fact, the steepness (or slope) of the ground in a particular area is $${\rm steepness} = {\rm rise\over run} = \frac{\rm change\,of\,elevation}{\rm distance}$$

We define the magnitude $E$ of the electric field vector in a similar way.

Between any two equipotential lines, the average electric field is equal to $$E_{avg}={\Delta V\over d} = {\hbox{change in potential}\over \hbox{distance between the lines}}$$
(no alternate text)

For example, on the left, where the 2V and 3V equipotential lines are 0.1 meters apart, the average electric field is $$E={3\u{V}-2\u{V}\over 0.1\u{m}} = 10\u{V/m}.$$ To the right, though, where the lines are farther apart, the electric field is $$E={3\u{V}-2\u{V}\over 0.2\u{m}} = 5\u{V/m}.$$ The electric field is stronger where the lines are closer together.