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Chapter 19 | Electric Potential and Electric Field 843
Making Connections: Uniform Fields
Figure 19.9 (a) A massive particle launched horizontally in a downward gravitational field will fall to the ground. (b) A positively charged particle launched horizontally in a downward electric field will fall toward the negative potential; a negatively charged particle will move in the opposite direction.
Recall from Projectile Motion (Section 3.4) that a massive projectile launched horizontally (for example, from a cliff) in a uniform downward gravitational field (as we find near the surface of the Earth) will follow a parabolic trajectory downward until it hits the ground, as shown in Figure 19.9(a).
An identical outcome occurs for a positively charged particle in a uniform electric field (Figure 19.9(b)); it follows the electric field “downhill” until it runs into something. The difference between the two cases is that the gravitational force is always attractive; the electric force has two kinds of charges, and therefore may be either attractive or repulsive. Therefore, a negatively charged particle launched into the same field will fall “uphill”.
Example 19.5 Field and Force inside an Electron Gun
(a) An electron gun has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. What is the electric field strength between the plates? (b) What force would this field exert on a piece of plastic with a charge that
gets between the plates?
Strategy
Since the voltage and plate separation are given, the electric field strength can be calculated directly from the expression
. Once the electric field strength is known, the force on a charge is found using . Since the electric field

is in only one direction, we can write this equation in terms of the magnitudes, .
Solution for (a)
The expression for the magnitude of the electric field between two uniform metal plates is
value for and the plate separation of 0.0400 m, we obtain
(19.32)
Solution for (b)

(19.31) Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this
The magnitude of the force on a charge in an electric field is obtained from the equation