(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 0.500 μC charge that gets between the plates? Strategy Since the voltage

Question

(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 0.500 μC 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 [latex]E=\frac{V_{ AB }}{d}[/latex]. Once the electric field strength is known, the force on a charge is found using F = q E . Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, F = q E .

Accepted Answer

Solution for (a)
The expression for the magnitude of the electric field between two uniform metal plates is

[latex]E=\frac{V_{ AB }}{d}[/latex]. (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 value for [latex]V_{ AB }[/latex] and the plate separation of 0.0400 m, we obtain

[latex]E=\frac{25.0 kV }{0.0400 m }=6.25 imes 10^{5} V / m[/latex]. (19.32)

Solution for (b)
The magnitude of the force on a charge in an electric field is obtained from the equation

F = qE. (19.33)

Substituting known values gives

[latex]F=\left(0.500 imes 10^{-6} C ight)\left(6.25 imes 10^{5} V / m ight)=0.313 N[/latex]. (19.34)

Discussion
Note that the units are newtons, since 1 V/m = 1 N/C . The force on the charge is the same no matter where the charge is located between the plates. This is because the electric field is uniform between the plates.

Question 19.5: (a) An electron gun has parallel plates separated by 4.00 cm...

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