   Chapter 16, Problem 7P

Chapter
Section
Textbook Problem

Oppositely charged parallel plates are separated by 5.33 mm. A potential difference of 600. V exists between the plates. (a) What is the magnitude of the electric field between the plates? (b) What is the magnitude of the force on an electron between the plates? (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.90 mm from the positive plate?

(a)

To determine
The magnitude of electric field.

Explanation

Given info: Potential difference is 600 V. The distance between the plates is 5.33 mm.

Explanation:

Formula to calculate the electric field between the plates is,

E=|ΔVΔx|

• Δx is the distance.
• ΔV is the potential difference.

Substitute 5.33 mm for Δx and 600 V for ΔV

E=|600V5

(b)

To determine
The electric force on the electron.

(c)

To determine
The work done on the electron.

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