   Chapter 15, Problem 2P

Chapter
Section
Textbook Problem

A charged particle A exerts a force of 2.62 N to the right on charged particle B when the particles are 13.7 mm apart. Particle B moves straight away from A to make the distance between them 17.1 mm. What vector force does particle B then exert on A?

To determine
The force exerted by particle B on A.

Explanation

Given info: The force exerted by particle A on B is 2.62 N when they are 13.7 mm apart. The distance of separation changes to 17.7 mm.

From Coulomb’s Law,

F=ke|q1||q2|r2

• ke is the Coulomb constant.
• q1 and q2 are the charges.
• r is the distance of separation.

The force is proportional to r2 . Therefore,

F1F2=(r2r1)2 (I)

• F1 and F2 are the forces

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