   Chapter 15, Problem 28P

Chapter
Section
Textbook Problem

A particle of mass 1.00 × 10−9 kg and charge 3.00 pC is moving in a vacuum chamber where the electric field has magnitude 2.00 × 103 N/C and is directed straight upward. Neglecting other forces except gravity, calculate the panicle’s (a) acceleration and (b) velocity after 2.00 s if it has an initial velocity of 5.00 in/s in the downward direction.

(a)

To determine
The particle’s acceleration.

Explanation

Given info: The charge of the particle is 3.00pC . The mass of the particle is 1.00×109kg . Electric field strength is 2.00×103N/C .

Formula to calculate the magnitude of acceleration of the proton is,

a=qEmg

• q is the charge of nucleus.
• E is the electric field.
• m is the mass of proton.
• g is the acceleration due to gravity.

Substitute 3.00pC for q, 2.00×103N/C for E and 1.00×109kg for m.

a=(3.00pC)(2.00×103N/C)1

(b)

To determine
The velocity after 2.00 s.

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