   Chapter 15, Problem 17P

Chapter
Section
Textbook Problem

A small object of mass 3.80 g and charge −18.0 μC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What is the magnitude and direction of the electric field?

To determine
The magnitude and direction of electric field.

Explanation

Given info: Mass of the object (m) is 3.80 g. The charge on the object (q) is 18.0μC .

The force on the object due to electric field is balanced by the weight.

Formula to calculate the force on the object due to electric field is,

F=|q|E (I)

• E is the electric field.

Formula to calculate the weight of the object

Fw=mg (II)

• g is the acceleration due to gravity.

From Equations (I) and (II)

|q|E=mg

On Re-arranging,

E=mg|q|

Substitute 3.80 g for m, 9.8ms2 for g and 18.0μC for q.

E=(3.80g)(9

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