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Chapter 25: Q.48 (page 711)

A room with 3.0-m-high ceilings has a metal plate on the floor with V = 0 V and a separate metal plate on the ceiling. A 1.0 g glass ball charged to +4.9 nC is shot straight up at 5.0 m/s. How high does the ball go if the ceiling voltage is (a) +3.0脳106Vand (b) -3.0脳106V?

Short Answer

Expert verified

The value of high for the ceiling voltage is +3.0脳106Vis 0.85cm

The value of high for the ceiling voltage is -3.0脳106Vis 2.55m

Step by step solution

01

Step: 1 The formulation of celling voltage 

The net force = Fe+Fg=(electricforce+gravitationalforce)

Fe=EqFg=mgFe+Fg=Eq+mg

from newton law formula

Fnet=ma

applying all data

ma=Eq+mga=Eq+mgmAsE=鈭�Vmd鈭�V=3脳106Vg=9.8m/s2m=1gd=3.0mq=4.9nCa=14.7m/s2

For Trigonometric equation;

vy2=voy2+2ayvoy=0.5vy=0y=0.85m

02

Step: 2 The definition of celling voltage 

The net force
Fe-Fg=(electricforce+gravitationalforce)

As F=ma

ma=Eq-mga=Eq-mgmAsE=鈭�Vmd鈭�V=3脳106Vg=9.8m/s2m=1gd=3.0mq=4.9nCa=-4.9m/s2

From trigonometric equation;

vy2=voy2+2ayvoy=0.5vy=0y=2.55m

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