170k views
2 votes
A 0.8 g object is placed in a 159 N/C uniform electric field. Upon being released from rest, it moves 72 m in 2.9 s. Determine the object's acceleration & charge magnitude. Assume the acceleration is due to the E-field (i.e., ignore all other forces). a =

User Netricate
by
5.9k points

1 Answer

3 votes

Answer:

The acceleration and charge are 17.122 m/s² and
8.6*10^(-5)\ C

Step-by-step explanation:

Given that,

Mass of object = 0.8 g

Electric field = 159 N/C

Distance = 72 m

Time = 2.9 s

We know that,

The electric force is


F = Eq....(I)

The newton's second law


F=ma

Put the value of F in the equation (I)


ma=Eq...(II)

We calculate the acceleration

Using equation of motion


s=ut+(1)/(2)at^2


a =(2s)/(t^2)


a=(2*72)/((2.9)^2)


a=17.122\ m/s^2

From equation (II)


q=(ma)/(E)


q=(0.8*10^(-3)*17.122)/(159)


q=0.000086148427673\ C


q=8.6*10^(-5)\ C

Hence, The acceleration and charge are 17.122 m/s² and
8.6*10^(-5)\ C

User Zincfan
by
5.9k points