Question

A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine the object's acceleration & charge magnitude. Assume the acceleration is due to the E-field (i.e., ignore all other forces).

Answer 1

Step-by-step explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

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

`s=0+(1)/(2)at^2`

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

`a=(2* 12\ m)/((1.2\ s)^2)`

a = 16.67 m/s²

Now put the value of a in equation (1) as :

`q=(ma)/(E)`

`q=(0.0008\ kg* 16.67\ m/s^2)/(534\ N/C)`

q = 0.0000249 C

or

`q=2.49* 10^(-5)\ C`

Hence, this is the required solution.