A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8560 N/C. The mass of the water drop is 3.80 10-9 kg.
Do you realize that you have twice posted an incomplete question?
The complete question has already been posted and answered.
http://www.jiskha.com/display.cgi?id=1294529904
To understand why the small drop of water is suspended motionless in air, we need to consider the forces acting on the drop.
In this case, the two main forces at play are the gravitational force acting downward and the electric force acting upward due to the electric field.
First, let's calculate the gravitational force:
Gravitational force (Fg) = mass (m) x acceleration due to gravity (g)
Here, the mass of the water drop is given as 3.80 x 10^-9 kg, and the acceleration due to gravity can be approximated as 9.8 m/s^2.
Fg = (3.80 x 10^-9 kg) x (9.8 m/s^2)
Fg = 3.724 x 10^-8 N
Next, let's calculate the electric force:
Electric force (Fe) = charge (q) x electric field (E)
In this case, the charge of the water drop can be calculated using the equation:
q = mass (m) x acceleration due to gravity (g)
q = (3.80 x 10^-9 kg) x (9.8 m/s^2)
q = 3.724 x 10^-8 C
Now we can calculate the electric force:
Fe = (3.724 x 10^-8 C) x (8560 N/C)
Fe = 3.187 x 10^-4 N
Since the electric force and the gravitational force are equal in magnitude but opposite in direction, the drop remains suspended motionless in air.
Therefore, the small drop of water is suspended motionless in the air due to the equilibrium between the gravitational force and the electric force.