To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.
Here,
m = mass
g = Acceleration due to gravity
Rearranging to find the charge,
Replacing,
Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is
Here,
n = Number of electrons
e = Charge of each electron
Replacing,
Therefore the number of electrons that reside on the drop is 
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where 
are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that 
, which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
