Parallel plate electric fields -- # of electrons transferred

[radiant_june]

Homework Statement

An oil droplet is suspended between two horizontal parallel plates with a separation of 0.4 cm. If the potential difference of 320V is applied to the plates, determine the number of electrons transferred to/from the droplet.

Given/Known Values
m_droplet = 5.2×10^-6 kg
d = 0.4 cm = 0.004 m (distance between plates)
V = 320 V
k = 9.0×10⁹ Nm²/C²

Homework Equations

Equations
F_E = (k⋅q₁⋅q₂)/r²
E = V/d
E = F_E/q_test
V = (k⋅q)/r
1 electron = 1.602×10^-19 C

The Attempt at a Solution

My assumption was that I could find the charge (q) of the oil droplet, using the V = (k⋅q)/r formula. Then I could divide q by 1.602×10^-19 to get the number of electrons transferred.

V = (k⋅q)/r
320 = (9.0×10⁹⋅q)/0.004
1.28/9.0×10⁹ = q
q = 1.422×10^-10

electrons = q/1.602×10^-19
= 1.422×10^-10/1.602×10^-19
= 887779164.9 electrons

Considering the excessive amount of electrons, I do not think I did this correctly. Any help would be appreciated!

 

[gneill]

The given voltage V is the potential difference between the two metal plates. This establishes an electric field between the plates. What forces are operating on the oil drop? Can you draw a Free Body Diagram for it?

 

[rude man]

As worded, looks like a trick question to me ...