How Does a Positively Charged Oil Drop Behave in a Uniform Electric Field?

[kbyws37]

A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates. If the electric force on the drop is found to be 8.60 × 10−16 N and the electric field magnitude is 153 V/m, what is the magnitude of the charge on the drop in terms of the elementary charge e?


The correct answer should be 35e, but I am not getting it.

My attempt:
I tried the equation
F = Eq
8.60 × 10−16 N = (153 V/m)(q)
q = 5.6 x 10-18 N

Then I used
e = q/k
e = 5.6 x 10-18 N / 8.99 x 10^9

which does not give me the right answer

 

[Hootenanny]

I tried the equation
F = Eq
8.60 × 10−16 N = (153 V/m)(q)
q = 5.6 x 10-18 N

Good

Then I used
e = q/k
e = 5.6 x 10-18 N / 8.99 x 10^9

Not so good. I don't know where you got that equation from, but to express your charge in terms of elementary charges, you have to divide through by the value of the elementary charge (1.6x10^-19 C).