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).

 

[m2003]

.

I would suggest double-checking your calculations and making sure you are using the correct units. The magnitude of the charge on the drop can be calculated using the equation q = F/E, where q is the charge, F is the electric force, and E is the electric field magnitude. Plugging in the given values, we get q = (8.60 × 10−16 N)/(153 V/m) = 5.6 × 10−18 C. To convert this to elementary charge, we can use the equation q = ne, where n is the number of elementary charges and e is the elementary charge. Rearranging, we get n = q/e. Plugging in the value for q and the known value for e (1.602 × 10−19 C), we get n = (5.6 × 10−18 C)/(1.602 × 10−19 C) = 35. Therefore, the magnitude of the charge on the drop is 35e.