Solve for charge(q), given radius, electric field, and density

[Lemon-Sam]

Homework Statement

In Millikan's experiment, an oil drop of radius 1.73 μm and density 0.865 g/cm3 is suspended in chamber C when a downward electric field of 1.44 × 105 N/C is applied. Find the charge on the drop, in terms of e. (Tolerance of solution is +/- 2%)

Known:

radius, r = 1.73 μm x 10^-6 m

density, σ = 0.865 x 10³ kg/m³

Electric field, E = 1.44 × 105 N/C

magnitude of charge of electron, e = 1.6 x 10^-19

gravitational constant, g = 9.8 m/s²

Homework Equations

1. Force due to electric field: F = -qE

2. Mass of the drop, m = (4pi/3)*r³*σ

3. Force of gravity, F = mg = (4/3)*pi*r³*σ*g

The Attempt at a Solution

(1) I rearranged equations (1) & (3) from above to solve for q, thus:

q = (-4*pi*r³*σ*g)/3E

(2) I then plugged in the given values:

q = [-4*pi*(1.73 x 10^-6m)³*0.865 x 10³ kg/m³*9.8 m/s²]/(3*1.44 x 10⁵ N/C)

= -1.28 x 10^-18 C

(3) Now, to put the solution in terms of e, I divided my result from (2) by the magnitude of charge of electron, e:

q = (-1.28 x 10^-18 C/1.6 x 10^-19 C)(1e)

= -8e C

However, my answer is incorrect, and I'm not sure why. I also tried to reverse the sign of my solution, but that was incorrect as well.

 

[issacnewton]

using the data I am getting

$$q=-3.83\times 10^{-18}\;\mathrm{C}$$

check math

 

[Lemon-Sam]

I checked the math again, and it turns out that the correct result was just -8, instead of -8e. Thank you for your reply, Issac.