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

## 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

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.

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using the data I am getting

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

check math

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.