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## 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 10

^{3}kg/m

^{3}

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/s

^{2}

## Homework Equations

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

2. Mass of the drop, m = (4pi/3)*r

^{3}*σ

3. Force of gravity, F = mg = (4/3)*pi*r

^{3}*σ*g

## The Attempt at a Solution

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

q = (-4*pi*r

^{3}*σ*g)/3E

(2) I then plugged in the given values:

q = [-4*pi*(1.73 x 10

^{-6}m)

^{3}*0.865 x 10

^{3}kg/m

^{3}*9.8 m/s

^{2}]/(3*1.44 x 10

^{5}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.