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

by Lemon-Sam
Tags: chargeq, density, electric, field, radius, solve
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Oct7-11, 09:18 PM
P: 4
1. The problem statement, all variables and given/known data

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%)


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

2. Relevant 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

3. 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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Oct7-11, 11:59 PM
P: 610
using the data I am getting

[tex]q=-3.83\times 10^{-18}\;\mathrm{C}[/tex]

check math
Oct9-11, 02:58 PM
P: 4
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.

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