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Solve for charge(q), given radius, electric field, and density 
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#1
Oct711, 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%) 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} 2. Relevant 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 3. 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. 


#2
Oct711, 11:59 PM

P: 610

using the data I am getting
[tex]q=3.83\times 10^{18}\;\mathrm{C}[/tex] check math 


#3
Oct911, 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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