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

by Lemon-Sam
Tags: chargeq, density, electric, field, radius, solve
 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 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*σ 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.
 P: 604 using the data I am getting $$q=-3.83\times 10^{-18}\;\mathrm{C}$$ check math
 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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