Charge of bead falling through an electric field

[fenixbtc]

Homework Statement

postively charged bead having a mass of .61g falls from rest in a vacuum from a height of 5.9 m in a uniform vertical electric field of magnitude 11600 N/C. The bead hits the ground at a speed of 19 m/s. acceleration of gravity is 9.8 m/s^2. find the charge on the bead, answer in units of uC (micro Coulombs)

Homework Equations

1. v^2=vi^2 + 2a(y-yi) used to find the acceleration.
2. F=ma used to find force of the bead
3. E=F/qo used to find the charge
4. 1N = (1kg)(1m/s^2) conversion

The Attempt at a Solution

using equation 1 i find that the acceleration is (19 m/s)^2 / (2 * 5.9m) = 30.5932 m/s^2
equation 2 F = .61e-3kg(30.5932m/s^2) = .01866186 N <-using equation 4.
equation 3 rearranged to qo=F/E = (1.866186e-2N) / (11600N/C) = 1.60878e-6C = 1.60878uC

this is a homework problem submitted online and it gives you a couple chances to get it right, each time decreasing your score. where am i going wrong since the program is telling me it's incorrect?

thanks!
David

 

[rl.bhat]

Net acceleration a = g - a', where a' is the acceleration of the charges bead due to electric field only.

 

[fenixbtc]

thank you! that worked.