If a particular oil droplet has a mass of 2.0 × 10-14 kg, and the droplet has fi

Question

If a particular oil droplet has a mass of 2.0 × 10-14 kg, and the droplet has five excess electrons on it, what is the magnitude and direction of the electric field necessary to hold the droplet at rest?

For part (a), I used the equation N=q/e to find the value of q...
N=q/e
5=q/e
5=q/(1.6x10^-19)
q=5x(1.6x10^-19)
q=8x10^-19

If the oil droplet acquired a sixth excess electron, what would be the magnitude of the droplet’s acceleration and how long would it take to travel through a distance of 5.0 cm?

Explanation / Answer

When the charge is balanced, the force of gravity is balanced by the electrostatic force. Then F_g=qE Solving for E: E=mg/q, where q=-5e coulombs and e is the elementary charge E=mg/q E=[(2x10^14)(-9.8 )]/(8x10^-19) = 245000J Part b) Now the charge q is -6e The electrostatic force could be calculated by F_e=qE where E equals the value found above Then by Newton' Second Law, qE-mg=ma. Solving for the acceleration; we get a=(qE-mg)/m a= [(9.6x10^19)(245000) - (2x10^-14)(9.8 )]/(2x10^-14) = 10.78m/s^2 to travel a distance of 5 cm time req : d=.5at^2 .05=.5(10.78 )t^2 t=9.63x10^-2s

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