An oil droplet with a mass of 3.50 x 10^-15 kg accelerates downward at a rate of 2.50 m/s2 when it is between two horizontal charged plates that are 1.00 cm apart. Assuming that the excess charge on the droplet is negative and the top plate is positive, how many excess electrons does the droplet carry if the potential difference between the plates is 533 V?

Is this the equation I should be using..?

E = v/d

q E = m g

If I use the above eqs. I get the wrong answer =(

Plz help!! and thank you in advance! =)

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asked by Marium
  1. qE = mg only if the droplet is stationary. Use the acceleration rate (a) to get

    qE -mg = ma

    You should be able to solve for q.

    You do have the correct formula for E

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    posted by drwls
  2. I just figured that out but thank you very much! :)

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    posted by Marium

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