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Homework Help: Homework help:A small plastic sphere with a mass of 2.60 X 10-15 kg

  1. Aug 19, 2014 #1
    1. The problem statement, all variables and given/known data

    "A small plastic sphere (with a mass of 2.60 X 10-15 kg) is suspended between
    two plates, with a potential difference of 265.4 V across the plates. The separation of the plates is
    0.500 cm. The upper plate is negative and the lower one is positive.
    a)What type of charge is on the sphere?Explain your reasoning.
    b)What is the magnitude of the charge on the sphere?
    c)How many excess or deficit electrons does the sphere have?"

    2. Relevant equations


    3. The attempt at a solution

    a) The charge on the sphere is positive. In order for the sphere to suspend, there must be a net force of zero on the object, where the magnitude of the upwards electrical force is equal to the downwards gravitational force. As the electrical force is pointed upwards (acting in the direction of the electrical field), it must be attracted to the negatively-charged plate. If the sphere’s charge were negative, the electrical force would be repelling from the negative plate and pointed downwards, where the net force acting on the object would no longer be zero.

    b) Fe= εq; ε= ∆V/d
    Fe= Fg
    (∆V/d)(q)= mg
    ∴q=mgd/∆V=2.60 x 10^-15 kg(9.8N/kg)(0.500cm)(1m/100cm)/265.4V=4.8 x 10^-4 C

    c) q=Ne
    ∴N= q/e= 4.8 x 10^-4 C /(1.6 x 10^-19 C) =3.0 x 10^15 electrons
    ∴Deficit of 3.0 x 10^15 electrons, as charge is positive.

    Please advise if this is correct?
  2. jcsd
  3. Aug 19, 2014 #2


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    Your method looks good. But check to see if you really get a power of -4 in the answer for the charge.
  4. Aug 21, 2014 #3
    Oh...calculation error...this should be 4.8 x 10^-19 C. TSny, thank you for letting me know.
  5. Aug 21, 2014 #4
    So then for c)
    ∴N= q/e= 4.8 x 10^-19 C /(1.6 x 10^-19 C) =3.0 electrons
    ∴Deficit of 3.0 electrons, as charge is positive.
  6. Aug 21, 2014 #5


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    2017 Award

    That looks good.
  7. Aug 21, 2014 #6
    Thank you for your help!
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