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Charge question

  1. May 30, 2005 #1
    I'm learning about charge right now and I have no idea how to start this one.. any help is appreciated.

    A helium balloon has a charge of q = 5.5 x 10^-8. It rises vertically into the air. d = 600m. from the surface of the earth to final position A. The electric field that normally exists in the atmosphere near the surface of the earth has a magnitude E = 150 N/C and is directed downward. What is the difference in electric potential between the two positions?
  2. jcsd
  3. May 30, 2005 #2

    Doc Al

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    Staff: Mentor

    What's the definition of electric potential? How does it relate to the field? (Hint: The charge on the balloon is irrelevant.)
  4. May 30, 2005 #3
    Is this true for d = 0m and d = 600m ?

    if so i think I could solve it using:

    V= Ed
  5. May 31, 2005 #4

    Doc Al

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    Staff: Mentor

    Yes, assume the field is uniform.


    A more accurate way to write this is:
    [tex]\Delta V = - E \Delta d[/tex]
    Last edited: May 31, 2005
  6. May 31, 2005 #5
    I note that, even though the electric field is pointing downwards, the positively charged balloon moves upwards, unless and until it decelerates and comes to rest but that a different case.
    Last edited: May 31, 2005
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