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Electric field question

  1. Sep 20, 2015 #1
    This is from the 2009 HSC exam. (I'm in Australia)
    I checked the answers and found that electric field strength was
    E = 100/0.10 = 1000
    My question is, what would the electric field strength be if the positively charged plate was +100V instead of 0V
    Would it be 200/0.1 = 2000?
    Is the formula E = The difference in volts between the two plates / distance.
    Please help this has been bugging me.
  2. jcsd
  3. Sep 20, 2015 #2


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    No it was not. The electric field strength is a dimensionful quantity and you simply cannot quote it as just a number without a unit. The field strength is 1000 V/m = 1 kV/m.

    It would be 2000 V/m.
  4. Sep 20, 2015 #3
    I just realized that I shouldn't have posted here. Sorry about that.
    But another question
    In this gif where
    E = V/d
    Shouldn't it be E = 2V/d
    as there is voltage going to the positive and negatively charged plate for example
    if the battery has 10V
    The negatively charged would be -10V and the positively charge would be 10V
  5. Sep 20, 2015 #4


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    why would you think that ?
    There isn't a 20V difference across the plates
    one terminal of the battery, the positive, is +10V relative to the 0V of the negative terminal

  6. Sep 20, 2015 #5
    Oh ok thanks. I just thought that for some reason. I didn't really know how the battery worked.
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