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Homework Help: Electric Field Strength Question

  1. Sep 20, 2015 #1
    I already posted this, but it was on the wrong section.
    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 = 1000V/m
    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 = 2000V/m?
    Is the formula E = The difference in volts between the two plates / distance.
    A person replied and said the two answer are correct

    But another question came up to my mind
    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
    Then E = 20/d
  2. jcsd
  3. Sep 20, 2015 #2


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    Gold Member

    If the battery is 10V then it creates a 10V difference. If you want to think of the two voltages as 0 and 10, or -10 and 0, or -5 and +5, it's up to you. They are all equivalent. But it is not -10 and +10 since that would be a 20V difference.
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