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Effect of radius changes on electric fields and potential difference?

  1. Jan 31, 2012 #1
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    Hello!

    Okay so I understand that electric potential:

    V = kQ/r

    ...must be influenced by the radius doubling because it would make the potential energy half of what it originally was because of the proportionality law, v is proportional to 1/r.

    With electric fields though, how can there possibly be no change? The formula is

    E = kQ/r^2

    ...meaning if should be influenced as well?

    The answer is c, btw.
     
  2. jcsd
  3. Jan 31, 2012 #2

    Simon Bridge

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    So do the math - calculate the field before and after the change in position.
    Note: E is a vector.
     
  4. Jan 31, 2012 #3
    Oh! Right!

    Since E is a vector, and the distance and charges on both sides are equal, then they always simply cancel out to zero. Whereas with electric potential, a scalar quantity, it is not affected by direction, merely magnitude, in which case both are positive, so the radius change will definitely affect it.

    Thanks bud!
     
  5. Jan 31, 2012 #4

    Simon Bridge

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    No worries - that "Oh! Right!" feeling is what I was going for :)
     
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