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Relation between E and Potential gradient.

  1. Mar 1, 2014 #1
    According to the theory,
    E= -dv/dx
    or E.dx = -dv
    So if both are positive, the potential drop should increase.
    But as we know, if a positive charge is placed, as the distance from it keeps on increasing, field strength starts decreasing and potential drop should increase But this is contradictory right?

    Similarly if we make a diagram of field lines along the x-axis and place a sample charge at (a,0,0) and move it to (b,0,0) such that b>a, how can we predict which is at greater potential : a or b? The field value isn't given
     
  2. jcsd
  3. Mar 1, 2014 #2

    UltrafastPED

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    You must take the sign of the charge into account.
     
  4. Mar 2, 2014 #3

    sophiecentaur

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    Well, yes but Electric Potential is defined in terms of a unit positive charge. The Potential Energy of a charge will depend upon its magnitude and sign.
    In situations like this, the thing to do is to follow the signs rigorously and not to trust to intuition. The system is quite consistent.
     
  5. Mar 2, 2014 #4

    UltrafastPED

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    Isn't that what I said?
     
  6. Mar 2, 2014 #5

    sophiecentaur

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    I didn't say I disagreed. I was just amplifying and emphasising.
     
  7. Mar 2, 2014 #6

    Dale

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    Yes, this is correct.

    As the field strength decreases the potential drop also decreases. I am not sure why you think otherwise.


    You don't need to calculate fields to determine the potential. Potentials are linear, so you can simply add them up or shift them around like any other linear function.
     
  8. Mar 4, 2014 #7
    Think of potential as the energy required to bring a charge from infinitely far away to the point in question. The closer you get to the point in question (say a positive charge), the higher your potential. Now think of the potential as a hilltop. At infinity it is flat, but as you get closer to your charge, the hill gets steeper. This "steepness" is the field.
     
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