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Potential difference: positive or negative?

  1. Oct 5, 2015 #1
    Why potential at a point can only be obtained by supposing/assuming a positive test charge there?
    For example
    There was a question
    In figure two points A and B are located in a region of electric field.,The potential difference ##VB##-##VA## is
    1-positive
    2-Negative
    3-zero
    4-none of the above
    the answer is 2-Negative
    Apparently the source charge is positive
    O.png
    we will only get the answer to be 2-Negative when we would assume positive test charge but if we assume negative test charge at A &B we are going to get answer 1-positive.So is it like we are bound to assume positive charge at points mentioned in question to get whether potential difference is positive or negative?
     
  2. jcsd
  3. Oct 5, 2015 #2

    Dale

    Staff: Mentor

    Yes. The "test charge" is always positive, by definition.
     
  4. Oct 5, 2015 #3

    ehild

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    Gold Member

    The figure shows the electric field lines.
    How is the electric field at a point defined?
     
  5. Oct 5, 2015 #4
    Force per unit charge.
    The electric field 5eb237ccb8c2716d347ab313cad7918e.png at a given point is defined as the (vectorial) force 183083a13a40d344ebc290a84579b0c3.png that would be exerted on a stationary test particle of unit charge by electromagnetic forces
     
  6. Oct 5, 2015 #5

    ehild

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    And what is that test particle?What does unit charge mean?
    http://www.thefreedictionary.com/electric+field+strength
     
    Last edited by a moderator: Apr 17, 2017
  7. Oct 6, 2015 #6
    And we are told to put negative charge there(in electric field ) in that case we can ,right?
     
  8. Oct 6, 2015 #7

    ehild

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    You can put a negative charge into any electric field. Do it. And what do you do after?
     
  9. Oct 6, 2015 #8
    But it can not be called "test charge",right?
     
  10. Oct 6, 2015 #9

    Dale

    Staff: Mentor

    Correct. A test charge is always positive.
     
  11. Oct 8, 2015 #10
    The potential difference between points A and B (or any two points) does not depend on the test charge used. The relation is standard

    V_B - V_A = - \int_A^B \overrightarrow{E}.d\overrightarrow{s}

    You can use a negative or positive test charge and multiply both sides of the above relation with the charge. The equation will change to a relation between the difference of potential energies, and the work done by the electrostatic force. In case my Latex is not clear, there is a dot product between the electric field and the infinitesimal displacement in the integral, and the integral is over any path joining points A and B. It does not matter whether you "call" it a test charge or not. The fact is that you can use either positive or negative charges.
     
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