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Understanding on potential near a charge

  1. Jun 15, 2012 #1
    From the fig 1 & 2 in the attachment.

    i) what is the sign of potential difference in fig.1 Vp – VQ
    ii) And in fig.2 VQ – VP

    For simplicity, lets take value at P,Q and R as in the figure(attach) for a unit +ve charge.

    Both the answer are positive. I can understand i) is positive but not ii), as that means VQ > VP. I know this is what we get when we use the formulae V = Q/4ε°r. But my understanding near the source charge, the potential to do work should always be greater i.e Vp > VQ .

    Electric potential at a place is defined as the potential of a unit +ve charge to do work when placed at that point and therefore point near the source would always have greater potential.


    What is wrong on my understanding?

    Also, can't we take unit -ve charge as test charge, since it too has the same potential to do work as the unit +ve charge, only direction of work done is opposite.


    Thanks.
     

    Attached Files:

  2. jcsd
  3. Jun 15, 2012 #2

    Simon Bridge

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    What is this supposed to be showing us? I am reading these as lines of charges without units for charge or distance given. I don't understand what the arrow is supposed to represent.
    If I take these points to be the locations of point charges than the PDs will be undefined.
    A +ve charge is attracted to a negative charge so you have to do work to take it away. Thus points close to a negative charge have a low potential compared with points a similar distance from the same strength positive charge.

    The sign of the potential depends on your reference point.

    Yes, that's quite valid ... the physics works out the same you just have to pay close attention to the signs. The shape of the potential vs space graph will just be upside down compared with the convention.
     
  4. Jun 15, 2012 #3
    these arrows are line of electric field of +ve and -ve charge Q.
    these are points where its potentials are Vp and Vq.
    This is where I am confused. Isn't the same work is done by the test charge if placed at that same point. That is same potential magnitude at same distance from the source. Only direction changes but since potential is a scalar quantity direction shouldn't matter.
     
    Last edited: Jun 15, 2012
  5. Jun 15, 2012 #4

    Simon Bridge

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    Ah - so there is one charge indicated by the circles with the + and - sign respectively. Note: the letter Q is commonly used to denote a charge.

    In one case work is done on the test charge and in the other work is done by the test charge. Direction matters because work is determined by the change in potential difference, not the magnitude of the potential.
     
  6. Jun 15, 2012 #5
    I'm little confused with work done still.

    Like when you say work is done ON THE TEST CHARGE - is this work done on the test charge BY AN EXTERNAL AGENT in pushing it against the electric field.

    And work done BY THE TEST CHARGE - is this work done on the test charge DUE TO SOURCE CHARGE i.e either repellsion or attraction.
     
  7. Jun 15, 2012 #6

    Simon Bridge

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    I think you need to re-examine the relationship between work and energy then revisit the electrostatic case.
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html

    The potential is defined as the work (per unit charge) done against the electric field. Thus changes with the field - in the direction the test charge wants to go by itself - would show up as negative.

    Note: scalar's can be negative numbers. The negative sign does not have to mean direction.
     
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