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Electric potential question

  1. Feb 12, 2012 #1
    If the pos charged particle moves from point 1 to 2 I believe that the electric potential energy will increase.

    My reasoning for this is because I know electric potential energy is found by knowing how much work it takes to move a charged particle from point 1 to point 2. Because a pos charge at point 1 is attracted to the two neg charges near point 2, inorder to move the pos charge to point 2 will not require as much work as it will to move the charged particle from point 2 back to point 1... is this reasoning correct?
     

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  3. Feb 12, 2012 #2

    SammyS

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    No. Well, the reasoning is not so bad, but the answer is wrong.

    If it takes less work to move the charged particle from point 1 to point 2, than it does to move it from point 2 to point 1,
    then why do you say the electric potential is higher at point 2 ?
     
  4. Feb 12, 2012 #3
    I say the electric potential at point to is highter because more work is required to move it from 2 back to 1.
     
  5. Feb 12, 2012 #4

    SammyS

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    So the electric potential is higher at point 1.

    (In fact, it takes negative work to move the particle from point 1 to point 2.)
     
  6. Feb 12, 2012 #5
    In my book I just read about the change in electric potential is equal to the product of Delta(V) and q. So I guess If i can determine what Vinitial and Vfinal is I will know if there is a decrease or an increase....
    But how can I relate Vi at point 1 where there are pos charges and Vf at point 2 where there are neg charges? Will Vi be a positive value and Vf be a negative value?
     
  7. Feb 12, 2012 #6

    SammyS

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    That's one way to calculate ΔV.

    The definition is Work per unit charge:
    If it takes an amount of work, W, to move a charge of Q from point A to point B, then the potential difference at point B relative to point A, is ΔV = VB - VA = W/Q .​
    In the case of electric potential, just as in the case of potential energy in general, what's important is the difference in potential at two locations,
    not some value in an absolute sense.
     
  8. Feb 12, 2012 #7
    Work can be defined as W = F d, where both are vectors and you take the scalar product - F = force able to do work, d = displacement on body at which work was applied. Looking at this definition, it takes the same magnitude of work getting from point A to B as from B to A. The displacement is the same, just in the other direction.

    In this case, electric potential energy decreases in going from A to B. This is because external work done = change in potential energy and the fact that negative work must be done. You can make sense of this fact by looking at the force that would be exerted on the positive particle by the electric fields throughout the displacement. A positive charged particle is going to experience a net force to the negative charged particles. In order for an external force to "move" the particle from A to B it (in net) has to be in the opposite direction of this electrostatic force. When ever the displacement is in the opposite direction of the force, it is said to be negative work. The electrostatic force does positive work on the particle equal in magnitude to the external work, assuming energy is not changed by other means like having a different final kinetic energy.

    So assume an external force is applied to the particle equal to and opposite in direction to the electrostatic force. In this state, no work is done as there is no displacement (assuming 0 kinetic energy.) Then imagine the external force is weakened for an small time interval so that the particle begins to accelerate to the right. The external force can then be increased again, equal to the electrostatic force having the particle now moving at a constant rate.

    At this point, the external force as done some negative work and the electrostatic force as done positive work. In fact, at this point it has done more positive work than the external force has done negative. That is, the particle now has kinetic energy and it is equal to the positive work done - the negative work done.

    Now assume it is about to approach point B. The external force must now increase itself for a small time interval to bring the particle to a stop. Once this is done, kinetic energy is decreased again, leaving that energy amount into the negative work done by the external force. In other words, it has taken the kinetic energy out as the other force put it in, in the beginning. So in the end, both forces did equal work, but, work applied = - work of conservative force (the electrostatic force.)

    Even though this seems to be a very specific example of what is going on, it is true no matter how the particle is moved from point A to B.

    It's also intuitive to think a positive charged particle has more potential when closer to positive charges as they repel each other. The closer they are, the more potential they have to spread away.
     
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