Electric Potential Energy

Discussion in 'Homework Help' started by screen1988, Mar 28, 2013.

  1. screen1988

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    Mar 7, 2013
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  2. WBahn

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    It is the work done to move a charge Q and a charge q from infinite separation (where it is defined that there is zero electric potential energy) to a separation distance of r.
     
  3. screen1988

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    Mar 7, 2013
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    Why we need to define q is in infinity? I can't gasp it because the concept infinity is very abstract to me. For example, I have a charge q in my pocket then I take it and put it in the vicinity of this source charge Q, the distance between Q and q is r. Then how can I apply the concept above because my charge q isn't in infinity?
     
  4. WBahn

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    If you start at something other than infinity, then you are starting out with a non-zero electric potential and the work to take them to their final separation of r has to be added to that initial potential.

    The reason than infinite separation is used to define the zero potential energy is because if you take to charges and place them in empty space at any finite distance from each other, then they will exert a force on each other and it will require work to move them to a different separation. The only time that the force between them goes to zero is in the limit that they are infinitely far apart.

    Now, you don't have to get them too far apart before the error associated with not taking into account the rest of the potential from there to infinity is negligible.
     
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  5. screen1988

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    Thanks, now I want to ask you one more question.
    For example, there is a charge Q that is placed at A point. Now I have an other charge q. F1 is coulomb force Q exert to q. F2 is the force that I use to bring the charge q from infinity to B that is R m from A. To caculate the work I have to use to bring the charge q from infinity to B, we don't need to compute this work directly. We can calculate the work of F1 force to move the charge q from B to infinity.
    \int _{\infty }^{R}\vec{F_{1}}\vec {dr}=\int _{R}^{\infty }\vec  {F_{2}}d\vec  {r}
    Now how can I know that these works are equal?
     
    Last edited: Mar 31, 2013
  6. WBahn

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    Because F1 and F2 are equal and opposite, thus F1 = -F2 and does nothing more than swap the limits of integration.

    Remember, the work done if under the constraint of being quasi-static, meaning that it is the force needed to move the charge without accelerating it. The "quasi" part is a loophole that let's you accelerate a bit at the beginning and declerate it the same amount at the end.
     
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  7. screen1988

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    I think this is the problem that I can't understand. If accelerate of the charge is zero then according to Newton law ƩF=0 => F1=-F2. But why we have to calculate the work done under the constrain of being quasi-static?
     
    Last edited: Apr 1, 2013
  8. WBahn

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    Because if it isn't then you have other types of energy that are changing, namely kinetic energy. You want all of the work done to go only into electric potential energy.
     
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  9. screen1988

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    Oh, I see it now but I have a new problem. If we mustn't accelerate the charge and initially, the charge is at rest, it's velocity is zero, then the velocity of the charge keeps at zero and we can't move the charge from infinity to a specific point?

    Ah, I have just re-read your post #6:
    Now I think I had the answer! You mean that we can initially accelerate a bit and therefore, its velocity is very small => kinetic energy is also very small.
    I am waiting for your confirm.
     
  10. WBahn

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    Yes, that is the point behind quasi-static.
     
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