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A lot of questions to ask about E

  1. Mar 1, 2013 #1
    What is electric potential energy actually?More generally,what is potential energy?
    F=-dU/dr
    How to interpret this equation?
    690ab6aada2735237ee2fba30dc1698b.png
    From the above eq,what is the physical meaning of V?
    Furthermore,E=-∇V
    What,again,is the physical meaning of this eq.
    If there is a point,let say S,with zero potential(e.g mid point of a line joining two charges with same magnitude but opposite signs)
    Does it mean we do no work to move a unit charge from the reference point to S?
    I am extremely confused when studying E&M
    thank you
     
  2. jcsd
  3. Mar 1, 2013 #2

    jtbell

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    Staff: Mentor

    When a positive charge q moves from point A to point B, if ΔV is positive (that is, if your integral evaluated with A as the lower limit and B as the upper limit, is positive), then you (or some other external agent) have to do net work W = qΔV on the charge in order to move it from A to B.

    This assumes the kinetic energy of the charge is the same at both points; if it's not, then the work must also include the change in KE as per the work-energy theorem.

    If either q or ΔV (but not both!) is negative, then W is negative: the charge does work on you. If both q and ΔV are negative, W is positive, and you do the work.
     
  4. Mar 1, 2013 #3
    then W is negative: the charge does work on you.
    what do u mean by the charge does work on you?It means some energy is transferred to me?
    It sounds weird
     
  5. Mar 2, 2013 #4
    think of it this way... if ΔV>0, then there is a field directed away from that point. So to move a +q there, you have to 'push' against it, and so you do the work.
    but if ΔV<0, then the field is directed towards that point. So a +q would 'want to go' there! so you don't have to provide it any energy and so you don't do any work in that case. the energy is provided by the source of ΔV. and so its said that 'work is done on you'... imagine that the charge 'pulls' you into that region! it does work on you.
     
  6. Mar 2, 2013 #5
    thanks
    but how about the other questions
     
  7. Mar 2, 2013 #6

    BruceW

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    Homework Helper

    "What is electric potential energy actually?More generally,what is potential energy?"
    we can appreciate that there is energy associated with motion. This is called kinetic energy. But there are also forms of energy that are not associated with motion. They come under the label 'potential energy'. Energy can be transferred from one type to another. So you can think of all the forms of potential energy as having the possibility to cause motion (i.e. can be turned from potential energy into kinetic energy). Therefore, one way to think about potential energy, is that it is 'stored' energy.
     
  8. Mar 2, 2013 #7

    BruceW

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    Homework Helper

    "From the above eq,what is the physical meaning of V?"
    V is the electric potential. essentially it is the (electrical) potential energy per charge. So if you have a particle of charge Q, then VQ is the potential energy of that particle. If you have a particle such as this, then you can multiply both sides of the equation you wrote, by the charge of the particle, and you get:
    [tex]- Q \ \Delta V = \int \vec{F} \cdot d \vec{l} [/tex]
    Where F is the force on the particle due to electric field. And on the left-hand side of the equation, we see this is just the loss of electric potential energy. As others have said, if the particle moves oppositely to the electric force, then the electric potential energy will increase. Either way, we can see this is an equation telling us how to calculate the transfer of energy to/from the electric potential energy.
     
  9. Mar 3, 2013 #8
    What is E=-∇V?? First of all search Wikipedia for 'gradient' (in mathematics of course). Then you'll understand that this is a shorthand for writing
    E= - i*dV/dx - j*dV/dy - k*dV/dz.
    What does this mean physically??
    Its the way of expressing how the potential changes with position in space. Note that potential has been defined as V= - P[itex]\vec{E}[/itex].[itex]\vec{dr}[/itex]...
    So, naturally [itex]\vec{E}[/itex] comes from taking the derivative of the potential. But there's another bit. The potential is a scalar quantity obtained from the dot-product of two vectors. So in order to get back the vector just by differentiating, the ∇ operator has been used, since it has the unit vectors embedded within it.
    So that equation gives us a vector from a scalar!
    And it tells us how the potential changes from one position to another, giving us the electric field!
    Note that in regions where V does not change, there is no electric field, like the inside of a charged conductor... think about it.
     
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