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Conservative Nature of Electric forces

  1. Nov 11, 2009 #1
    My textbook says for every electric field due to a static charge distribution, the force exerted by that field is conservative.

    If that force is used to accelerate a charged particle, wouldn't the particle disapate some of its energy, as EMR? therefore you cannot say U_a + K_a = U_b + K_b as some of the particles energy would have gone

    can anyone explain what that above statement means with respect to the motion of the particle?

    thankyou :)
     
  2. jcsd
  3. Nov 11, 2009 #2

    gabbagabbahey

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    I've highlighted the key word in your textbook's statement....In electrostatics charges don't radiate.
     
  4. Nov 11, 2009 #3
    i understand that, the charge distribution is static, but there was a question, where a charge is accelerated by a static charge distribution.

    Say a charge is being accelerated by a static charge distribution, wouldn't that charge lose some energy?

    such that is incorrect to say Ua + Ka = Ub + Kb, since the charge accelerates
     
  5. Nov 11, 2009 #4

    gabbagabbahey

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    Sure, the accelerating charge will radiate some energy away. However, this is unimportant. Energy conservation and conservative forces are not the same thing. What is the actual definition of a conservative force?
     
  6. Nov 11, 2009 #5
    A conservative force is a force that conserves energy, such that the work done by a conservative force can be released and is reversible.

    So i suppose as the electron moves away some of its kinetic energy is dissipated as EMR but its Kinetic energy is also turned into potential energy which can be recovered.

    But it is still wrong to say, Ua + Ka = Ub + kb as some energy is lost in the movement from a to b, i think the question assumes that this loss is negligible don't you think?
     
  7. Nov 11, 2009 #6

    gabbagabbahey

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    That's a somewhat inaccurate definition.

    This is a much better definition....From this definition, you can say 3 mathematically equivalent things:

    (1)The work done by the force over any simple, closed path is zero (otherwise it would be an irreversible process)

    [tex]W=\oint\textbf{F}\cdot d\textbf{r}=0[/tex]

    (2)The curl of [itex]\textbf{F}[/itex] must vanish (you can derive this easily using Stokes's Theorem and property 1)

    [tex]\mathbf{\nabla}\times{\textbf{F}}=0[/tex]

    (3)The force can be written as the (negative by convention) gradient of a single-valued, contiuous potential function

    [tex]\textbf{F}=-\mathbf{\nabla}U[/tex]


    Now, surely any electrostatic force satisfies property 2 (and hence the other two properties as well), right?
     
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