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Energy change for a charged particle

  1. Aug 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Without introducing any coordinates or basis vectors, show that, when a charged particle interacts with electric and magnetic fields, its energy changes at a rate $$\frac{dE}{dt}=\vec{v}\cdot \vec{E} $$

    2. Relevant equations
    ##E_{kin} + E_{pot}= En =## const (1)
    ##E_{pot}=\vec{r}\cdot\vec{E}## (2)...suppose homogenic electric field and also that magnetic field does not affect magnitude of velocity. ##E## describes electric field

    3. The attempt at a solution
    First of all I suppose they ask on kinetic energy. Total energy is conserved. Then I put (2) in (1) and differentiate to get:
    $$\frac{dE_k}{dt}=-\frac{d}{dt}(\vec{r}\cdot\vec{E}). $$
    I hope, a minus sign vanish in opposite definition of vectors \vec{r}. I hope in homogenic electric field is ##\vec{E}=##const. than
    $$\frac{dE_k}{dt}=\vec{v}\cdot\vec{E}. $$
    Is it allright?
     
  2. jcsd
  3. Aug 2, 2017 #2

    TSny

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    You don't need to assume a uniform E field, and therefore you don't need to assume that -r⋅E is the potential energy.

    The rate at which the energy of the particle increases is the rate at which the electromagnetic force does work on the particle. So, consider the electromagnetic force and the amount of work it does when the particle undergoes a small displacement dr.
     
  4. Aug 2, 2017 #3
    Aha I see, so ##\frac{dE}{dt}=\frac{d}{dt}\int{\vec{E}d\vec{r}}+\frac{d}{dt}\int{(\vec{v}\times\vec{B})d\vec{r}}##, but now ##\vec{E}## is not constant and integration cannot be done so easy. How may I now get a result? Honestly now I'm not sure why is the latter term zero (how prove it).
     
  5. Aug 2, 2017 #4

    TSny

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    I don't understand why you are integrating. To get the rate of change of energy, you just need to consider an infinitesimal displacement dr of the particle during an infinitesimal time dt. If you have a force F that acts on a particle, how much work does F do on the particle when the particle moves through a displacement dr?
     
  6. Aug 2, 2017 #5
    What do you know about magnetic fields and work? What is the direction of ##\vec{v}\times\vec{B}## and ##\vec{r}##?
     
  7. Aug 2, 2017 #6
    Ou yes ##dE=dW=\vec{E}d\vec{r}## sorry I have to thing.
     
  8. Aug 2, 2017 #7
    Hmm again, more thinging. Direction of ##\vec{v}## is same as ##\vec{r}##. Cross product ##\vec{v}\times\vec{B}## results in ortogonal to both of them i.e. also to ##\vec{r}## i.e. dot product is zero.
    Thank you both of you.
     
  9. Aug 2, 2017 #8

    TSny

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    Right. The magnetic force doesn't do any work. So, the change in energy is due solely to the work done by the electric force.
     
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