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Help with derivation of torque equation T = p x E

  1. Jul 19, 2013 #1
    [itex]\vec{\tau}[/itex]=[itex]\vec{F}[/itex]x[itex]\vec{r}[/itex]

    [itex]\vec{F}[/itex]=[itex]\vec{E}[/itex]q

    [itex]\vec{p}[/itex]=q[itex]\vec{d}[/itex]

    [itex]\vec{\tau}[/itex]=([itex]\vec{E}[/itex]q)xr=[itex]\vec{E}[/itex]x[itex]\vec{p}[/itex]

    so then how do we end up with [itex]\vec{\tau}[/itex]=[itex]\vec{p}[/itex]x[itex]\vec{E}[/itex]?
     
  2. jcsd
  3. Jul 19, 2013 #2
    You need to tell us what you are trying to accomplish. What do your variables represent, what is your reasoning behind each step, etc.
     
  4. Jul 19, 2013 #3

    jtbell

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    Staff: Mentor

    No, the definition is [itex]\vec \tau = \vec r \times \vec F[/itex]. At least, that's the definition in every textbook I've ever used.
     
  5. Jul 20, 2013 #4

    Philip Wood

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    Gold Member

    The torque, [itex]\vec{\tau} [/itex], about a point, O, of a force, [itex]\vec{F} [/itex], acting at point P which is located at position [itex]\vec{r} [/itex] from O is DEFINED AS
    [tex]\vec{\tau} = \vec{r} \times \vec{F}.[/tex]
    As with all quantities in Physics, we would't define it as we do, unless what we've defined is useful. In this case, there is a simple relationship between the torque about O of a force acting on a particle, and the rate of change of the particle's angular momentum, [itex]\vec{J} [/itex] about O.

    Namely
    [tex]\vec{\tau} = \frac{d\vec{J}}{dt}.[/tex]
    [itex]\vec{J} [/itex] is defined exactly analogously to [itex]\vec{\tau} [/itex], but substituting the particle’s linear momentum, [itex]\vec{p} [/itex], for the force, [itex]\vec{F} [/itex], on the particle, so...
    [tex]\vec{J} = \vec{r} \times \vec{p}.[/tex]
     
  6. Jul 21, 2013 #5
    thanks this answers everything


    now how do i close threads?
     
    Last edited: Jul 21, 2013
  7. Jul 21, 2013 #6

    ZapperZ

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    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    You don't.

    Zz.
     
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