Help with derivation of torque equation T = p x E

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  • #1
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[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]?
 

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  • #2
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You need to tell us what you are trying to accomplish. What do your variables represent, what is your reasoning behind each step, etc.
 
  • #3
jtbell
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[itex]\vec{\tau}[/itex]=[itex]\vec{F}[/itex]x[itex]\vec{r}[/itex]
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.
 
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  • #4
Philip Wood
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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]
 
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  • #5
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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.
thanks this answers everything


now how do i close threads?
 
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  • #6
ZapperZ
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thanks this answers everything


now how do i close threads?
You don't.

Zz.
 

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