Why is torque τ = p × E instead of τ = E × p?

[iScience]

$\vec{\tau}$=$\vec{F}$x$\vec{r}$

$\vec{F}$=$\vec{E}$q

$\vec{p}$=q$\vec{d}$

$\vec{\tau}$=($\vec{E}$q)xr=$\vec{E}$x$\vec{p}$

so then how do we end up with $\vec{\tau}$=$\vec{p}$x$\vec{E}$?

 

[Mmm_Pasta]

You need to tell us what you are trying to accomplish. What do your variables represent, what is your reasoning behind each step, etc.

 

[jtbell]

$\vec{\tau}$=$\vec{F}$x$\vec{r}$

No, the definition is $\vec \tau = \vec r \times \vec F$. At least, that's the definition in every textbook I've ever used.

 

[Philip Wood]

The torque, $\vec{\tau}$, about a point, O, of a force, $\vec{F}$, acting at point P which is located at position $\vec{r}$ from O is DEFINED AS
$$\vec{\tau} = \vec{r} \times \vec{F}.$$
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, $\vec{J}$ about O.

Namely
$$\vec{\tau} = \frac{d\vec{J}}{dt}.$$
$\vec{J}$ is defined exactly analogously to $\vec{\tau}$, but substituting the particle’s linear momentum, $\vec{p}$, for the force, $\vec{F}$, on the particle, so...
$$\vec{J} = \vec{r} \times \vec{p}.$$

 

[iScience]

No, the definition is $\vec \tau = \vec r \times \vec F$. At least, that's the definition in every textbook I've ever used.

thanks this answers everythingnow how do i close threads?

 

[ZapperZ]

thanks this answers everything


now how do i close threads?

You don't.

Zz.