Help with derivation of torque equation T = p x E

[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.