What is the needle’s dipole moment?

[tony873004]

A needle suspended from a string hangs horizontally. The electric field at the needle’s location is horizontal with a magnitude 3.7*103 N/C and is at an angle of 30° with the needle. There is no net electrical force acting on the needle, but the string exerts a torque of 3.7*10-3 to hold the needle in equilibrium. What is the needle’s dipole moment?

I know that <br /> \overrightarrow \tau = \overrightarrow {\rm{p}} \times \overrightarrow {\rm{E}} \,\,<br />

But I can't simply re-write it as <br /> \frac{{\overrightarrow \tau }}{{\overrightarrow {\rm{E}} }} = \overrightarrow {\rm{p}} \,\,\,\,<br />
, can I? I don't think you can do this to a cross-product. How can I solve for the dipole moment?

 

[Winzer]

You will only be able to that if p and E are perpendicular, in which case they are not. Do you remember how to find the magnitude of a cross product? Sin(x) maybe?mm?

 

[tony873004]

I know its |A x B| = |A||B| sin(x). But in this problem, I don't know what A is. That's what I'm trying to solve for.

|A x B|
------- = |A|. This looks like a dead end.
|B|

 

[Winzer]

how about
p=\frac{\tau}{sin(\theta) E} ?

 

[tony873004]

If I do that, then I'm dividing units of Nm by N/C, which gives me mC. The book doesn't talk about this, and my class notes are not very good, but I think that is the right unit for dipole moment, meters*coulombs?

If so, I get 3.7e-4/sin(30) *3.7e3 = 2.738 mC for the answer.

If this is correct, don't go away, because I'm still not confident I understand this.

 

[Winzer]

m*C are the correct units.
I see some inconsistencies in your signs and values for your exponents to the ones you stated originally.

 

[tony873004]

Yes, my original question should have been 3.7*10^3 N/C and 3.7*10^-4 for torque. (I imagine you knew that 103 meant 10^3 :) )

 

[tony873004]

t is a vector, and so is E. But in the division that got me 2.738 mc, I divided only their magnitudes. Was this right? I'm guessing they want p to be a vector with direction. It will either be into or out of the board. How do I know which?

 

[tony873004]

Ok, now I follow your chain of logic. Thanks for your help. I've got a scalar as an answer, and I still need to assign it a direction. Any thoughts?

<br /> \begin{array}{l}<br /> \overrightarrow \tau = \overrightarrow {\rm{p}} \times \overrightarrow {\rm{E}} \, \\ <br /> \\ <br /> \left| {\overrightarrow {\rm{p}} \times \overrightarrow {\rm{E}} } \right| = \left| {\overrightarrow {\rm{p}} } \right|\left| {\overrightarrow {\rm{E}} } \right|\sin \theta \\ <br /> \\ <br /> \overrightarrow \tau = \left| {\overrightarrow {\rm{p}} } \right|\left| {\overrightarrow {\rm{E}} } \right|\sin \theta \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| {\overrightarrow {\rm{p}} } \right|\, = \frac{{\,\overrightarrow \tau }}{{\left| {\overrightarrow {\rm{E}} } \right|\sin \theta }}\,\,\, \\ <br /> \\ <br /> \left| {\overrightarrow {\rm{p}} } \right|\, = \frac{{3.7 \times 10^{ - 4} {\rm{N}} \cdot {\rm{m}}}}{{\sin \theta \cdot 3.7 \times 10^3 {\rm{N/C}}}} = 2.7{\rm{ C}} \cdot m \\ <br /> \,\,\,\, \\ <br /> \end{array}<br />