Volume distribution of charge

[Ripperbat]

Homework Statement

A rod of dielectric material is spun about it's axis with angular velocity $$\omega$$. A uniform magnetic field B exists in a direction along the axis of the bar. Determine a charge distribution which produces the same electric field as does the rotating rod. The electric susceptibility of the material is x_E

Homework Equations

The Attempt at a Solution

The force on a charge q at a distance r is F=q*$$\omega$$*r*B and the electric field is E=r*B*$$\omega$$.
The polarization of the rod at distance r is P=x_E*$$\epsilon$$₀*$$\omega$$*B*r.
So the volume distribution of charge should be -div P = -($$\partial$$P_r) / ($$\partial$$*r). But according to the book it should be
-div P = -($$\partial$$P_r) / ($$\partial$$*r) - (P_r) / r .


Could somebody please explain why it should be that way?? thx

 

[Redbelly98]

So the volume distribution of charge should be -div P = -($$\partial$$P_r) / ($$\partial$$*r). But according to the book it should be
-div P = -($$\partial$$P_r) / ($$\partial$$*r) - (P_r) / r .


Could somebody please explain why it should be that way?? thx

That is the divergence in cylindrical coordinates.

http://hyperphysics.phy-astr.gsu.edu/Hbase/diverg.html#c3

 

[Ripperbat]

Thank you thank you thank you! :!)