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yssi83
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Homework Statement
Link to assignment
The fact that it is imperfectly conducting is supplied so that the charges in the sphere will move with the same angular velocity.
The B-field induced by the moving charges will be disregarded.
Homework Equations
[tex]\vec{F}[/tex]=Q[[tex]\vec{E}[/tex]+[tex]\vec{v}[/tex][tex]\times[/tex][tex]\vec{B}[/tex]]
[tex]\vec{F}[/tex]=0 => [tex]\vec{E}[/tex]=-[tex]\vec{v}[/tex][tex]\times[/tex][tex]\vec{B}[/tex]
The equations above apply inside the sphere. And lead to the E-field inside.
The Attempt at a Solution
I have found the E-field inside, and the volume charge density inside.
E=B[tex]\omega[/tex]x[tex]\hat{x}[/tex]+B[tex]\omega[/tex]y[tex]\hat{y}[/tex]
[tex]\rho[/tex]=-2B[tex]\omega[/tex][tex]\epsilon_{o}[/tex]
This gives the potential inside (have set the potential at r=0 to V_0)
V(r)= V_0 - [tex]\frac{1}{2}[/tex]B[tex]\omega[/tex]r^2 (sin [tex]\theta[/tex])^2
The total charge inside and the total charge on the surface are excactly equal but opposite as one would expect. total charge = 0
Q_inside = -[tex]\frac{8}{3}[/tex][tex]\pi[/tex][tex]\epsilon_{0}[/tex]B[tex]\omega[/tex]R^3
Q_surface = [tex]\frac{8}{3}[/tex][tex]\pi[/tex][tex]\epsilon_{0}[/tex]B[tex]\omega[/tex]R^3
Then the problem is to find the potential and E_field outside the sphere. I can't seem to figure out if the charges on the surface are the only ones contributing.
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