91Ó°ÊÓ

A phonograph record carries a uniform density of "static electricity" $\mathrm{σ}$and rotates at angular velocity $\mathrm{Ó\neg }$.

A uniformly charged solid sphere, of radius Rand total charge Q,is centered

at the origin and spinning at a constant angular velocity $\mathrm{Ó\neg }$about the zaxis.

The surface current density of a surface charge density $\mathrm{σ}$moving with a speed v is

$\mathbf{K}\mathbf{=}\mathbf{σ\pm ¹}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

The volume current density of a volume charge density $\mathrm{ÒÏ}$moving with a speed v is

$\mathit{J}\mathbf{=}\mathit{ÒÏ}\mathbf{v}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

The speed of the charge density rotating with angular velocity$\mathrm{Ó\neg }$in a circle of radius r is

$v=\mathrm{Ó\neg }r$

Substitute this in equation (1) to get

$K=\mathrm{σ}\mathrm{Ó\neg }r$

Thus, the surface current density is $\mathrm{σ}\mathrm{Ó\neg }r$.

The volume charge density of a sphere of radius R uniformly charged with a charge Q is

$$\begin{gathered} \mathrm{ÒÏ}=\frac{Q}{{\displaystyle \frac{4}{3}\mathrm{Ï€}{R}^3}} \\ =\frac{3Q}{4\mathrm{Ï€}{R}^3} \end{gathered}$$

The speed at any point in the sphere rotating with angular velocity $\mathrm{Ó\neg }$about the z at a radial distance r and making angle $\mathrm{θ}$with the z axis is

$v=\mathrm{Ó\neg }r\sin \mathrm{θ}$

Substitute these in equation (2) to get

$$\begin{gathered} J=\frac{3Q}{4\mathrm{Ï€}{R}^2}×\mathrm{Ó\neg }r\sin \mathrm{θ} \\ =\frac{3Q\mathrm{Ó\neg }r\sin \mathrm{θ}}{4\mathrm{Ï€}{R}^3} \end{gathered}$$

Thus, the volume current density islocalid="1657774147962" $\frac{3Q\mathrm{Ó\neg }r\sin \mathrm{θ}}{4\mathrm{Ï€}{R}^3}$.