91Ó°ÊÓ

In a hollow metal sphere has inner radius is$a,$outer radius is$b,$and the electric field strength inside the metal as a function of the radius is $r$ from the center.

The current density as a function of $r$. The current is the exact spherical cross-section, as

$I=J4\mathrm{Ï€}{r}^2$

Then finally,

$J=\frac{I}{4\mathrm{Ï€}{r}^2}$

So, the electric field is

$E=\frac{1}{\mathrm{σ}}J$

Therefore, an expression for the electric field strength inside the metal as a function of the radius$r$from the center is:

$E=\frac{I}{4\mathrm{π}\mathrm{σ}{r}^2}$.

The electric field strength at the inner surfaces of a copper sphere is$a=1.0cm$ and the outer surface of a copper sphere is $b=2.5cm$.

For copper,$\mathrm{σ}=6.0×{10}^7{\mathrm{Î\copyright }}^{−1}{\mathrm{m}}^{−1}$
The inner surfaces of a copper sphere$r=a$. so,
$E_{\text{in}}=\frac{I}{4\mathrm{π}\mathrm{σ}{a}^2}$

$=\frac{25A}{4×\mathrm{Ï€}×6.0×{10}^7{\mathrm{Î\copyright }}^{−1}{\mathrm{m}}^{−1}×{(0.01)}^{2}m}$

$=3.3×{10}^{−4}\mathrm{V}/\mathrm{m}$

The outer surfaces of a copper sphere $r=b$. so,

$E_{\text{out}}=\frac{I}{4\mathrm{π}\mathrm{σ}{b}^2}$

$=\frac{25A}{4×\mathrm{Ï€}×6.0×{10}^7{\mathrm{Î\copyright }}^{−1}{\mathrm{m}}^{−1}×{(0.025)}^{2}m}$

$=5.3×{10}^{−5}\mathrm{V}/\mathrm{m}$

The electric field strength at the inner surfaces of a copper sphere is $3.3×{10}^{−4}\mathrm{V}/\mathrm{m}$and the electric field strength at the outer surfaces of a copper sphere is $5.3×{10}^{−5}\mathrm{V}/\mathrm{m}$.