A hollow metal sphere has$6\mathrm{cm}$and $10\mathrm{cm}$inner and outer radii, respectively. The surface charge density on the inside surface is $-100\mathrm{nC}/{\mathrm{m}}^2$. The surface charge density on the exterior surface is $+100\mathrm{nC}/{\mathrm{m}}^2$. What are the strength and direction of the electric field at points $4$,$8$and$12\mathrm{cm}$ from the center?

The sphere's surface charge density is equal to the sphere's charge divided by its area.

$\mathrm{\eta }=\frac{\mathrm{Q}}{\mathrm{A}}$

The inner surface's surface charge islocalid="1648918075927" ${\mathrm{\eta }}_{\text{in}}=-100\mathrm{nC}/{\mathrm{m}}^2$.

Radius is localid="1648918222133" $\mathrm{r}=6\mathrm{cm}$

On the inner surface, there is a charge,

${\mathrm{Q}}_{\text{in}}=4{\mathrm{\pi r}}_{\mathrm{i}}^2{\mathrm{\eta }}_{\text{in}}$

$=4\mathrm{\pi }(0.06\mathrm{m}{)}^2\left(-100\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2\right)$

$=-4.5\times {10}^{-9}\mathrm{C}$

The outside sphere's surface charge islocalid="1648918082193" ${\mathrm{\eta }}_{\text{ext}}=100\mathrm{nC}/{\mathrm{m}}^2$.

Radius is localid="1648918228090" $\mathrm{r}=10\mathrm{cm}$.

The outside sphere's charge is,

${\mathrm{Q}}_{\text{ext}}=4{\mathrm{\pi r}}_{\mathrm{i}}^2{\mathrm{\eta }}_{\text{in}}$

$=4\mathrm{\pi }(0.10\mathrm{m}{)}^2\left(100\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2\right)$

$=1.25\times {10}^{-8}\mathrm{C}$

Electric flux,

${\mathrm{\Phi }}_{\mathrm{e}}=\mathrm{EA}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{ϵ}}_{\mathrm{o}}}$

$\mathrm{E}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{r}}^2}$

For point localid="1648918243870" $\mathrm{r}=4\mathrm{cm}$:

The charge enclosed islocalid="1648918332859" $4.5\times {10}^{-9}\mathrm{C}$.

Because the negative charges are positioned on the inner surface at pointlocalid="1648918251163" $\mathrm{r}=6\mathrm{cm}$, it is positive, not negative.

The electric field is,

${\mathrm{E}}_{\mathrm{r}=4\mathrm{cm}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{{\mathrm{Q}}_{\mathrm{r}=4\mathrm{cm}}}{{\mathrm{r}}^2}$

$=\frac{1}{4\mathrm{\pi }\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)}\frac{\left(4.5\times {10}^{-9}\mathrm{C}\right)}{(0.04\mathrm{m}{)}^2}$

$=2.5\times {10}^4\mathrm{N}/\mathrm{C}$

Because the electric field is positive, it is directed outward.

For point localid="1648918259540" $\mathrm{r}=8\mathrm{cm}$:

The electric field inside the conductor is zero.

There is no net charge since there is none$\mathrm{Q}=0$.

The electric field is zero,

${\mathrm{E}}_{\mathrm{r}=8\mathrm{cm}}=0$

For point $\mathrm{r}=12\mathrm{cm}$:

For the outer surface, the enclosed charge is the same$1.25\times {10}^{-8}\mathrm{C}$. Because everything is neutral inside the sphere.

The electric field is

${\mathrm{E}}_{\mathrm{r}=12\mathrm{cm}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{{\mathrm{Q}}_{\mathrm{r}=12\mathrm{cm}}}{{\mathrm{r}}^2}$

$=\frac{1}{4\mathrm{\pi }\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)}\frac{\left(1.25\times {10}^{-8}\mathrm{C}\right)}{(0.12\mathrm{m}{)}^2}$

$7.9\times {10}^4\mathrm{N}/\mathrm{C}$

Because the electric field is positive, it is directed outward.