91影视

Figure is,

Electric flux is,

${\mathrm{桅}}_{\mathrm{e}}=\mathrm{EA}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{系}}_{\mathrm{o}}}$

$\mathrm{E}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{系}}_{\mathrm{o}}\mathrm{A}}..1$

The charge of the sphere divided by the volume of the sphere is the volume charge density of the surface.

${\mathrm{Q}}_{\text{in}}=\mathrm{蚁痴}=\mathrm{蚁}\left(\frac{4}{3}{\mathrm{蟺谤}}^3\right)$

$\mathrm{E}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{系}}_{\mathrm{o}}\mathrm{A}}$

$=\frac{\mathrm{蚁}\left(\frac{4}{3}{\mathrm{蟺谤}}^3\right)}{{\mathrm{系}}_{\mathrm{o}}\left(4{\mathrm{蟺谤}}^2\right)}$

role="math" localid="1648766646051" $\mathrm{E}=\frac{\mathrm{蚁r}}{3{\mathrm{系}}_{\mathrm{o}}}....3$

For the inner sphere at distance$\mathrm{r}<\mathrm{a}$:

The charge density is localid="1648766990163" $\mathrm{蚁}=-\mathrm{Q}/(4/3){\mathrm{蟺补}}^3$.

The electric field is,

${\mathrm{E}}_{\mathrm{r}<\mathrm{a}}=\frac{\mathrm{蚁r}}{3{\mathrm{系}}_{\mathrm{o}}}$

$=\left(-\mathrm{Q}/(4/3){\mathrm{蟺补}}^3\right)\left(\frac{\mathrm{r}}{3{\mathrm{系}}_{\mathrm{o}}}\right)$

$=\frac{-\mathrm{Qr}}{4{\mathrm{蟺系}}_{\mathrm{o}}{\mathrm{a}}^3}$

Outside of the little sphere$(\mathrm{a}<\mathrm{r}<\mathrm{b})$:

The net charge was contained by the Gaussian surfacelocalid="1648918692131" $-\mathrm{Q}$.

Area is,

$\mathrm{A}=4{\mathrm{蟺谤}}^2$

This charge is enclosedlocalid="1648918728705" ${\mathrm{Q}}_{\text{in}}$in the gaussian surface is,

${\mathrm{Q}}_{\text{in}}=-\mathrm{Q}$

Substitute all values in equationlocalid="1648918736750" $1$,

we get,

${\mathrm{E}}_{\mathrm{a}<\mathrm{r}<\mathrm{b}}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{系}}_{\mathrm{o}}\mathrm{A}}$

$=\frac{-\mathrm{Q}}{{\mathrm{系}}_{\mathrm{o}}\left(4{\mathrm{蟺谤}}^2\right)}$

$=\frac{1}{4{\mathrm{蟺系}}_{\mathrm{o}}}\frac{-\mathrm{Q}}{{\mathrm{r}}^2}$

At $(\mathrm{b}<\mathrm{r}<\mathrm{c})$the electric field is zero.

Because the conductor has no net chargelocalid="1648918753302" ${\mathrm{Q}}_{\text{in}}=0$.

The electric field is zero since it is dependent on the quantity of charge within the bulk.

${\mathrm{E}}_{\mathrm{b}<\mathrm{r}<\mathrm{c}}=0\mathrm{N}/\mathrm{C}$

For $\mathrm{r}>\mathrm{c}$(outside the sphere):

The charge enclosed is$-\mathrm{Q}$and$2\mathrm{Q}$.

${\mathrm{Q}}_{\text{in}}=2\mathrm{Q}+(-\mathrm{Q})=\mathrm{Q}$

The gaussian surface has an area of,

$\mathrm{A}=4{\mathrm{蟺谤}}^2$

Substitute all values in equationlocalid="1648918802887" $1$,

We get,

${\mathrm{E}}_{\mathrm{r}>\mathrm{c}}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{系}}_{\mathrm{o}}\mathrm{A}}$

$=\frac{\mathrm{Q}}{{\mathrm{系}}_{\mathrm{o}}\left(4{\mathrm{蟺谤}}^2\right)}$

$=\frac{1}{4{\mathrm{蟺系}}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{r}}^2}$