Question

aA uniformly charged ball of radius$\mathrm{a}$and charge $-\mathrm{Q}$is at the center of a hollow metal shell with inner radius $\mathrm{b}$and outer radius $\mathrm{c}$. The hollow sphere has net charge$+2\mathrm{Q}$. Determine the electric field strength in the four regions$\mathrm{r}\le \mathrm{a}$,$\mathrm{a}<\mathrm{r}<\mathrm{b}$,$\mathrm{b}\le \mathrm{r}\le \mathrm{c}$,and$\mathrm{r}>\mathrm{c}$.

Short answer

Electric field strength in ${\mathrm{E}}_{\mathrm{r}<\mathrm{a}}$is$\frac{-\mathrm{Qr}}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{a}}^3}$.

Electric field strength in ${\mathrm{E}}_{\mathrm{a}<\mathrm{r}<\mathrm{b}}$is$\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{-\mathrm{Q}}{{\mathrm{r}}^2}$.

Electric field strength in ${\mathrm{E}}_{\mathrm{a}<\mathrm{r}<\mathrm{b}}$is $0$.

Electric field strength in${\mathrm{E}}_{\mathrm{r}>\mathrm{c}}$is$\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{r}}^2}$.

Step-by-step solution

Figure for electric field strength in four region

Figure is,

Formula for electric field

Electric flux is,

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

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

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

${\mathrm{Q}}_{\text{in}}=\mathrm{\rho V}=\mathrm{\rho }\left(\frac{4}{3}{\mathrm{\pi r}}^3\right)$

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

$=\frac{\mathrm{\rho }\left(\frac{4}{3}{\mathrm{\pi r}}^3\right)}{{\mathrm{ϵ}}_{\mathrm{o}}\left(4{\mathrm{\pi r}}^2\right)}$

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

Calculation for electric field at distancer<a 

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

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

The electric field is,

${\mathrm{E}}_{\mathrm{r}<\mathrm{a}}=\frac{\mathrm{\rho r}}{3{\mathrm{ϵ}}_{\mathrm{o}}}$

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

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

Calculation for electric field at distance (a<r<b)

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,

$\mathrm{A}=4{\mathrm{\pi r}}^2$

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,

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

$=\frac{-\mathrm{Q}}{{\mathrm{ϵ}}_{\mathrm{o}}\left(4{\mathrm{\pi r}}^2\right)}$

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

Calculation for electric field at distance (b<r<c)

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.

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

Calculation for electric field at distance r>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,

$\mathrm{A}=4{\mathrm{\pi r}}^2$

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

We get,

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

$=\frac{\mathrm{Q}}{{\mathrm{ϵ}}_{\mathrm{o}}\left(4{\mathrm{\pi r}}^2\right)}$

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