91Ó°ÊÓ

(a).

Density of volume charge is,

$\mathrm{ÒÏ}=\frac{\mathrm{Q}}{\mathrm{V}}.......1$

Charge on the ball is$80\mathrm{nC}$.

${\mathrm{Q}}_{\text{in}}=\left(80\mathrm{nC}\right)\left(\frac{1×{10}^{-9}\mathrm{C}}{\mathrm{nC}}\right)=80×{10}^{-9}\mathrm{C}$

The volume of the ball equals since it is formed like a sphere.

$\mathrm{V}=\frac{4}{3}{\mathrm{Ï€°ù}}^3=\frac{4}{3}\mathrm{Ï€}(0.20\mathrm{m}{)}^3=3.35×{10}^{-2}{\mathrm{m}}^3$

$\mathrm{ÒÏ}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{80×{10}^{-9}\mathrm{C}}{3.35×{10}^{-2}{\mathrm{m}}^3}=2.4×{10}^{-6}\mathrm{C}/{\mathrm{m}}^3$

(b).

The charge is,

$\mathrm{Q}=\mathrm{ÒÏV}=\mathrm{ÒÏ}\left(\frac{4}{3}{\mathrm{Ï€°ù}}^3\right)=\frac{4}{3}{\mathrm{Ï€ÒÏr}}^3$

For radius $\mathrm{r}=5\mathrm{cm}=0.05\mathrm{m}$:

Charge is,

${\mathrm{Q}}_{\mathrm{r}=5\mathrm{cm}}=\frac{4}{3}{\mathrm{Ï€ÒÏr}}^3$

$=\frac{4}{3}\mathrm{π}\left(2.4×{10}^{-6}\mathrm{C}/{\mathrm{m}}^3\right)(0.05\mathrm{m}{)}^3$

$=1.25×{10}^{-9}\mathrm{C}$

$=1.25\mathrm{nC}$

For radius $\mathrm{r}=10\mathrm{cm}=0.10\mathrm{m}$:

Charge is,

${\mathrm{Q}}_{\mathrm{r}=10\mathrm{cm}}=\frac{4}{3}{\mathrm{Ï€ÒÏr}}^3$

$=\frac{4}{3}\mathrm{π}\left(2.4×{10}^{-6}\mathrm{C}/{\mathrm{m}}^3\right)(0.10\mathrm{m}{)}^3$

$=10×{10}^{-9}\mathrm{C}$$=10\mathrm{nC}$

For radius $\mathrm{r}=20\mathrm{cm}=0.20\mathrm{m}$:

Charge is,

${\mathrm{Q}}_{\mathrm{r}=20\mathrm{cm}}=\frac{4}{3}{\mathrm{Ï€ÒÏr}}^3$

$=\frac{4}{3}\mathrm{π}\left(2.4×{10}^{-6}\mathrm{C}/{\mathrm{m}}^3\right)(0.20\mathrm{m}{)}^3$

$=80.4×{10}^{-9}\mathrm{C}=80.4\mathrm{nC}$

(c).

The current of electricity is,

${\mathrm{Φ}}_{\mathrm{e}}=\mathrm{EA}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{Ï\mu }}_{\mathrm{o}}}$

$\mathrm{E}=\frac{{\mathrm{Q}}_{\text{in}}}{4{\mathrm{Ï€Ï\mu }}_{\mathrm{o}}{\mathrm{r}}^2}$

field of electricity,

In terms of distance$\mathrm{r}=5\mathrm{cm}$:

${\mathrm{E}}_{\mathrm{r}=5\mathrm{cm}}=\frac{1}{4{\mathrm{Ï€Ï\mu }}_{\mathrm{o}}}\frac{{\mathrm{Q}}_{\mathrm{T}=5\mathrm{cm}}}{{\mathrm{r}}^2}$

$=\frac{1}{4\mathrm{π}\left(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)}\frac{\left(1.25×{10}^{-9}\mathrm{C}\right)}{(0.05\mathrm{m}{)}^2}$

$=5×{10}^3\mathrm{N}/\mathrm{C}$

In terms of distance $\mathrm{r}=10\mathrm{cm}$:

${\mathrm{E}}_{\mathrm{r}=10\mathrm{cm}}=\frac{1}{4{\mathrm{Ï€Ï\mu }}_{\mathrm{o}}}\frac{{\mathrm{Q}}_{\mathrm{r}=10\mathrm{cm}}}{{\mathrm{r}}^2}$

$=\frac{1}{4\mathrm{π}\left(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)}\frac{\left(10×{10}^{-9}\mathrm{C}\right)}{(0.10\mathrm{m}{)}^2}$

$=9×{10}^3\mathrm{N}/\mathrm{C}$

In terms of distance $\mathrm{r}=20\mathrm{cm}$:

${\mathrm{E}}_{\mathrm{r}=20\mathrm{cm}}=\frac{1}{4{\mathrm{Ï€Ï\mu }}_{\mathrm{o}}}\frac{{\mathrm{Q}}_{\mathrm{r}=20\mathrm{cm}}}{{\mathrm{r}}^2}$

$=\frac{1}{4\mathrm{π}\left(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)}\frac{\left(80.4×{10}^{-9}\mathrm{C}\right)}{(0.20\mathrm{m}{)}^2}$

$=18×{10}^3\mathrm{N}/\mathrm{C}$