Question

A $20-\mathrm{cm}$radius ball is uniformly charged to$80\mathrm{nC}$.

a. What is the ball's volume charge density $(\mathrm{C}/{\mathrm{m}}^3)$localid="1648741376835" $?$

b. How much charge is enclosed by spheres of radiilocalid="1648741380973" $5$,localid="1648741279896" $10$andlocalid="1648741787973" $20\mathrm{cm}$localid="1648741405448" $?$

c. What is the electric field strength at points localid="1648741424743" $5$,localid="1648741429590" $10$andlocalid="1648741433205" $20$localid="1648741437392" $\mathrm{cm}$from the centerlocalid="1648741447708" $?$

Short answer

a.The volume charge density of the ball $\mathrm{\rho }$is $2.4\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^3$.

b.Charge is contained within spheres of radius localid="1648741499012" $5$is localid="1648741492923" $1.25\mathrm{nC}$,localid="1648741506064" $10$islocalid="1648741511169" $10\mathrm{nC}$and localid="1648741515337" $20$islocalid="1648741520392" $80.4\mathrm{nC}$.

c.The strength of the electric field at points localid="1648741533167">5
islocalid="1648741540359" $5\mathrm{kN}/\mathrm{C}$,localid="1648741549994" $10$islocalid="1648741524494" $9\mathrm{kN}/\mathrm{C}$andlocalid="1648741558174" $20$islocalid="1648741545101" $18\mathrm{kN}/\mathrm{C}$.

Step-by-step solution

Calculation for ball's volume density (part a)

(a).

Density of volume charge is,

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

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

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

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

$\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}\mathrm{\pi }(0.20\mathrm{m}{)}^3=3.35\times {10}^{-2}{\mathrm{m}}^3$

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

Calculation for charge of radius5,10and20cm

(b).

The charge is,

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

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

Charge is,

${\mathrm{Q}}_{\mathrm{r}=5\mathrm{cm}}=\frac{4}{3}{\mathrm{\pi \rho r}}^3$

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

$=1.25\times {10}^{-9}\mathrm{C}$

$=1.25\mathrm{nC}$

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

Charge is,

${\mathrm{Q}}_{\mathrm{r}=10\mathrm{cm}}=\frac{4}{3}{\mathrm{\pi \rho r}}^3$

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

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

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

Charge is,

${\mathrm{Q}}_{\mathrm{r}=20\mathrm{cm}}=\frac{4}{3}{\mathrm{\pi \rho r}}^3$

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

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

Calculation for Electric field strength at points 5,10and20 cm.(part c)

(c).

The current of electricity is,

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

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

field of electricity,

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

${\mathrm{E}}_{\mathrm{r}=5\mathrm{cm}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{{\mathrm{Q}}_{\mathrm{T}=5\mathrm{cm}}}{{\mathrm{r}}^2}$

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

$=5\times {10}^3\mathrm{N}/\mathrm{C}$

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

${\mathrm{E}}_{\mathrm{r}=10\mathrm{cm}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{{\mathrm{Q}}_{\mathrm{r}=10\mathrm{cm}}}{{\mathrm{r}}^2}$

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

$=9\times {10}^3\mathrm{N}/\mathrm{C}$

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

${\mathrm{E}}_{\mathrm{r}=20\mathrm{cm}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{{\mathrm{Q}}_{\mathrm{r}=20\mathrm{cm}}}{{\mathrm{r}}^2}$

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

$=18\times {10}^3\mathrm{N}/\mathrm{C}$