A spherically symmetric charge distribution produces the electric field E u = 15000r2 2nr N/C, where r is in m.

a. What is the electric field strength at r = 20 cm?

b. What is the electric flux through a 40-cm-diameter spherical surface that is concentric with the charge distribution?

c. How much charge is inside this 40-cm-diameter spherical surface?

Given:

Electric field vector $\overrightarrow{E}=\left(5000r^2\right)\hat{r}N/C$

Radius r=20 $\mathrm{cm}=0.20\mathrm{m}$

Formula used:

The electrical field is given by the formula

$\overrightarrow{E}=\left(5000r^2\right)\hat{r}$

Here,

r is in meter

Calculation:

The electric field intensity is calculated as

$\overrightarrow{E}=\left(5000r^2\right)\hat{r}$

Inserting the values

$\overrightarrow{E}=\backslash \mathrm{left}(5000\backslash \mathrm{left}{(}^{0.20}$)$r^{^}$

$=200\hat{r}\mathrm{N}/\mathrm{C}$

$\overrightarrow{E}=\left(5000r^2\right)\hat{r}N/C$

$d=\text{diameter}=40\mathrm{cm}$

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

Formula used:

Area of the sphere is calculated by the formula

$A=4\pi r^2$

Electric flux is calculated by the formula

$\varphi =\overrightarrow{E}·\overrightarrow{A}$

$\varphi$is the electric flux

E is the electric field

A is the area

The electric field intensity is calculated as

$\overrightarrow{E}=\left(5000r^2\right)\hat{r}$

Inserting the values

$\overrightarrow{E}=\left(5000({)}^{0.20}\right)\hat{r}$

$=200\hat{r}\mathrm{N}/\mathrm{C}$

Area of the surface is given as

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

$=4(3.14)({)}^{0.20}\hat{r}$

$=0.5024\hat{r}{\mathrm{m}}^2$

Electric flux is given as

$\varphi =\overrightarrow{E}·\overrightarrow{A}$

$=\left(200\hat{r}\right)(0.5024\hat{r}$

$=100.5\mathrm{N}-{\mathrm{m}}^2/\mathrm{C}$