91影视

It is given that electric field in some region is found to be$E=K{r}^r\hat{r}$..The charge density and total charge inside a sphere of radius R has to be evaluated.

If there is a surface area enclosing a volume, possessing a charge$q$inside the volume then the electric field due to the surface or volume charge is given as

$$\begin{gathered} \mathbf{鈭�}\mathbf{.}\mathit{E}\mathbf{=}\frac{\mathbf{P}}{{\mathbf{蔚}}_{\mathbf{0}}}\mathbf{;} \\ \mathbf{鈭�}\mathit{E}\mathbf{.}\mathit{d}\mathit{a}\mathbf{=}\frac{\mathbf{q}}{{\mathbf{蔚}}_{\mathbf{0}}} \end{gathered}$$

Here${}^{\mathbf{P}}$is the charge density, is the total charge inside the volume, and

(a)

The divergence of radial electric field vector in spherical coordinates is written as

Substitute$K{r}^3\hat{r}$for ${}^{E_r}$into${}^{鈭�路E=\frac{1}{r^2}\frac{a}{ar}\left(r^2{E}_r\right)}$

Substitutelocalid="1654329764734" ${}^{5k{r}^2}$forlocalid="1654329780082" $鈭�路E$into differential form of gauss law.

localid="1654329799794" $$\begin{gathered} P={蔚}_0\left(5K{r}^2\right) \\ =5{蔚}_{0}K{r}^2 \end{gathered}$$

Thus, the charge density is obtained as localid="1654329806624" $P=5{蔚}_{0}K{r}^2$.

Obtain the total charge

(b)

Apply Gauss law on the Gaussian surface, which is a sphere of radius R The electric field at the radiuslocalid="1654329823739" ${}^R$becomeslocalid="1654329840710" ${}^{E=K{r}^3}\hat{r}$and the surface area becomeslocalid="1654329860396" ${}^{4{\mathrm{蟺谤}}^2}$

Substitutelocalid="1654329884232" ${}^{K{r}^3}\hat{r}$forlocalid="1654329928194" ${}^E$andlocalid="1654329939411" ${}^{4{\mathrm{蟺谤}}^3}$forlocalid="1654329913829" ${}^{da}$intolocalid="1654329899560" ${}^{鈭�E.da=\frac{q_{enclosed}}{{蔚}_0}}$

localid="1654329960503" $$\begin{gathered} 鈭�E.da{=}_{\stackrel{炉}{{蔚}_0}} \\ \left(K{R}^3\right)\left(4{\mathrm{蟺搁}}^2\right)=\frac{{\mathrm{q}}_{\mathrm{enclosed}}}{{\mathrm{蔚}}_0} \\ {\mathrm{q}}_{\mathrm{enclosed}}{=}^{4{\mathrm{蟺蔚}}_0{\mathrm{KR}}^5} \end{gathered}$$

Thus, total charge inside the sphere of radius R is localid="1654329978559" ${}^{q_{enclosed}=4{\mathrm{蟺蔚}}_0{\mathrm{KR}}^5}$

Another way to obtain the total charge by the integrating the differential charge.

Consider a spherical shell of radiuslocalid="1654330011802" ${}^r$and thicknesslocalid="1654329992475" ${}^{dr}$inside the sphere of radiuslocalid="1654330032331" ${}^R$.

The differential charge present due to the spherical shell islocalid="1654330055399" ${}^{dq=pd\mathcal{s}}$

Thus, total charge inside the sphere, obtained by integration, is$q_{enclosed}{=}^{4{\mathrm{蟺蔚}}_0{\mathrm{KR}}^5}$