Q9P - Introduction to Electrodynamics

Question

Suppose the electric field in some region is found to be${}^{\mathbf{E}\mathbf{=}\mathbf{ }\mathbf{K}{\mathbf{r}}^{\mathbf{3}}}\hat{\mathbf{r}}$

in spherical coordinates (k is some constant).

(a) Find the charge density ${}^{\mathbf{P}}$ 

(b) Find the total charge contained in a sphere of radius ${}^{\mathit{R}}$centered at the origin.(Do it two different ways.)

Step-by-Step Solution

Answer

(a)The charge density is obtained as ${}^{P=5{\epsilon }_{0}K{r}^2}$.

(b)The total charge inside the sphere, obtained using gauss law is${}^{q_{enclosed}=4{\mathrm{\pi \epsilon }}_0{\mathrm{KR}}^5}$The total charge inside the sphere, obtained by integration, is ${}^{{}^{q_{enclosed}=4{\mathrm{\pi \epsilon }}_0{\mathrm{KR}}^5}}$

Step 1: Describe the given information

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.

Step 2: Define the Gauss law in differential form and integral form

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

$$\begin{gathered} \mathbf{\nabla }\mathbf{.}\mathit{E}\mathbf{=}\frac{\mathbf{P}}{{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{;} \\ \mathbf{\oint }\mathit{E}\mathbf{.}\mathit{d}\mathit{a}\mathbf{=}\frac{\mathbf{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}} \end{gathered}$$

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

Step 3: Obtain the charge density

(a)

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

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

Substitute${}^{5k{r}^2}$for$\nabla ·E$into differential form of gauss law.

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

Thus, the charge density is obtained as $P =5{\epsilon }_{0}K{r}^2$.

Obtain the total charge

(b)

 

Apply Gauss law on the Gaussian surface, which is a sphere of radius R The electric field at the radius${}^R$becomes${}^{E=K{r}^3}\hat{r}$and the surface area becomes${}^{4{\mathrm{\pi r}}^2}$

Substitute${}^{K{r}^3}\hat{r}$ for${}^E$ and${}^{4{\mathrm{\pi r}}^3}$ for${}^{da}$ into${}^{\oint E. da =\frac{q_{enclosed}}{{\epsilon }_0}}$

$$\begin{gathered} \oint E.da{=}_{\overline{{\epsilon }_0}} \\ \left(K{R}^3\right)\left(4{\mathrm{\pi R}}^2\right)=\frac{{\mathrm{q}}_{\mathrm{enclosed}}}{{\mathrm{\epsilon }}_0} \\ {\mathrm{q}}_{\mathrm{enclosed}}{=}^{4{\mathrm{\pi \epsilon }}_0{\mathrm{KR}}^5} \end{gathered}$$

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

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

Consider a spherical shell of radius${}^r$ and thickness${}^{dr}$inside the sphere of radius${}^R$.

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

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