Question

A charge distribution that is spherically symmetric but not uniform radially produces an electric field of magnitude$\mathit{E}\mathbf{=}\mathit{K}{\mathit{r}}^{\mathbf{4}}$, directed radially outward from the center of the sphere. Here $\mathit{r}$is the radial distance from that center, and$\mathit{K}$is a constant. What is the volume density r of the charge distribution?

Short answer

The volume charge density is$6{\mathrm{K\epsilon }}_0{\mathrm{r}}^3$

Step-by-step solution

Listing the given quantities

$\mathrm{E}={\mathrm{Kr}}^4$

Understanding the concept of charge density and electric field

Using the formula of volume charge density and electric field determine the volume charge density

Explanation

We use,

$$\begin{gathered} E=\frac{\left|q_{enc}\right|}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}{\int }_0^1\mathrm{\rho }\left(\mathrm{r}\right)4{\mathrm{\pi r}}^2\mathrm{dr} \end{gathered}$$

To solve for $\mathrm{\rho }\left(\mathrm{r}\right)$and obtain,

role="math" localid="1657353658777" $$\begin{gathered} \mathrm{\rho }\left(\mathrm{r}\right)=\frac{{\mathrm{\epsilon }}_0}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}{\mathrm{r}}^2\mathrm{E}\left(\mathrm{r}\right) \\ =\frac{{\mathrm{\epsilon }}_0}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{Kr}}^6\right) \\ =6{\mathrm{K\epsilon }}_0{\mathrm{r}}^3 \end{gathered}$$

The volume charge density is$6{\mathrm{K\epsilon }}_0{\mathrm{r}}^3$