2.9 Suppose the electric field in some region is found to be \( \vec{E} = k r^3 \hat{r} \), in spherical coordinates and \(k\) is some constant. Using \( \rho = r \) in spherical coordinates.

Gauss’ Law

$$\oint_S \vec{E} \cdot d \vec{A} = \oint_S \vec{E} \cdot \hat{n} \, dA = \frac{Q_{enc}}{ \epsilon_0 }$$

$$\oint_S \vec{E} \cdot d \vec{A} = \oint_S \vec{E} \cdot \hat{r} \, dA = \int_0^{\pi} \int_0^{2 \pi} k \rho^3 \rho^2 \sin \phi \, d \theta \, d \phi = \frac{Q_{enc}}{ \epsilon_0 }$$

$$Q_{enc} = 2 \pi k \rho^5 \epsilon_0 \int_0^{\pi} \sin \phi \, d\phi = 4 \pi k \rho^5 \epsilon_0$$

Also,

$$\nabla \cdot \vec{E} = \frac{ \rho }{ \epsilon_0 }$$

$$\nabla \cdot \vec{E} = \frac{1}{\rho^2} \frac{\partial}{\partial \rho} ( \rho^2 E_{\rho} ) + \frac{1}{\rho \sin \theta} \frac{ \partial }{ \partial \theta} ( E_{\theta} \sin \theta ) + \frac{1}{\rho \sin \theta} \frac{ \partial E_{\phi} }{ \partial \phi}$$

$$\nabla \cdot \vec{E} = \frac{ 1 }{ r^2 } ( 5 k r^4 ) = \frac{ \rho }{ \epsilon_0 }$$

$$\rho = 5 k r^2 \epsilon_0$$

In spherical coordinates…remembering \( \rho \) is used as a variable radial value and not to be confused with the volume charge density

$$Q_{enc} = \int_V \rho \, dV$$

$$Q_{enc} = \int_0^r \int_0^{\pi} \int_0^{2 \pi} 5 k \rho^2 \epsilon_0 \rho^2 \sin \phi \, d\theta \, d\phi \, d\rho$$

$$Q_{enc} = 20 \pi k \epsilon_0 \int_0^r \rho^4 \, d\rho = 4 \pi k \rho^5 \epsilon_0$$