If volume charge density is given as \(\rho_{v}=(\cos \omega t) / r^{2} \mathrm{C} / \mathrm{m}^{2}\) in spherical coordinates, find \(\mathbf{J}\). It is reasonable to assume that \(\mathbf{J}\) is not a function of \(\theta\) or \(\phi\).

Given the volume charge density \(\rho_{v}=(\cos \omega t) / r^{2} \mathrm{C} / \mathrm{m}^{2}\) in spherical coordinates, we are tasked with finding the current density vector \(\mathbf{J}\). Since \(\rho_{v}\) is given in spherical coordinates, the current density vector \(\mathbf{J}\) will also be in spherical coordinates, which can be expressed as \(\mathbf{J} = J_r \mathbf{\hat{r}} + J_\theta \mathbf{\hat{\theta}} + J_\phi \mathbf{\hat{\phi}}\).

Since \(\mathbf{J}\) does not have angular dependence (not a function of \(\theta\) or \(\phi\)), we only need to find the radial component \(J_r\). To do so, we use the continuity equation, which is \(\nabla \cdot \mathbf{J} = -\frac{\partial \rho_{v}}{\partial t}\). The continuity equation in spherical coordinates for the radial component is: \(\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 J_r) = -\frac{\partial \rho_{v}}{\partial t}\) Now, evaluate \(\frac{\partial \rho_{v}}{\partial t}\): \(\frac{\partial \rho_{v}}{\partial t} = -\frac{\omega \sin \omega t}{r^2} \mathrm{C} / \mathrm{m}^{2}\) Substitute this into the continuity equation: \(\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 J_r) = \frac{\omega \sin \omega t}{r^2} \mathrm{C} / \mathrm{m}^{2}\)

To find \(J_r\), we need to integrate both sides with respect to \(r\). Focus on the left side, then the right side: (Left Side) \(\int\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 J_r) dr = \int\frac{\partial}{\partial r}(r^2 J_r) dr\) \(\Longrightarrow\) \(\Rightarrow\) \(r^2 J_r = F_1(t) + C_1\) (Right Side) \(\int\frac{\omega \sin \omega t}{r^2} dr = \omega \sin \omega t \int\frac{1}{r^2}dr\) \(\Longrightarrow\) \(\Rightarrow\) \(-\frac{\omega \sin \omega t}{r} = F_2(t) + C_2\) Now equating the left side with the right side: \(r^2 J_r = -\frac{\omega \sin \omega t}{r} + C\) Solve for \(J_r\): \(J_r = -\frac{\omega \sin \omega t}{r^3} + \frac{C}{r^2}\)

Now that we have determined the radial component \(J_r\), we can write the final expression for the current density vector in spherical coordinates. The other components, \(J_\theta\) and \(J_\phi\) are zero, so: \(\mathbf{J} = -\frac{\omega \sin \omega t}{r^3} \mathbf{\hat{r}}\)