0, $) coordinates. That is show,following the same method as discussed in lectures, the relationships for gradient and divergence in spherical coordinates. \text { 1. } \nabla \psi=\hat{\mathbf{r}} \frac{\partial \psi}{\partial r}+\hat{\phi} \frac{1}{r} \frac{\partial \psi}{\partial \phi}+\hat{\theta} \frac{1}{r \sin \phi} \frac{\partial \psi}{\partial \theta} \text { where we have } \psi(x, y, z) \Longrightarrow \psi(r, \theta, \phi) \text {. } \text { If the vector } \overrightarrow{\mathbf{A}}=\hat{\mathbf{r}} \mathrm{A}_{r}(r, \theta, \phi)+\hat{\theta} \mathrm{A}_{\theta}(r, \theta, \phi)+\hat{\phi} \mathrm{A}_{\phi}(r, \theta, \phi) \text { as is expressed in spherical system of coordinates, } then we can obtain the divergence in spherical system of coordinates as: then we can obtain the divergence in spherical system of coordinates as: \nabla \cdot \overrightarrow{\mathbf{A}}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \mathrm{~A}_{r}\right)+\frac{1}{r \sin \phi} \frac{\partial}{\partial \phi}\left(\mathrm{A}_{\phi} \sin \phi\right)+\frac{1}{r \sin \phi} \frac{\partial \mathrm{A}_{\theta}}{\partial \theta} \nabla \cdot \overrightarrow{\mathbf{A}}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \mathrm{~A}_{r}\right)+\frac{1}{r \sin \phi} \frac{\partial}{\partial \phi}\left(\mathrm{A}_{\phi} \sin \phi\right)+\frac{1}{r \sin \phi} \frac{\partial \mathrm{A}_{\theta}}{\partial \theta}

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