Question Number 58698 by arnabmaiti550@gmail.com last updated on 28/Apr/19
$$\overset{\rightarrow} {\bigtriangledown}\centerdot\left(\frac{{e}^{{br}} }{{r}^{\mathrm{2}} }\:\overset{\wedge} {{e}}_{{r}} \right)=?\:\:\:\:{b}\:{is}\:{a}\:{constant}. \\ $$
Commented by tanmay last updated on 28/Apr/19
$${pls}\:{mention}\:{co}−{ordinate}\:{system} \\ $$$$\left.\mathrm{1}\right)\:{csrtesian}\:\mathrm{3}{d}\: \\ $$$$\left.\mathrm{2}\right){polar}\:\left({r},\theta\right) \\ $$$$\left.\mathrm{3}\right){sperical}\left({r},\theta,\phi\right) \\ $$$$\left.\mathrm{4}\right){cylindrical} \\ $$
Commented by arnabmaiti550@gmail.com last updated on 28/Apr/19
$$\mathrm{spherical}\:\mathrm{coordinate}\:\mathrm{system} \\ $$
Commented by tanmay last updated on 28/Apr/19
$${ok}…{let}\:{me}\:{try}\:{to}\:{solve}… \\ $$
Commented by tanmay last updated on 28/Apr/19
Answered by tanmay last updated on 28/Apr/19
$${if}\:{it}\:{is}\:{in}\:{spherical}\:{coordinate} \\ $$$${then}\:\overset{\rightarrow} {\bigtriangledown}.\overset{\rightarrow} {{A}}=\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\frac{\partial}{\partial{r}}\left({r}^{\mathrm{2}} {A}_{{r}} \right) \\ $$$$=\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\frac{\partial}{\partial{r}}\left({r}^{\mathrm{2}} ×\frac{{e}^{{br}} }{{r}^{\mathrm{2}} }×\overset{\Lambda} {{e}}_{{r}} \right) \\ $$$$=\frac{\mathrm{1}}{{r}^{\mathrm{2}} }×\frac{\partial}{\partial{r}}\left({e}^{{br}} \:\overset{\Lambda} {{e}}_{{r}} \right) \\ $$$$=\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\left\{\overset{\Lambda} {{e}}_{{r}} \frac{\partial\left({e}^{{br}} \right)}{\partial{r}}+{e}^{{br}} \frac{\partial\left(\overset{\Lambda} {{e}}_{{r}} \right)}{\partial{r}}\right\} \\ $$$$=\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\left\{\overset{\Lambda} {{e}}_{{r}} ×{e}^{{br}} ×{b}+{e}^{{br}} ×\frac{\partial\overset{\Lambda} {{e}}_{{r}} }{\partial{r}}\right\} \\ $$$${i}\:{am}\:{not}\:{sure}\:{is}\:{it}\:{correct}\:{or}\:{not}… \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{i}}\:\boldsymbol{{think}}\:\boldsymbol{{question}}\:\boldsymbol{{is}}\:\boldsymbol{{in}}\:\boldsymbol{{polar}}\:\boldsymbol{{coirdinate}} \\ $$$$\boldsymbol{{system}} \\ $$$$ \\ $$