▽^→ ∙((e^(br) /r^2 ) e_r ^∧ )=? b is a constant.


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Question Number 58698 by arnabmaiti550@gmail.com last updated on 28/Apr/19

$\vec{▽} \cdot (\frac{e^{b r}}{r^{2}} {\overset{\land}{e}}_{r}) = ? b i s a c o n s t a n t .$
Commented by tanmay last updated on 28/Apr/19

$p l s m e n t i o n c o - o r d i n a t e s y s t e m$ $1) c s r t e s i a n 3 d$ $2) p o l a r (r, θ)$ $3) s p e r i c a l (r, θ, ϕ)$ $4) c y l i n d r i c a l$
Commented by arnabmaiti550@gmail.com last updated on 28/Apr/19

$spherical coordinate system$
Commented by tanmay last updated on 28/Apr/19

$o k . . . l e t m e t r y t o s o l v e . . .$
Commented by tanmay last updated on 28/Apr/19

	Answered by tanmay last updated on 28/Apr/19
	$if it is in spherical coordinate then ▽^→ .A^→ =(1/r^2 )(∂/∂r)(r^2 A_r ) =(1/r^2 )(∂/∂r)(r^2 ×(e^(br) /r^2 )×e_r ^Λ ) =(1/r^2 )×(∂/∂r)(e^(br) e_r ^Λ ) =(1/r^2 ){e_r ^Λ ((∂(e^(br) ))/∂r)+e^(br) ((∂(e_r ^Λ ))/∂r)} =(1/r^2 ){e_r ^Λ ×e^(br) ×b+e^(br) ×(∂e_r ^Λ /∂r)} i am not sure is it correct or not... but i think question is in polar coirdinate system$
	$i f i t i s i n s p h e r i c a l c o o r d i n a t e$ $t h e n \vec{▽} . \vec{A} = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} A_{r})$ $= \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \times \frac{e^{b r}}{r^{2}} \times {\overset{Λ}{e}}_{r})$ $= \frac{1}{r^{2}} \times \frac{\partial}{\partial r} (e^{b r} {\overset{Λ}{e}}_{r})$ $= \frac{1}{r^{2}} {{\overset{Λ}{e}}_{r} \frac{\partial (e^{b r})}{\partial r} + e^{b r} \frac{\partial ({\overset{Λ}{e}}_{r})}{\partial r}}$ $= \frac{1}{r^{2}} {{\overset{Λ}{e}}_{r} \times e^{b r} \times b + e^{b r} \times \frac{\partial {\overset{Λ}{e}}_{r}}{\partial r}}$ $i a m n o t s u r e i s i t c o r r e c t o r n o t . . .$ $b u t i t h i n k q u e s t i o n i s i n p o l a r c o i r d i n a t e$ $s y s t e m$
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