e-br-r-2-e-r-b-is-a-constant-

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 \dots l e t m e t r y t o s o l v e \dots

Commented by tanmay last updated on 28/Apr/19

Answered by tanmay last updated on 28/Apr/19

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 \dots

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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