A positive charge of magnigude 2.5μC is at the centre of an uncharged spherical conducting shell of inner radius 60cm and an outer radius of 90cm. (i)find the charge densities on thd inner and outer surfaces of the shell and the total charge on each surface. ii)find the electricity field everywhere.


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Question Number 39792 by NECx last updated on 11/Jul/18

$A p o s i t i v e c h a r g e o f m a g n i g u d e$ $2.5 μ C i s a t t h e c e n t r e o f a n u n c h a r g e d$ $s p h e r i c a l c o n d u c t i n g s h e l l o f i n n e r$ $r a d i u s 60 c m a n d a n o u t e r r a d i u s$ $o f 90 c m .$ $(i) f i n d t h e c h a r g e d e n s i t i e s o n t h d$ $i n n e r a n d o u t e r s u r f a c e s o f t h e$ $s h e l l a n d t h e t o t a l c h a r g e o n e a c h$ $s u r f a c e .$ $i i) f i n d t h e e l e c t r i c i t y f i e l d$ $e v e r y w h e r e .$
Commented by NECx last updated on 11/Jul/18

$p l e a s e h e l p$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jul/18

	$i) s u r f a c e c h a r g e d e n s i t y (σ) o n t h e i n n e r$ $s u r f a c e s h e l l σ_{i} = \frac{- 2.5 \times 10^{- 6}}{4 Π \times {(60 \times 10^{- 2})}^{2}}$ $s u r f a c e c h a r g e d e n s i t y (σ) o n t h e o u t e r$ $s u r f a c e s h e l l σ_{o} = \frac{2.5 \times 10^{- 6}}{4 Π {(90 \times 10^{- 2})}^{2}}$ $e l e c t r i c f i e l d i n s i d e t h e s h e l l i s z e r o . . . b e c a u s e$ $c h a r g e a t c e n t r e a n d i n d u c e d c h a r g e d o n t h e$ $i n n e r w a l l o f s h e l l b a l a n c e e a c h o t h e r$ $E \times 4 Π r^{2} = \frac{2.5 μ C + (- 2.5 μ C)}{ϵ_{0}}$ $s o E = 0$ $f o r o u t s i d e t h e s h e l l w a l l . .$ $E \times 4 Π R^{2} = \frac{2.5 μ C + (- 2.5 μ C) + (2.5 μ C)}{ϵ_{0}}$ $E = \frac{2.5 \times 10^{- 6}}{4 Π ϵ_{0} {(90 \times 10^{- 2})}^{2}}$
	Commented by NECx last updated on 11/Jul/18

	$T h i s i s c o o l . T h a n k y o u s o m u c h$
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