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Question Number 33450 by ajfour last updated on 16/Apr/18

Commented by ajfour last updated on 16/Apr/18

$F i n d e l e c t r i c f l u x o f a s e m i i n f i n i t e$ $t h r e a d o f c h a r g e h a v i n g l i n e a r$ $c h a r g e d e n s i t y λ, t h r o u g h a c i r c u l a r$ $d i s c o f r a d i u s R i f o n e e n d o f t h e$ $t h r e a d i s a t t h e c e n t r e o f t h i s$ $c i r c u l a r a r e a .$
	Answered by 33 last updated on 18/Apr/18
	$consider an element of lenght dx at dist x from centre of ring dφ = (dq/(2ε_0 ))(1−(x/(√(R^2 +x^2 )))) φ = ∫_( 0) ^( L) (((λdx))/(2ε_0 )) (1−(x/(√(R^2 + x^2 )))) ⇒ φ = (λ/(2ε_0 ))[x− (√(x^2 + R^2 )) ]_0 ^L φ = (λ/(2ε_0 ))lim_(L→∞) {L−(√(L^2 +R^2 )) +R} φ = ((λR)/(2ε_0 ))$
	$c o n s i d e r a n e l e m e n t$ $o f l e n g h t d x a t d i s t x$ $f r o m c e n t r e o f r i n g$ $d ϕ = \frac{d q}{2 ϵ_{0}} (1 - \frac{x}{\sqrt{R^{2} + x^{2}}})$ $ϕ = \underset{0}{\int^{L}} \frac{(λ d x)}{2 ϵ_{0}} (1 - \frac{x}{\sqrt{R^{2} + x^{2}}})$ $\Rightarrow ϕ = \frac{λ}{2 ϵ_{0}} {[x - \sqrt{x^{2} + R^{2}}]}_{0}^{L}$ $ϕ = \frac{λ}{2 ϵ_{0}} l i \underset{L \to \infty}{m} {L - \sqrt{L^{2} + R^{2}} + R}$ $ϕ = \frac{λ R}{2 ϵ_{0}}$
	Commented by ajfour last updated on 17/Apr/18

	$p l e a s e e x p l a i n t h e c o n c e p t o f$ $f i r s t l i n e i t s e l f .$ $a n s w e r i s c o r r e c t .$
	Commented by 33 last updated on 19/Apr/18

	$y o u a r e w e l c o m e s i r :)$
	Commented by 33 last updated on 18/Apr/18

	$f o r a p o i n t c h a r g e p l a c e d$ $a t t h e a x i s o f r i n g a t d i s t^{'} x^{'}$ $f l u x t h r o u g h t h e d i s k c a n b e$ $c a l c u l a t e d$ $w e c a n m a k e a n i m a g i n a r y$ $c u r v e d s u r f a c e (a p a r t o f s p h e r e) w i t h b o u n d a r y$ $a s t h e r i n g a n d i t s$ $c e n t r e c o i n c i d i n g w i t h t h e$ $p o s i t i o n o f p o i n t c h a r g e .$ $ϕ = (t o t a l f l u x) \frac{a r e a o f s u r f a c e}{a r e a o f s p h e r e}$
	Commented by 33 last updated on 18/Apr/18

	Commented by 33 last updated on 18/Apr/18

	$x \neq 0$
	Commented by 33 last updated on 18/Apr/18

	$i s t h a t o k s i r ?$
	Commented by ajfour last updated on 18/Apr/18

	$t h a n k y o u v e r y m u c h S i r .$
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