In a region an electric field exist in a given direction and it passes through a circle of radius R normally. The magnitude of electric field is given as : E = E_0 (1− (r/R)). where r is the distance from centre of circle .Find electric flux through plane of circle within it.


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Question Number 33435 by rahul 19 last updated on 16/Apr/18

$I n a r e g i o n a n e l e c t r i c f i e l d e x i s t i n a g i v e n$ $d i r e c t i o n a n d i t p a s s e s t h r o u g h a c i r c l e$ $o f r a d i u s R n o r m a l l y . T h e m a g n i t u d e$ $o f e l e c t r i c f i e l d i s g i v e n a s :$ $E = E_{0} (1 - \frac{r}{R}) . w h e r e r i s t h e d i s t a n c e$ $f r o m c e n t r e o f c i r c l e . F i n d e l e c t r i c$ $f l u x t h r o u g h p l a n e o f c i r c l e w i t h i n i t .$
	Answered by ajfour last updated on 16/Apr/18

	$ϕ = \oint \bar{E} . d \bar{A}$ $= \int_{0}^{R} E_{0} (1 - \frac{r}{R}) (2 π r d r)$ $= 2 π E_{0} \int_{0}^{R} (r - \frac{r^{2}}{R}) d r$ $= 2 π E_{0} (\frac{R^{2}}{2} - \frac{R^{2}}{3}) = 2 π E_{0} (\frac{R^{2}}{6})$ $ϕ = E_{0} (\frac{π R^{2}}{3}) .$
	Commented by rahul 19 last updated on 16/Apr/18

	$S i r, u h a v e d o n e b y c o n s i d e r i n g$ $e l e m e n t a l r i n g a n d t h e n i n t e g r a t i n g$ $f r o m 0 \to R . B u t w i l l i t n o t f o r m a d i s c ?$ $a n d h e r e c i r c l e i s g i v e n (i m e a n h o l l o w$ $f r o m i n s i d e) .$
	Commented by ajfour last updated on 16/Apr/18

	$b y m e n t i o n i n g a c i r c l e f l u x i s$ $a s k e d t o b e c a l c u l a t e d t h r o u g h t h e$ $c i r c u l a r f l a t p l a n e a r e a w h o s e$ $b o u n d a r y i s t h i s c i r c l e .$ $F l u x i s a s u r f a c e i n t e g r a l .$
	Commented by rahul 19 last updated on 16/Apr/18

	$S o h e r e i f i n p l a c e o f c i r c l e, d i s c$ $i s m e n t i o n e d a n s w e r w i l l r e m a i n s a m e,$ $r i g h t ?$
	Commented by ajfour last updated on 16/Apr/18

	$y e s!$
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