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

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!

Facebook Tweet Pin