| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 20:03:26 IST
|
|
|
A rod of length L has a total charge Q distributed uniformly along its length.It is bent in the shape of a semicircle.Find the magnitude of the electric field at the centre of curvature of the semicircle.
|
|
|
|
20 Apr 2008 20:33:42 IST
|
|
|
Re:Electric field
|
this reply: 5 points
(with 1 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
THE VERICAL COMPONENTS WILL ADD AND HORIZONTAL COMPONENTS WILL CANCELL OUT. SO IN THIS CASE U NEED INTEGRATION OF dE SIN THETA
HOPE U UNDERSTAND
|
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
20 Apr 2008 20:41:54 IST
|
|
|
I have rated u but please explain in details
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
20 Apr 2008 21:17:53 IST
|
|
|
Tell me in which step you got a problem.
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
20 Apr 2008 21:26:19 IST
|
|
|
the answer must be: E = Q/ 2  2 L 2 |
this reply: 2 points
(with 0
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
20 Apr 2008 21:37:05 IST
|
|
|
|
Will nip in at times to solve problems :)
|
this reply: 10 points
(with 2
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
20 Apr 2008 22:03:28 IST
|
|
|
Answer is q/2e0L2
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
20 Apr 2008 22:11:28 IST
|
|
|
Yes, it does come to that when you integrate the above. I've made a typo towards the end.
|
Will nip in at times to solve problems :)
|
this reply: 10 points
(with 2
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
![[Thumb - untitled.JPG]](/upload/2008/4/20/be87b39177fc907b22f8562e8b7c3f1d_42779.JPG_thumb)
