| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Mar 2008 19:34:42 IST
|
|
|
what is the self potential energy of a circular disc of surface charge density sigma,radiusR.? I'm havin' sum confusion........
|
|
|
|
13 Mar 2008 19:52:54 IST
|
|
|
infinite
|
adversities cause some men to break other to break records............i m of the other type....... :-) |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
13 Mar 2008 19:54:11 IST
|
|
|
i think it works out to (  R/4  )^2 is the oermitivity of free space |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
13 Mar 2008 19:54:11 IST
|
|
|
NOPES!!!
U hav got it absolutely wrong my freind!!!!...... it's very much a finite quantity.......
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
13 Mar 2008 19:55:14 IST
|
|
|
@suresan.....please check out dimensions of ur ans.....
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
13 Mar 2008 20:09:02 IST
|
|
|
sorry i read potential not potential energy
|
adversities cause some men to break other to break records............i m of the other type....... :-) |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
13 Mar 2008 20:45:35 IST
|
|
|
I am really sorry I am working it out pretty careless
it works out something like this.
(pi X(sigma)^2 R^2) / 4X epsilonzero.
Look i am not sure even the concept used is right. Just thought it might work out.
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
14 Mar 2008 11:29:38 IST
|
|
|
no yaar stillthe ans isn't correct......I am getting it as (pie*(sigma)^2*R^3) / 6*epsilonzero.....bt the fiitjee sol.n gives it something else....... can any1 please tell me the ans alongwith the method????
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
14 Mar 2008 17:37:17 IST
|
|
|
I am really sorry my friend, please forgive me , I was doing it in mind and i missed some thing naturall.
then answer would be some thing like this
(pi X sigma ^ 2 X R ^ 3)/ 6 epsilonzero.
I am not sure of the concept i used i will send the detail just now. please wait
|
this reply: 2 points
(with 0
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
14 Mar 2008 17:56:36 IST
|
|
|
Draw a circle to reprresent the disc of radius R. Draw to circles inside with radii r and r+dr. Now assume that the charges are distrubuted uniformly over the disc of radius R. Then charges inside the circle of radiusr q1 =  R ^ 2 this can be cosidered concentrated at the centre for all calculation the charges inside the circular strip of radius r and width dr = 2 r dr then the potential energy of the sysem = integral from 0 to R (q1 q2) / 4 re where e is the permitivity of free space. I don't know whether you got the idea or whether you approve this. Please check it if you find it iteresting and see whether you get the same answer or not. Anyway thank you for that interesting question. best wishes suresan
|
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
14 Mar 2008 18:58:23 IST
|
|
|
thanks suresan for answering.....I too have followed the same method and getting the same ans as u.......it means that the FIITJEE sol.n are wrong.......
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
14 Mar 2008 20:08:22 IST
|
|
|
i don understand how can u assume the charge to be at the centre its not a spherical object
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
15 Mar 2008 10:12:48 IST
|
|
|
getting it as 2R3/6o
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
1
