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13 Mar 2008 19:34:42 IST

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532

self potential energy

None

what is the self potential energy of a circular disc of surface charge density sigma,radiusR.?

I'm havin' sum confusion........

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Comments (12)

ankur khurana

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13 Mar 2008 19:52:54 IST

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infinite

suresan

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13 Mar 2008 19:54:11 IST

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i think it works out to

(

R/4

)^2

is the oermitivity of free space

zamuka chen

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13 Mar 2008 19:54:11 IST

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NOPES!!!
U hav got it absolutely wrong my freind!!!!......

it's very much a finite quantity.......

zamuka chen

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13 Mar 2008 19:55:14 IST

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@suresan.....please check out dimensions of ur ans.....

ankur khurana

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13 Mar 2008 20:09:02 IST

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sorry i read potential not potential energy

suresan

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13 Mar 2008 20:45:35 IST

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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.

zamuka chen

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14 Mar 2008 11:29:38 IST

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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????

suresan

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14 Mar 2008 17:37:17 IST

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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

suresan

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14 Mar 2008 17:56:36 IST

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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

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

zamuka chen

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14 Mar 2008 18:58:23 IST

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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.......

sarthak

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14 Mar 2008 20:08:22 IST

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i don understand how can u assume the charge to be at the centre its not a spherical object

Abhishek Ray

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15 Mar 2008 10:12:48 IST

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getting it as

²R³/6

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