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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 14:40:46 IST
Accepted Answer [?]
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let capacitance of conduc. be Cc & that of plalte be Cp.
now when fisrt operation is done -----
conduc. has q charge & plate is left with Q - q .
as 2 r in contact potential is same , hence we get---
Q - q/q = Cp/Cc ----------------- 1
now when conduc. has aquired max(say qmax) chage & plate is now brought to contact no chage flow. but again potentials r same.
hecne---
Q/qmax = Cp/Cc ------------------ 2
from 1&2
Q - q/q = Q/qmax or
qmax = Qq/Q - q
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this reply: 7 points
(with 1 
in 2 votes ) [?]
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