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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 May 2007 20:44:13 IST
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For the last one,...
apply gauss law on a surface passing through the thickness of the shell..
since it is conducting...so there can't be electric field in all points inside it...so E for the gauss law is zero...Since lhs is 0 so rhs is also 0..
so charge enclosed is 0
So a -q is induced at the inside surface of the shell
But by conservation of charge ...shell should remain uncharged..
so a +q is induced at outer surface..
NOTE: charges always reside on surface of a CONDUCTOR
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31 May 2007 18:00:28 IST
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ans of Q1
for maximun electric force df/dq = 0
where f = q(Q-q) = qQ - q^2
df/dq = Q - 2q=0
=>. q = Q/2
then the charge were half half divide
& maximum distanes solved by formula
F=Kq2q1/r^2
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