capacitor one..[7]
ye silly wala pata nahi kyun nahi bana..
this was one of the four question jo nahi bana paper 2 me...
[2]
yaar poora diagram dikha do..
the method should be to take equate the force
Jitna mere ko samajh aaya..
kq2/h2=2.T.2pi.R.(h/r) (h/r is approximate for sin theta) (The 2 comes because there are 2 surfaces. and there will be tension due to each surface..)
(Solution by eureka..)
so I think this should solve..
so isnt h3 = q2/(8pi2Tε)
which is the 4th option.. ?
Even I am getting D it seems :(
h/R is approximate of cos theta naa...
and isse to D hi aega.. jo mera aya...
they ahve taken Kq2/h2=2 T2Ï€R(h/R)
ek extra 2 kaha se aya..
wahi theta.. I took the other angle to be theta..
and din see ur post... so changed the above option..
for the capacitor one,the charges will redistribute to 3Q/2 each,then since the charges are now equal and the caps are in series,an effective capacitor of C/2 and charge 3q/2 gives the answer (A)
1/2Li2= (3q/2)2/(2*(C/2))
i = 3q/√2LC