Question
distance b/w plate of capacitor C &q charge is increased to double . work done=??
as distance d is double thus C= Ae/2d thus capacitance will be halved so wd shld be (1/2 )q2/(C/2) ie 1/4 q2/C
BUT Sir the ans given is=( 1/2)q2/C Plzz do me a little favour by solving this query
Joshi sir comment
(1/2 )q2/(C/2) = q2/C not 1/4 q2/C
so work done = new energy - previous energy
Read 1 Solution.
hey, initialy energy stored in capacitor =(1/2)q2/c
finally..........................................................=(1/2)q /c/2=q2/c
hence, work done= change i energy= Ef - Ei = q2/c - q2/2c = q2/2c unit
RAJAT AWASTHI 11 year ago
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