1please answer .
a parallel plate capacitor ( area = 20 cm2) ( distance between the plates = 1mm) is connected to a battery of 12 V . this separation is then increased to 2mm. We see that no heat was produced. why ?? we can see this by checking out the values for initial / final energy and work done ....but ...there was a battery ,, conducting wires the how come was no energy lost ???
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4 Answers
3e(20)(10)/1 = C
e(20)(10)/2 = c1
therefoere energy lost = 1/2[e(20)(10)/2]144
now new charge = 12 = q2/C1
q2 = C1 12
INITIAL CHARGE = C 12
-e(20)(10)/2 144 = net energy requiered to mobilise the charge !!!!!!!!!!!
so we gain energy by mobilising lesser charge thatz wat it means ........
the potential engy lost is lesser than this ........
hence no heat loss!!!!!!!!
1we can see this by checking out the values for initial / final energy and work done ....but ...there was a battery ,, conducting wires the how come was no energy lost ???