Multiple again

accha got it is it 3f???

For such problems, solve by numbering the plates as 1,2,3,4... this will avoid all sort of confusions... n u'll not commit any error (unless u do some silly mistake) !!!

for example, in d above problem numbering the plates can be done as 1,2,1,3 from top to bottom (first n third plate are given same no. as they are connected)

we need to find d net capacitance b/w A & B i.e., b/w 2 & 3

now, two capacitors exist between 1 and 2 and one capacitor exists between 1 and 3... first two capacitors are parallelly connected, so net capacitance will be 2C and this is connected in series to the third one... hence net capacitance = 2C/3...

therefore C_AB=2C/3 ........

dun laugh at me for such a detailed explanation... this is juzz for ur clear understanding so dat u can solve all such similar problems :)

Thanks a lot race......i needed the explanation too :)