as per the image we name the plates
now we simplify it as follows by naming the junctions
here C' is the capacitance
C'=S*E/d (read epsilon)
now simplification is easy
Each plate has plate area S. Find the capacitance between A and B.
Can anyone convert this to a series parallel combination ? I did by charge distribution and wrote Q=CV for all of them.(that takes at least 10 mins) Can u solve by faster method?
a simpler version of this problem is given in irodov.(3.113)
as per the image we name the plates
now we simplify it as follows by naming the junctions
here C' is the capacitance
C'=S*E/d (read epsilon)
now simplification is easy
nasiko can u pls solve it using your method ...atleast the labelled diagram with 2-3 steps..please im weak in this