Consider the following arrangement of four plates interconnected as follows . Each plate has the area A and the polate separation is d . Here plates 1 and 3 are interconnected . Find the equivalent capacitance between a and b

Explanation

Note that this kind of arrangement is done when we want to constant more capacitors using less number of plates . In the above arrangement we have in total three capacitors using only four plates (however ordinarily you require six plates to construct three capacitors ). This has been done by sharing the plate 2 with plate 1 and 3 has been shared by plate 2 as well as plate 4 . The method of solving this kind of problem is to sketchthe complete network of three capacitors showing each of three capacitors separately.

Here plate 2 and plate 3 are being shared by twop adjacent plates so in the diagram they will appear twice . Remember you have to show three capacitors using plate groups 1-2,2-3 and ,3-4 . First start with point a and you can see it is connected to plate 2 and plate has to appear twice in your network because it is being shared with plate 1 and 3 . So draw it as shown :
Now since plate 1 is on one side of plate 2 and plate 3 on the other side. draw them in front of plate 2 as shown :

Now you see plate 1 and 3 are inter connected so join them as follow :

Then you see that even plate 3 is being shared by two plates viz . plate 2 and 4 . So plate 3 should also appear twice in your diagram as shown below:

Now you note that plate 4 is in front of plate 3 and 4 is connected to point b . so draw it as follows :

So your final diagram has three capacitors each of capacitance C`(C=(epsilon_(0)A)/(d))` in which two are connected in parallel and in series with the third one as shown :

Hence the final capacitance is `(2C'C)/(2C+C)=(2C)/(3) or (2)/(3) (epsilon_(0)A)/(d)`