An air capacitor of capacitance C plate area A and plate separation d has been filled with three dielectrics of constant `k_(1)=3 , k_(2)=6` and `k_(3)=4 ` as shown in figure . What is the new capacitance ?

Explanation

Shortcut method :
Before solving this problem try to understand the following concept . When a capacitor of capacitance C is divided into two parts in such a manner that each capacitor plate has area half of the earller vaue than each capacitor is understood to be connected in parallel and each one has value C/2 as shown below :

But if you divided the gap of the capacitor in two equal parts parallel to the plates then each one will have the capacitance 2 C (as d has been halved ) connected in series to each oither as shown below:

Now in the given question you can see the capacitor has been divided into three chambers viz. left half and the right half which is further into two part i.e upper half and lower half .

So all you have to do first is that write down the value of these separated chamgers in terms of C . Left half is `(C )/(2)` and right half `(C)/(2) ` is multipled by 2 each for upper and lower half .

( That means right side C and C are in series and they are in parallel to C/2 making the entire capacitance C)
Now put the respective dielectrics in the chambers so each chamber will have its new value of capacitance K ( dielectric constant ) times the old value .

Now the right side 6 C and 3 C and in series so their equivalent value is `(6Cxx3C)/(6Cxx+3C)` =2C and this 2C is in parallel to the left side chamber capacitance 2 C . So the total capacitance becomes 4 C.