Find the effective capacitance between the point P and Q in the given figure

Hint:The given circuit is a Wheatstone bridge circuit and here , since all the capacitances have the same value, the values are same on either side of the formula. Reconstruct the diagram and deduce the value.

Formulas used:
If C1 and C2 are two capacitors that are connected in series then their net capacitance would be
${{C}_{{}}}=\dfrac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}$
When C1 and C2 are connected in parallel then its net capacitance would be
$C={{C}_{1}}+{{C}_{2}}$

Complete step by step answer:
The circuit can be reconstructed as a Wheatstone bridge circuit:

Where all the capacitors hold the same value, that is $4\mu F$
The capacitor in the middle can be neglected, as there is no flow of charge through it.
Now, the capacitors near P are in series. If they are taken as C1 and C2,
Its resultant capacitance would be $\dfrac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\dfrac{16}{8}=2\mu F$
Similarly the capacitors near Q are also in series and their resultant capacitance would also be $2\mu F$.Now , both these newly generated capacitance values are in series.

Hence the total capacitance of the system would be the sum of these values,i.e, $2\mu F+2\mu F=4\mu F$

Note:Whenever we are solving questions that involve finding the net value of capacitance, first the students must construct a Wheatstone’s circuit to simplify the solving procedure. It is also given in the question that it is a wheatstone’s bridge circuit which is an electric circuit used to measure an unknown capacitance by balancing two legs of a bridge circuit , one leg of which includes the unknown. Hence , this concept is used.