(c) The arrangement shows a Wheatstone bridge.
As $\frac{C_{1}}{C_{3}} = \frac{C_{4}}{C_{5}} = 1,\ $therefore the bridge is balanced.
$\frac{1}{C_{s_{1}}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2},C_{s_{1}} = 2\mu\ F$
Similarly, ${C_{s}}_{2} = 2\mu\ F$
$\therefore$ effective capacitance
$= C_{p} = {C_{s}}_{1} + {C_{s}}_{2} = 2 + 2 + = 4\mu\ F$