# The expression for the equivalent capacitance of the system shown above is (A is the cross-sectional area of one of the plates)

Let’s find the capacity of each capacitor
$C_1 = \large\frac{\in_0 A}{d}$
$C_2 = \large\frac{\in_0 A}{2d}$
$C_3 = \large\frac{\in_0 A}{3d}$
The three capacitors are in parallel.
Equivalent capacitance $C = C_1+ C_2+C_3$
$\qquad= \large\frac{\in_0 A}{d} +\frac{\in_0 A}{2d}+\frac{ \in_0 A}{3d}$
$\qquad= \large\frac{\in_0 A}{d} \bigg( 1+ \large\frac{1}{2} +\frac{1}{3} \bigg)$
$\qquad= \large\frac{11}{6} . \bigg( \large\frac{\in_0 A}{d} \bigg)$