Five identical capacitor plates each of area A are arranged such that adjacent plates are at d distance apart. Plates are connected to a source of emf V as shown. If charge on plate 1 is $\frac{n{\in }_{0}AV}{d}.$ Find n = ?

These five plates constitute four identical capacitors in positive each of capacity $\frac{{\epsilon }_{0}A}{d}.$ Now as plate 1 is connected to positive terminal of battery and is a part of one capacitor only, so charge $q_1=+\left(\frac{{\epsilon }_{0}AV}{d}\right)=1\left(\frac{{\epsilon }_{0}AV}{d}\right)$