Five identical capacitor paltes, each of area A, are arranged such that adjacent plates are at a distance d apart, the plates are connected to a source of emf V as shown in the figure

The charge on plate 1 is ………..and on plate 4 is.............

Explanation

`((epsilon_oAV)/d,(-2epsilon_0AV)/d)`
In the circuit shown in figure,there is a capacitor plates 1 and 2 the capacity of which is `c_1=(epsilon_0AV)/d` potential and difference between its plates is V.Therefore,charge stored in its `=c_1V=(epsilon_0AV)/d`since,plate 1 is connected with possitve terminal,hence this charge q will be possitive .
Plate 4 is making two capacitors ,one with 3 and other with 5.Hence ,charge on it will be `2q or (-2epsilon_0AV)/d`
Charge on its is negative because this is connected with negative plate.Charges on both sides of the plates are shown below.