if each parallel conducting plates have area 1, then another area 2 is added to each plates, if i look at this formula Q=CV=epsilon(A1+A2)V/d, then shouldnt charge increases then charge density also increases

To understand how the addition of an extra area affects the charge and charge density in parallel conducting plates, let's break down the formula Q = CV = ε(A₁ + A₂)V/d:

Q represents the total charge stored on the plates.
C is the capacitance, which is determined by the geometry and material properties of the plates and the medium separating them.
V is the potential difference between the plates.
ε is the permittivity of the medium between the plates.
A₁ is the initial area of each plate.
A₂ is the additional area added to each plate.
d is the separation distance between the plates.

When the area of each plate is increased from A₁ to (A₁ + A₂), it means we have effectively increased the total area available for charge to accumulate. This implies that more charge can be stored on the plates, given the same potential difference and separation.

So, if we assume that the potential difference and the separation distance remain constant, adding the extra area will indeed lead to an increase in the total charge (Q). This is because the capacitance (C) is directly proportional to the total area (A₁ + A₂), as per the formula C = ε(A₁ + A₂)/d.

However, the charge density (σ) is defined as the charge per unit area. This means that as you increase the total area, the charge density will actually decrease. Since the charge is distributed across a larger area, the charge density decreases even though the total charge increases.

Therefore, even though the charge increases when adding area to each plate, the charge density decreases as the charge is spread out over a larger surface area.