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
Whether charge increases or not depends upon whether the capacitor remains connected to a constant-voltage source when the area is added. If it is, the n C increases by a factor (A1 +A2)/A1. Area increases by the same factor, and charge density (Q/A) would remain the same.
