if each parallel conducting plates have area 1, then another area 2 is added to each plates, is the charge density and charge still be the same
no, the original charge disperses to reduce charge density.
if i look at this formula Q=CV=epsilon(A1+A2)V/d, then shouldnt charge increases then charge density also increases
To determine if the charge density and charge remain the same when an additional area is added to parallel conducting plates, we need to consider the relationship between charge density, charge, and area.
Charge density (ρ) is defined as the amount of charge (Q) per unit area (A). Mathematically, it is expressed as ρ = Q/A.
Given that the initial area of each plate is 1, let's assume that the charge density is ρ₁. Consequently, the charge on each plate can be calculated as Q₁ = ρ₁ * A₁ = ρ₁ * 1 = ρ₁.
Now, if an additional area (2) is added to each plate, the total area of each plate becomes A₁ + 2. The charge density for the new configuration can be denoted as ρ₂.
To maintain the same charge density ρ₁, the total charge (Q₂) for the new configuration would be Q₂ = ρ₂ * (A₁ + 2).
Therefore, for the charge density and charge to remain the same, we need to ensure that ρ₁ = ρ₂ and Q₁ = Q₂.
So, the charge density and charge will not be the same if an additional area (2) is added to each plate, unless the charge density is adjusted accordingly to maintain equilibrium.