A parallel-plate capacitor possessing a plate area of 25.0 cm ^2 and a plate separation of 2.00 mm is connected to a 12.0 V battery. If the separation between the plates is subsequently reduced to 1.00 mm , determine the extra charge supplied by the battery to the positive plate.

Step-by-step solution

Initial capacitance C₁ = ₀ A d₁ Final capacitance C₂ = ₀ A d₂ Since the battery remains connected, the potential difference V across the capacitor is constant. Initial charge Q₁ = C₁ V = ₀ A V d₁ Final charge Q₂ = C₂ V = ₀ A V d₂ Extra charge supplied Q = Q₂ - Q₁ = ₀ A V ( 1 d₂ - 1 d₁ ) Substituting the given values: A = 25.0 10⁻⁴ m ^2 d₁ = 2.00 10⁻³ m d₂ = 1.00 10⁻³ m V = 12.0 V ₀ = 8.85 10⁻¹² F/m Q = 8.85 10⁻¹² 25.0 10⁻⁴ 12.0 ( 1 1.00 10⁻³ - 1 2.00 10⁻³ ) Q = 265.5 10⁻¹⁵ (1000 - 500) Q = 265.5 10⁻¹⁵ 500 = 1.3275