A parallel plate capacitor with area 200cm² and separation between the plates 1.5cm, is connected across a battery of emf V. If the force of attraction between the plates is 25 × 10⁻⁶ N, the value of V is approximately: (ε₀ = 8.85 × 10⁻¹² C²N⁻¹.m⁻²)

A = 200cm²
d = 1.5cm
F = 25 × 10⁻⁶ N
ε = ε₀
E = σ/2ε₀ = Q/2Aε₀
F = QE = Q²/2Aε₀
But Q = CV = ε₀A(V)/d
∴ F = (ε₀AV/d)²/2Aε₀ = (ε₀A)² × V²/d² × 2Aε₀ = (ε₀A)V²/d² × 2
25 × 10⁻⁶ = (8.85 × 10⁻¹²) × (200 × 10⁻⁴) × V²/2.25 × 10⁻⁴ × 2
V = √(25 × 10⁻⁶ × 2.25 × 10⁻⁴ × 2/(8.85 × 10⁻¹² × 200 × 10⁻⁴))
Here, on solving, V ≈ 250V