A parallel-plate capacitor connected to 500 V has plate separation 1 mm. When immersed in water (relative permittivity 81), the change in electric pressure between its plates is kε₀×10¹³ N m⁻². Find k.
Physics · Mixed Chapter
JEE Mains

Question

A parallel-plate capacitor connected to 500 V has plate separation 1 mm. When immersed in water (relative permittivity 81), the change in electric pressure between its plates is kε₀×10¹³ N m⁻². Find k.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Electric field between plates: The supply remains connected, so E = V/d = 500/(1×10⁻³) = 5×10⁵ V m⁻¹. Step 2 — Electrostatic pressure: At fixed V and d, the magnitude of electric pressure in a medium of relative permittivity εᵣ is p = ½εᵣε₀E². Step 3 — Change after filling with water: Δp = ½(εᵣ − 1)ε₀E². For εᵣ = 81, Δp = ½(80)ε₀(5×10⁵)² = 10¹³ε₀ N m⁻². Step 4 — Compare coefficients: Δp = k×10¹³ε₀ N m⁻² ⇒ k = 1. Answer: 1.