A parallel-plate capacitor is connected to 1 kV. Its dielectric has relative permittivity K = 6π and breakdown strength 10⁷ V m⁻¹; for safety the field cannot exceed 10% of the breakdown field. For capacitance 50 pF, the minimum plate area is p mm². Find p/75. Use ε₀ = (1/36π)×10⁻⁹ F m⁻¹.
Physics · Mixed Chapter
JEE Mains

Question

A parallel-plate capacitor is connected to 1 kV. Its dielectric has relative permittivity K = 6π and breakdown strength 10⁷ V m⁻¹; for safety the field cannot exceed 10% of the breakdown field. For capacitance 50 pF, the minimum plate area is p mm². Find p/75. Use ε₀ = (1/36π)×10⁻⁹ F m⁻¹.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Safe electric field: Dielectric strength = 10⁷ V m⁻¹; allowed field = 10% of that value, so Eₛₐ𝒻ₑ = 10⁶ V m⁻¹. Step 2 — Minimum separation: For 1 kV = 1000 V, E = V/d; therefore dₘᵢₙ = V/Eₛₐ𝒻ₑ = 1000/10⁶ = 10⁻³ m. Step 3 — Capacitor formula: C = Kε₀A/d ⇒ A = Cd/(Kε₀). Here C = 50×10⁻¹² F, K = 6π and ε₀ = 10⁻⁹/(36π) F m⁻¹. Step 4 — Plate area: A = [50×10⁻¹²×10⁻³]/[(6π)(10⁻⁹/36π)] = 3×10⁻⁴ m² = 300 mm². Step 5 — Requested numerical answer: p = 300 ⇒ p/75 = 300/75 = 4. Answer: 4.