Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 10⁶ V/m. The plate area is 10⁻⁴ m². What is the dielectric constant if the capacitance is 15 pF?

Correct option is D. 8.5
A = 10⁻⁴ m²
Emax = 10⁶ V/m
C = 15 pF = 15 × 10⁻¹² F
C = κε₀A/d
where:
C = Capacitance
k = dielectric constant
ε₀ = permittivity of free space (8.86 × 10⁻¹² F/m)
A = area of plates
d = distance between plates
V = Ed = 500 V
d = V/Emax = 500 V / 10⁶ V/m = 5 × 10⁻⁴ m
Substituting values in the capacitance formula:
15 × 10⁻¹² F = k × (8.86 × 10⁻¹² F/m) × (10⁻⁴ m²) / (5 × 10⁻⁴ m)
k = (15 × 10⁻¹² F) × (5 × 10⁻⁴ m) / ((8.86 × 10⁻¹²) × (10⁻⁴ m²))
k = (75 × 10⁻¹⁶) / (8.86 × 10⁻¹⁶)
k = 8.465
k ≈ 8.5