A capacitance transducer uses two quartz diaphragms of area 750 mm² separated by a distance 3.5 mm. The capacitance is 370 pF. When a pressure of 900 kN / m² is applied, the deflection is 0.6 mm. The capacitance at this pressure would be

Capacitance transducer: It uses the capacitance for converting the mechanical movement into an electrical signal. The input quantity causes the change of the capacitance which is measured by the transducer. Capacitors measure both the static and dynamic changes.

\(C = \frac{{\varepsilon A}}{d}\) 

That means, capacitance C ∝ Area

\(C \propto \frac{1}{{distance}}\) 

\(\frac{{{C_1}}}{{{C_2}}} = \frac{{{d_2}}}{{{d_1}}}\)

Given that: Capacitance transducer uses two quartz diaphragms:

Area of diaphragms (A) = 750 mm² = 7.5 × 10^-4 m

Distance (d) = 3.5 mm = 3.5 × 10^-3 m

Capacitance (C) = 370 PF = 3.7 × 10^-10 F

Applied Pressure (P) = 900 kN/m² = 9 × 10⁵ N/m²

Deflection (D) = 0.6 mm = 0.6 × 10^-3 m

To find capacitance at 9 × 10⁵ N/m² pressure- we know that,

Distance between plates d₂ = d₁ – D

= 2.9 mm

\(\Rightarrow {C_2} = {C_1} \cdot \frac{{{d_1}}}{{{d_2}}} = 370 \times {10^{ - 12}} \times \frac{{3.5 \times {{10}^{ - 3}}}}{{2.9 \times {{10}^{ - 3}}}}\) 

= 446.6 × 10^-12

⇒ C₂ ≈ 447 PF