Question

A parallel plate capacitive transducer uses plates of area 500 mm${}^2$ which are separated by a distance of 0.2 mm. The dielectric medium between the plates is air having permittivity of $8.85\times {10}^{-12}F/m.$ If a linear displacement reduces the distance between the plates to 0.18 mm, then the change in capacitance is___________pF.

• A. 2.458
• B. 22.125
• C. 24.58
• D. 245.8

Answer 1

The correct option is A 2.458

Initial capacitance

$C=\frac{{ϵ}_{o}A}{d}$

$=\frac{8.85\times {10}^{-12}\times 500\times {10}^{-6}}{0.2\times {10}^{-3}}=22.125pF$

Capacitance after application of displacement

$C+\Delta C=\frac{8.85\times {10}^{-12}\times 500\times {10}^{-6}}{0.18\times {10}^{-3}}=24.583pF$

Change in capacitance,

$\Delta C=24.583-22.125=2.458pF$