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Question

A parallel plate capacitive transducer uses plates of area 500 mm2 which are separated by a distance of 0.2 mm. The dielectric medium between the plates is air having permittivity of 8.85×1012F/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
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B
22.125
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C
24.58
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D
245.8
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Solution

The correct option is A 2.458
Initial capacitance

C=ϵoAd

=8.85×1012×500×1060.2×103=22.125 pF

Capacitance after application of displacement

C+ΔC=8.85×1012×500×1060.18×103=24.583pF

Change in capacitance,

ΔC=24.58322.125=2.458pF

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