Question

A capacitance displacement transducer consists of two triangular plates separated by uniform air gap of $1mm$ between fixed and moving plates. The value of capacitance $\left(C_{ab}\right)$ between plates of this transducer in $pF$ is
(Note: Plates are kept exactly one above the other)


Given data
$x=6cm,z=6cmand{\epsilon }_0=8.85\times {10}^{-12}\left(F/m\right)$

• A. 15.93

Answer 1

The correct option is A 15.93
$C=\frac{{\epsilon }_{o}A}{d}=\frac{8.85\times {10}^{-12}\times (\frac{1}{2}\times 6\times 6\times {10}^{-4})}{1\times {10}^{-3}}=15.93pF$

Here $A$ is the common area means only overlapping area of the two triangular plates.