Question

A capacitive type displacement transducer consists of two triangular plates, placed side by side, with negligible gap in between them and a rectangular plate moving laterally with an uniform air gap of 1 mm between the fixed plates and the moving plate. The schematic diagram is shown below with appropriate dimension.



With the position of the moving plate shown in the figure above, the values of the capacitances $C_{1}and{C}_2$ thus formed are

• A. $C_1=2.14pF,C_2=1.29pF$
• B. $C_1=14.17pF,C_2=21.25pF$
• C. $C_1=339pF,C_2=212pF$
• D. $C_1=18.16npF,C_2=23.2nF$

Answer 1

The correct option is B $C_1=14.17pF,C_2=21.25pF$

$C_1=\frac{A_1ϵ}{D},C_2=\frac{A_2ϵ}{d}$

we have, d =1 mm

$calaulating{A}_1:$


$A_1=\text{Area of ABCD + Area of OAD}$

in $\Delta QXY$

$tan\theta =\frac{XY}{QY}=\frac{10}{20}$

$\therefore tan\theta =\frac{1}{2}$

$\text{In}\Delta \text{OQP,}$

$\tan \theta =\frac{OP}{QP}$

$\frac{1}{2}=\frac{OP}{10}$

$OP=5cm$

$in\Delta OMD$,

$tan\theta =\frac{DM}{OM}$

$\frac{1}{2}=\frac{DM}{4}$

$DM=2cm$

$DM=AO=2cm$

$BP=AB+OA+PO$

$10=AB+2+5$

$\therefore AB=3cm$

$A_1\text{= Area of ABCD + Area of}\Delta OAD$

$=3\times 4+\frac{1}{2}\times 2\times 4$

$=12+4=16c{m}^2$

$A^2=\text{Area of strip BCNP}-A_1$

$=10\times 4-16$

$=40-16=24c{m}^2$

$C_1=\frac{A_1ϵ}{d}=\frac{16\times 8.84\times {10}^{-12}}{{10}^{-13}}=14.17pF$

$C_2=\frac{A_2ϵ}{d}=\frac{24\times 8.84\times {10}^{-12}}{{10}^{-13}}=21.25pF$