Question

The expression for the capacitance (C in pF) of a parallel plate capacitor is given by : $C=6.94\times {10}^{-3}\left(d^2/S\right)$. The diameter (d) of each plate is 20 mm and the spacing between the plates (S) is 0.25 mm. The displacement sensitivity of the capacitor is approximately.

• A. 44.4 pF/mm
• B. -44.4 pF/mm
• C. 11.1 pF/mm
• D. -11.1 pF/mm

Answer 1

The correct option is B -44.4 pF/mm

$C=6.94\times {10}^{-3}\left(\frac{d^2}{S}\right)$

$Sensitivity=\frac{\partial C}{\partial S}=-6.94\times {10}^{-3}\frac{d^2}{S^2}$

$=-6.94\times {10}^{-3}\frac{(20{)}^2}{(0.25{)}^2}$

$=-44.4pF/mm$

Note: negative sign indicates that as displacement increases then capacitance decreases.