A variable separation displacement sensor (Sensor A) and a variable area displacementsensor (Sensor B) are being used to measure the same input displacement of x = 0.5 mm.The two sensors are identical in their original geometry, each consisting of two square metalplates 10 cm × 10 cm, separated by a gap of 3 mm, and the space between the plates iscompletely filled by a slab with a dielectric constant of 4.5 (the permittivity of free space is 8.85pF/m). For measuring displacement x, answer the following questions.a) Calculate the capacitance of the variable separation displacement sensor, CA, in pF (withtwo digits of accuracy).b) Calculate the capacitance of a variable area displacement sensor, CB, in pF (with two digitsof accuracy).c) Based on the absolute value of sensitivity, which sensor is more sensitive at thismeasurement point (type either Sensor A or Sensor B in the box)?

a) The capacitance of a parallel plate capacitor is given by the formula C = εA/d, where ε is the permittivity of the material between the plates, A is the area of one of the plates, and d is the separation between the plates.

For Sensor A, the separation changes with displacement, so the new separation is d - x = 3mm - 0.5mm = 2.5mm = 2.5*10^-3 m. The area A = 10cm * 10cm = 0.01m * 0.01m = 0.0001 m^2. The permittivity ε = ε0 * εr = 8.85pF/m * 4.5 = 39.825 pF/m.

Substituting these values into the formula, we get:

C_A = εA/d = 39.825 pF/m * 0.0001 m^2 / 2.5*10^-3 m = 1.59 pF.

b) For Sensor B, the area changes with displacement, so the new area is A + 2xd = 0.0001 m^2 + 20.5mm10cm = 0.0001 m^2 + 0.00001 m^2 = 0.00011 m^2. The separation d = 3mm = 3*10^-3 m remains the same.

Substituting these values into the formula, we get:

C_B = εA/d = 39.825 pF/m * 0.00011 m^2 / 3*10^-3 m = 1.46 pF.

c) The sensitivity of a sensor is defined as the change in output (in this case, capacitance) per unit change in input (in this case, displacement). For Sensor A, the change in capacitance for a 0.5mm change in displacement is 1.59 pF - 1.46 pF = 0.13 pF. For Sensor B, the change in capacitance for the same change in displacement is 1.46 pF - 1.59 pF = -0.13 pF. The absolute value of the sensitivity is the same for both sensors, so neither sensor is more sensitive at this measurement point.