A variable dielectric capacitive displacement sensor consists of two square metal plates of5 cm in each side, separated by a 1 mm gap. A 1-mm-thick sheet of a dielectric material canalso slide between the two plates (same area as the plates). Given that the dielectric constant(i.e. relative permittivity) of air is 1 and that of the dielectric material is 4, for each case,calculate the capacitance of the sensor for the input displacements of x = 0.0, 2.5 and 5.0 cm.

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)?

The capacitance of a passive capacitance transducer depends on the following factors, except?1 pointDistance between the two parallel platesThe area of the two parallel platesRelative dielectric constantThe mass of the two parallel plates

An air capacitor has a capacitance12µF. When a dielectric is introduced between the plates, its capacitance becomes 18µF, the relative permittivity of the dielectric is:

Solve the given problem below. Submit your illustration and solutions as PDF format as an attachment to your final answer (60 points). A parallel plate capacitor is constructed of two (2) square conducting plates with side length l = 10.0 cm. The distance between the plates is d = 0.250 cm. A dielectric with a dielectric constant k = 15.0 and thickness 0.250 cm is inserted between the plates. The dielectric is 10.0 cm wide and is 5.0 cm long.Provide an illustration that best represents the scenario (20 points).Derive and provide the formula that shall best address the given scenario (20 points).Determine the capacitance of this system (20 points).

A variable reluctance sensor consists of a core, a variable air gap and an armature. The core is a 1 cm-diameter steel rod with a relative permeability of 100 that forms a 4 cm-diameter semi-circle. A coil of 500 turns is wound onto the core. The armature is a steel plate of thickness 0.5 cm with a relative permeability of 100. Determine the inductance of the sensor for air gaps of 1 mm and 3 mm (ε0 = 8.85×10-12 F m−1)