Question: A variable dielectric capacitive displacement...

Question: A variable dielectric capacitive displacement sensor consists of two square metal plates of side 5 cm, separated by a gap of 1 mm. A sheet of dielectric material 1 mm thick and of the same area as the plates can be slid between them. (a) (15 points) Derive the equation representing the relation between the capacitance and the displacement (b) (10 points) Given that the dielectric constant of air is 1 F/m and that of the dielectric material 4 F/m, calculate the capacitance of the sensor when the input displacement x = 2.5 cm. C = ?0?A / d where ?0 is the permittivity of free space (vacuum) of magnitude 8.85 pF/m, ? is the relative permittivity or dielectric constant of the insulating material, A m² is the area of overlap of the plates, and d m is their separation.

Transcript

00:01 Hello, in the question we have given that a variable dielectric capacitive displacement sensor consists of two square metal plates of side.

00:10 So, the width is given that is equals to 5 centimeters.

00:15 So, the length of this will be 5 only.

00:17 So, it is a square plate and separated by a gap of 1 millimeter.

00:23 So, d is 1 millimeters.

00:26 We will convert it when we will plug it in the formulas, then a sheet of dielectric material 1 millimeter thick of same area that of plates can be slide between them.

00:41 So, we have to derive.

00:43 So, we have to derive the equation representing the relation between the capacitance and the displacement.

00:49 Now, suppose if we draw one diagram.

00:53 So, suppose these are the plates over here, the capacitive plates, sorry, the conducting plates.

01:05 So, these are square plates.

01:06 I am just drawing the front view.

01:09 So, now here we have this dielectric material.

01:13 So, whose thickness is 1 millimeter itself.

01:17 So, sorry, so over here, so this is the dielectric material which is inserted.

01:29 So, this dielectric material of some dielectric constant epsilon r 2, let's say.

01:36 So, it is inserted.

01:38 Now, let's say this is the displacement to which it is inserted.

01:42 So, from the plate, let's say this distance is x.

01:46 So, this distance will be l minus x and they have given that the distance between the plate is d.

01:55 Now, so if we see now this will act as one capacitor, so which has the medium air and we have another capacitor over here which has the medium of dielectric.

02:11 So, now these two capacitors are parallel.

02:15 So, let's say this has the capacitance of c1 and this has the capacitance of c2.

02:21 So, we can say that c1 is parallel to c2.

02:25 So, the equivalent capacitance will be c1 plus c2.

02:29 Now, we know that the capacitance is given by, so i'll just write over here.

02:35 So, we know that c is given by epsilon naught a divided by d for the parallel plate capacitor.

02:43 Now, we have the dielectric material.

02:45 So, c1 we are calling it for the air...

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