Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates. Typically, they are squares 3.0 cm on a side and separated by 5.0 mm, with very thin air in between.

What is the capacitance of these deflecting plates and hence of the oscilloscope. (This capacitance can sometimes have an effect on the circuit you are trying to study and must be taken into consideration in your calculations.)

Question

contestada

Respuesta :

mahendra678 mahendra678 · Answer 1 · 2020-04-08T10:34:03+02:00

Answer:

The capacitance of the deflecting plates is [tex]C=1.59 pF[/tex].

Explanation:

The expression for the capacitance of the capacitor in terms of area and distance is as follows;

[tex]C=\frac{\varepsilon _{0}A}{d}[/tex]

Here, C is the capacitance, A is the area, d is the distance and [tex]\varepsilon _{0}[/tex] is the absolute permittivity.

Convert the side of the square from cm to m.

s= 3.0 cm

s= 0.030 m

Calculate the area of the square.

[tex]A= s_^{2}[/tex]

Put s= 0.030 m.

[tex]A=(0.030)_^{2}[/tex]

[tex]A=9\times10^{-4}m^{-2}[/tex]

Convert distance from mm to m.

d= 5.0 mm

[tex]d=5\times10^{-3}m[/tex]

Calculate the capacitance of the deflecting plates.

[tex]C=\frac{\varepsilon _{0}A}{d}[/tex]

Put [tex]d=5\times10^{-3}m[/tex], [tex]A=9\times10^{-4}m^{-2}[/tex] and [tex]\varepsilon _{0}=8.85\times 10^{-12}Fm^{-1}[/tex].

[tex]C=\frac{(8.85\times 10^{-12})(9\times10^{-4})}{5\times10^{-3}}[/tex]

[tex]C=1.59\times 10^{-12}F[/tex]

[tex]C=1.59 pF[/tex]

Therefore, the capacitance of the deflecting plates is [tex]C=1.59 pF[/tex].