Question

Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates. Typically, they are squares 3.00cm on a side and separated by 5.00mm , 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.)

Answer 1

Answer:

1.6 pF

Explanation:

The capacitance of a parallel-plate capacitor in air is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the separation between the plates

In this problem we have:

- Separation between the plates: d = 5.00 mm = 0.005 m

- Area of the plates: $A = 3.00 cm \cdot 3.00 cm = 9.00 cm^2 = 9\cdot 10^{-4}m^2$

Therefore, the capacitance is

$C=\frac{(8.85\cdot 10^{-12}F/m)(9\cdot 10^{-4} m^2)}{0.005 m}=1.6\cdot 10^{-12} F=1.6 pF$