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24.4. Capacitance of an Oscilloscope. 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 \(\mathrm{cm}\) on a side and separated by \(5.0 \mathrm{mm},\) with vacuum in between. What is the capacitance of these deflecting plates and hence of the oscilloscope? (Note: This capacitance can sometimes have an effect on the circuit you are trying to study and must be taken into consideration in your calculations.)

Short Answer

Expert verified
The capacitance of the oscilloscope's deflecting plates is 1.593 pF.

Step by step solution

01

Identify the formula for capacitance

The capacitance for parallel plates is given by the formula: \( C = \frac{\varepsilon_0 A}{d} \), where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \text{ F/m} \)), \( A \) is the area of the plates, and \( d \) is the separation distance between them.
02

Calculate the area of the plates

The area \( A \) of the square plates is given by the formula for the area of a square: \( A = \text{side}^2 \). Each side is 3.0 cm, so convert to meters, \( 3.0 \text{ cm} = 0.03 \text{ m} \). Therefore, \( A = (0.03 \text{ m})^2 = 0.0009 \text{ m}^2 \).
03

Convert separation distance to meters

The separation distance \( d \) between the plates is given as 5.0 mm, which needs to be converted to meters: \( 5.0 \text{ mm} = 0.005 \text{ m} \).
04

Substitute and solve for capacitance

Now, substitute the values into the capacitance formula: \( C = \frac{8.85 \times 10^{-12} \times 0.0009}{0.005} \), which calculates to \( C = 1.593 \times 10^{-12} \text{ F} \) or 1.593 pF.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oscilloscope
An oscilloscope is a vital tool in the electronic world, acting as a window into the behavior of electronic circuits. It visually represents how voltage changes over time on its screen. Engineers and technicians find oscilloscopes incredibly useful for troubleshooting circuits or for observing signals. Inside, the oscilloscope has deflecting plates which manipulate an electron beam to create a visible waveform on the display. This electron beam follows the electrical signals and the oscilloscope translates these signals into graphical plots. The efficiency of an oscilloscope depends on its ability to accurately display these plots without distortion. It helps users analyze frequency, amplitude, and other properties of electronic signals, ensuring that everything from simple circuits to complex communication systems functions properly. The capacitance of the deflecting plates can affect these signals, so it's crucial to calculate and understand it.
Parallel plate capacitor
A parallel plate capacitor consists of two large plates separated by a small distance, typically filled with a dielectric substance, or as in the case of the oscilloscope, vacuum. These plates can store electric charge. The plates in the oscilloscope are referred to as deflecting plates, and they have a specific role in directing the path of the electron beam. These plates have a capacitance, which is their ability to store charge. Calculating the capacitance involves knowing the size of the plates, their separation, and the material in between them. In oscilloscopes, the formula for capacitance is often used as \( C = \frac{\varepsilon_0 A}{d} \), with \( A \) as the area of the plates, \( d \) the distance between them, and \( \varepsilon_0 \) the permittivity of free space.
Electron beam deflection
Electron beam deflection is a process used in oscilloscopes to create images of electrical signals. When an electron beam passes between charged plates in an oscilloscope, the electric field causes the beam to bend. This bending helps in creating the trace of the electrical signal on the oscilloscope screen. By carefully calculating the angles and deflection, oscilloscopes can precisely represent fluctuating voltages over time. The deflection plates must have a specific capacitance to ensure that the electron beam responds correctly to the input signals. Adjusting the voltage on these plates can sharpen the image and adjust the path the beam takes, adapting the display to capture a wide range of frequencies and signal strengths.
Permittivity of free space
Permittivity of free space, symbolized as \( \varepsilon_0 \), is a fundamental constant representing the ability of the vacuum to permit electric field lines. In simpler terms, it measures how easily electric fields can flow through space. It's a key factor in calculating the capacitance of a capacitor. For a parallel plate capacitor, \( \varepsilon_0 \) is used to determine how much electric field can cross through the space between the plates without interfering materials. This constant is approximately \( 8.85 \times 10^{-12} \text{ F/m} \). Understanding this concept is crucial because it helps define how energy is stored in electric fields within oscilloscopes and other devices, impacting how accurately they can measure and display the signals they are assessing.

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Most popular questions from this chapter

24.11. A cylindrical capacitor has an inner conductor of radius 1.5 \(\mathrm{mm}\) and an outer conductor of radius 3.5 \(\mathrm{mm}\) . The two conductors are separated by vacuum, and the entire capacitor is 2.8 \(\mathrm{m}\) long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 \(\mathrm{mV}\) higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

24.41. The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of \(1.60 \times\) \(10^{7} \mathrm{V} / \mathrm{m}\) . The capacitor is to have a capacitance of \(1.25 \times 10^{-9} \mathrm{F}\) and must be able to withstand a maximum potential difference of 5500 \(\mathrm{V}\) . What is the minimum area the plates of the capacitor may have?

24.25. A \(5.80-\mu F\) , parallel-plate, air capacitor has a plate separation of 5.00 \(\mathrm{mm}\) and is charged to a potential difference of 400 \(\mathrm{V}\) . Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3}\) .

24.29. A parallel-plate vacuum capacitor with plate area \(A\) and separation \(x\) has charges \(+Q\) and \(-Q\) on its plates. The capacitor is disconnected from the source of charge, so the charge on each plate remains fixed. (a) What is the total energy stored in the capacitor? (b) The plates are pulled apart an additional distance \(d x\) . What is the change in the stored energy? (c) If \(F\) is the force with which the plates attract each other, then the change in the stored energy must equal the work \(d W=F d x\) done in pulling the plates apart. Find an expression for \(F\) (d) Explain why \(F\) is not equal to \(Q E\) , where \(E\) is the electric field between the plates.

24.48 . A parallel-plate capacitor has plates with area 0.0225 \(\mathrm{m}^{2}\) separated by 1.00 \(\mathrm{mm}\) of Tefion. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 \(\mathrm{V}\) . (b) Use Gauss's law (Eq. 24.23 ) to calculate the electric field inside the Teflon. (c) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

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