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Chapter 23: Problem 66

Deffecting Plates of an Oscilloscope. The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but opposite charges. Typical dimensions are about 3.0 \(\mathrm{cm}\) on a side, with a separation of about 5.0 \(\mathrm{mm} .\) The potential difference between the plates is 25.0 \(\mathrm{V} .\) The plates are close enough that we can ignore fringing at the ends. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

Short Answer

Expert verified
(a) Charge on each plate is approximately \(2.0 \times 10^{-12} \text{ C}\), (b) Electric field strength is \(5.0 \times 10^3 \text{ V/m}\), (c) Electron speed is approximately \(2.97 \times 10^6 \text{ m/s}\).

Step by step solution

01

Calculate the Capacitance

The capacitance for parallel plates is given by the formula \(C = \varepsilon_0 \frac{A}{d}\), where \(A\) is the area of the plates and \(d\) is the separation between them. The plate area \(A\) can be calculated as \(A = (3.0 \times 10^{-2} \text{ m})^2 = 9.0 \times 10^{-4} \text{ m}^2\) and \(d = 5.0 \times 10^{-3} \text{ m}\). Using \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2)\), calculate the capacitance.
02

Calculate the Charge on Each Plate

Use the formula \(Q = C \cdot V\) to find the charge \(Q\) on each plate. The potential difference \(V\) is given as 25.0 V. Substitute the capacitance \(C\) from the previous step and multiply with \(V\) to find \(Q\).
03

Calculate the Electric Field

The electric field \(E\) between parallel plates is given by \(E = \frac{V}{d}\). Use the given potential difference \(V = 25.0 \text{ V}\) and the separation \(d = 5.0 \times 10^{-3} \text{ m}\) to compute \(E\).
04

Calculate the Speed of the Electron

First, determine the kinetic energy gained by the electron as it moves from one plate to the other, using the potential energy change \(\Delta U = e \cdot V\), where \(e = 1.6 \times 10^{-19} \text{ C}\) is the charge of an electron and \(V = 25.0 \text{ V}\). The kinetic energy \(K = \frac{1}{2} m v^2\), where \(m = 9.11 \times 10^{-31} \text{ kg}\) is the mass of the electron. Equating \(\Delta U\) and \(K\), solve for the velocity \(v\) of the electron.

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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 device used to visualize and analyze the waveforms of electronic signals. It displays the voltage of an electronic signal over time, allowing users to observe the behavior of circuits and electrical systems. In this context, the oscilloscope utilizes deflection plates to control the movement of the electron beam, creating a visible pattern on the screen. These deflection plates are critical components:
  • Vertical deflection plates control vertical movement.
  • Horizontal deflection plates control horizontal movement.

The electron beam inside the oscilloscope passes between these plates, allowing the resulting image to be moved up or down, left or right, on the oscilloscope screen. This visualization helps in understanding various characteristics of electronic signals, including amplitude, frequency, and any distortions present.
Electric Field
The electric field is a vector field surrounding electric charges that exerts a force on other charges in the field. In a classroom oscilloscope, the electric field between the vertical deflecting plates is crucial for manipulating the electron beam. The strength of this electric field can be calculated using the formula:
  • \[ E = \frac{V}{d} \]

Here, \(E\) represents the electric field strength, \(V\) is the potential difference across the plates (25.0 V), and \(d\) is the separation between the plates (5.0 mm). This electric field causes electrons to accelerate as they travel between the plates, impacting the deflection of the beam on the oscilloscope screen.
Capacitance
Capacitance is a measure of a system's ability to store charge per unit voltage. For the vertical deflecting plates in an oscilloscope, the capacitance \(C\) can be calculated using the formula:
  • \[ C = \varepsilon_0 \frac{A}{d} \]

Where \(\varepsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2\)), \(A\) is the area of a plate (\(9.0 \times 10^{-4} \text{ m}^2\)), and \(d\) is the separation between the plates (5.0 mm). This capacitance affects the amount of electric charge each plate can hold when a voltage is applied, influencing the electric field strength and, consequently, the behavior of the electron beam inside the oscilloscope.
Electron Velocity
The velocity of an electron moving between the deflecting plates in an oscilloscope can be computed by considering the energy changes involved. The electron starts with potential energy, which gets converted into kinetic energy as it accelerates toward the positive plate. Using the relationship:
  • \( \Delta U = e \times V \)

This potential energy change \(\Delta U\) (with \(e\) as the electron charge, \(1.6 \times 10^{-19}\) C) equals the gain in kinetic energy \(K\) when the electron reaches the positive plate:
  • \( K = \frac{1}{2} mv^2 \)

By equating these energies and solving for velocity \(v\), one can determine how fast the electron moves upon reaching the positive plate. This calculation is vital for understanding the dynamics of electron movement in the presence of an electric field inside the oscilloscope.

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Most popular questions from this chapter

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) -direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{m} )\) and point \(b\) (at \(x=0.90 \mathrm{m} )\) is 240 \(\mathrm{V}\) . (a) Which point, \(a\) or \(b,\) is at the higher potential? (b) Calculate the value of \(E.\) (c) A negative point charge \(q=-0.200 \mu C\) is moved from \(b\) to \(a\) . Calculate the work done on the point charge by the electric field.

The electric potential \(V\) in a region of space is given by $$V(x, y, z)=A\left(x^{2}-3 y^{2}+z^{2}\right)$$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\vec{E}\) at any point in this region. (b) The work done by the field when a \(1.50-\mu \mathrm{C} \quad\) test charge moves from the point \((x, y, z)=\) \((0,0,0.250 \mathrm{m})\) to the origin is measured to be \(6.00 \times 10^{-5} \mathrm{J}\). Determine \(A .\) (c) Determine the electric field at the point \((0,0,\) 0.250 \(\mathrm{m} ) .\) (d) Show that in every plane parallel to the \(x z\) -plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V=1280 \mathrm{V}\) and \(y=2.00 \mathrm{m} ?\)

A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a,\) and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{J} .\) When the second charge is moved to point \(b,\) the electric force on the charge does \(-1.9 \times 10^{-8} \mathrm{J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\) ?

In a certain region, a charge distribution exists that is spherically symmetric but nonuniform. That is, the volume charge density \(\rho(r)\) depends on the distance \(r\) from the center of the distribution but not on the spherical polar angles \(\theta\) and \(\phi .\) The electric potential \(V(r)\) due to this charge distribution is $$V(r)=\left\\{\begin{array}{l}{\frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a} \\ {0} & {\text { for } r \geq a}\end{array}\right.$$ where \(\rho_{0}\) is a constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) and \(a\) is a constant having units of meters. (a) Derive expressions for \(\vec{E}\) for the regions \(r \leq a\) and \(r \geq a .[\)Hint\(:\) Use Eq. \((23.23) .]\) Explain why \(\vec{\boldsymbol{E}}\) has only a radial component. (b) Derive an expression for \(\rho(r)\) in each of the two regions \(r \leq a\) and \(r \geq a .[\)Hint t: Use Gauss's law for two spherical shells, one of radius \(r\) and the other of radius \(r+d r .\) The charge contained in the infinitesimal spherical shell of radius \(d r\) is \(d q=4 \pi r^{2} \rho(r) d r . ](c)\) Show that the net charge contained in the volume of a sphere of radius greater than or equal to \(a\) is zero. [Hint: Integrate the expressions derived in part (b) for \(\rho(r)\) over a spherical volume of radius greater than or equal to \(a.\)) Is this result consistent with the electric field for \(r>a\) that you calculated in part (a)?

Electric charge is distributed uniformly along a thin rod of length \(a,\) with total charge \(Q .\) Take the potential to be zero at infinity. Find the potential at the following points (Fig. P23.79): (a) point \(P,\) a distance \(x\) to the right of the rod, and (b) point \(R,\) a distance \(y\) above the right- hand end of the rod. (c) In parts (a) and \((\mathrm{b}),\) what does your result reduce to as \(x\) or \(y\) becomes much larger than \(a\) ?
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