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Chapter 18: Problem 13

Oscilloscope. Oscilloscopes are found in most science laboratories. Inside, they contain deflecting plates consisting of more-or-less square parallel metal sheets, typically about 2.5 \(\mathrm{cm}\) on each side and 2.0 \(\mathrm{mm}\) apart. In many experiments, the maximum potential across these plates is about 25 \(\mathrm{V}\) . For this maximum potential, (a) what is the strength of the electric field between the plates, and (b) what magnitude of acceleration would this field produce on an electron midway between the plates?

Short Answer

Expert verified
(a) 12500 V/m, (b) 2.2 x 10^{15} m/s^2

Step by step solution

01

Understand the Problem

We have parallel plates with certain dimensions and a maximum potential difference across them. We need to calculate two things: the electric field strength between the plates and the acceleration of an electron placed midway between them. These calculations require knowledge of potential difference, distance between plates, electron charge, and mass.
02

Calculate Electric Field Strength

The electric field strength between two parallel plates is given by the formula: \[ E = \frac{V}{d} \] where \( V = 25 \, \text{V} \) is the potential difference, and \( d = 2.0 \, \text{mm} = 0.002 \, \text{m} \) is the distance between the plates. Substitute these values into the formula: \[ E = \frac{25}{0.002} = 12500 \, \text{V/m} \] Thus, the electric field strength is \( 12500 \, \text{V/m} \).
03

Calculate Electron's Acceleration

The acceleration of the electron can be found using the formula: \[ a = \frac{F}{m} \] where \( F \) is the force experienced by the electron and \( m \) is its mass. The force can be calculated using \( F = qE \), where \( q = 1.6 \times 10^{-19} \, \text{C} \) is the electron's charge and \( E = 12500 \, \text{V/m} \). Substitute these values into the force formula: \[ F = (1.6 \times 10^{-19})(12500) = 2.0 \times 10^{-15} \, \text{N} \] Now, using the electron's mass \( m = 9.11 \times 10^{-31} \, \text{kg} \), calculate the acceleration: \[ a = \frac{2.0 \times 10^{-15}}{9.11 \times 10^{-31}} \approx 2.2 \times 10^{15} \, \text{m/s}^2 \] Therefore, the acceleration of the electron is approximately \( 2.2 \times 10^{15} \, \text{m/s}^2 \).

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Key Concepts

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

Electric Field Strength
The concept of electric field strength is crucial in understanding the interaction between charged particles and electric fields.
The electric field strength, represented by the symbol \( E \), is a measure of the force experienced by a unit positive charge placed in the field.
It's given by the formula:
  • \( E = \frac{V}{d} \)
where \( V \) is the potential difference between the plates, and \( d \) is the distance separating them.
In our case, the potential difference (\( V \)) is 25 Volts, and the distance (\( d \)) is 2.0 millimeters (or 0.002 meters).
Substituting these values in the formula, we find that the electric field strength is 12,500 V/m.
This means that each meter in the space between the plates offers an electric potential change of 12,500 Volts.
  • Electric field strength determines how much a charged particle is accelerated by the field.
  • Stronger electric fields exert a greater force on charged particles.
Electron Acceleration
The acceleration of an electron in an electric field is a fascinating topic in physics.
When an electron, which carries a negative charge, enters an electric field, it experiences a force given by the equation:
  • \( F = qE \)
where \( F \) is the force, \( q \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \text{ C} \)), and \( E \) is the electric field strength.
Once the force is known, the acceleration \( a \) of the electron can be calculated using Newton鈥檚 second law:
  • \( a = \frac{F}{m} \)
where \( m \) is the mass of the electron, about \( 9.11 \times 10^{-31} \text{ kg} \).
Plugging in these values, the electron's force is \( 2.0 \times 10^{-15} \text{ N} \), and solving for \( a \), we get approximately \( 2.2 \times 10^{15} \text{ m/s}^2 \).
This immense acceleration showcases just how influential electric fields can be on small charged particles like electrons.
  • Understanding this acceleration is crucial in designing devices like oscilloscopes where precise control of electron paths is required.
  • Electrons can achieve such high accelerations because they are much lighter than most particles.
Parallel Plates
Parallel plates are a fundamental component in the study of electric fields and are commonly used in devices like oscilloscopes.
They consist of two conductive plates placed parallel to each other, creating a uniform electric field when a potential difference is applied.
This setup is ideal for controlled experiments and has real-life applications:
  • Oscilloscopes use parallel plates to direct electron beams.
  • Capacitors, which store electrical energy, often use parallel plates.
The distance between these plates directly influences the electric field and, thus, the acceleration of charged particles.
This concept allows scientists and engineers to manipulate electron paths with precision.
The uniformity of the electric field between parallel plates is key; it ensures consistency in experiments and applications.
  • Electric fields between parallel plates are used to simulate and study various physical phenomena.
  • By understanding how parallel plates function, we can develop better electronic devices.

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Most popular questions from this chapter

Two oppositely charged identical insulating spheres, each 50.0 \(\mathrm{cm}\) in diameter and carrying a uniform charge of magnitude \(175 \mu \mathrm{C},\) are placed 1.00 \(\mathrm{m}\) apart center to center Fig. 18.53 ). (a) If a voltmeter is connected between the nearest points \((a\) and \(b)\) on their surfaces, what will it read? (b) Which point, \(a\) or \(b,\) is at the higher potential? How can you know this without any calculations?

You are working on an electronics project requiring a variety of capacitors, but have only a large supply of 100 nF capacitors available. Show how you can connect these capacitors to produce each of the following equivalent capacitances: (a) \(50 \mathrm{nF},\) (b) \(450 \mathrm{nF},(\mathrm{c}) 25 \mathrm{nF},\) (d) 75 \(\mathrm{nF.}\)

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 \(\mathrm{mm}\) . If the separation is decreased to \(1.15 \mathrm{mm},\) what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.55\(\mu \mathrm{C}\) . (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation?

Electric eels. Electric eels and electric fish generate large potential differences that are used to stun 9 enemies and prey. These potentials are produced by cells that each can generate 0.10 V. We can plausibly model such cells as charged capacitors. (a) How should these cells be connected \((\mathrm{in}\) series or in parallel) to produce a total potential of more than 0.10 \(\mathrm{V} ?\) (b) Using the connection in part (a), how many cells must be connected together to produce the 500 \(\mathrm{V}\) surge of the electric eel?
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