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Chapter 23: Problem 64

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 cm on a side, with a separation of about 5.0 mm. The potential difference between the plates is 25.0 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) The charge on each plate is approximately \( 1.33 \times 10^{-12} \) C. (b) The electric field strength is \( 5,000 \, \text{V/m} \). (c) The electron's speed is about \( 2.97 \times 10^6 \, \text{m/s} \).

Step by step solution

01

Calculate Capacitance

The capacitance of two parallel plates is given by the formula \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the separation between the plates. First, calculate the area \( A = (0.03)^2 \) m虏, since the side of the plate is 3.0 cm. The separation \( d = 0.005 \) m. Substitute these into the formula to find \( C \).
02

Calculate Charge on Each Plate

The charge \( Q \) on each plate can be found using the formula \( Q = CV \), where \( V = 25.0 \) V is the potential difference. Using the capacitance \( C \) found in Step 1, calculate \( Q \).
03

Calculate Electric Field Strength

The electric field \( E \) between two parallel plates is given by \( E = \frac{V}{d} \). Use the potential difference \( V = 25.0 \) V and separation \( d = 0.005 \) m to calculate \( E \).
04

Determine Electron's Final Velocity

The kinetic energy gained by the electron as it moves from the negative to the positive plate is equal to the work done by the electric field, calculated as \( KE = eV \), where \( e = 1.6 \times 10^{-19} \) C is the charge of an electron. The kinetic energy is \( \frac{1}{2}mv^2 \). Set \( eV = \frac{1}{2}mv^2 \) and solve for \( v \), where \( m = 9.11 \times 10^{-31} \) kg is the electron mass.

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Key Concepts

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

Parallel Plate Capacitance
Capacitance plays a crucial role in the behavior of parallel plate capacitors. A capacitor's ability to store charge is defined by its capacitance, noted by the symbol \( C \). For two parallel plates, the capacitance is calculated using the formula:
  • \( C = \frac{\varepsilon_0 A}{d} \) where:
  • \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) is the permittivity of free space.
  • \( A \) is the area of one of the plates.
  • \( d \) is the separation distance between the plates.
To find the area \( A \), we square the side length of the plate: \( A = (0.03)^2 \) m虏 for given 3.0 cm sides. The separation, \( d \), is 0.005 m.
This relationship shows how capacitance increases with the plate area and decreases with the plate separation. Thus, larger or closer plates can store more electric charge.
Potential Difference
Potential difference, often referred to as voltage, is the energy difference between two points in an electric field. In a parallel plate capacitor, it is defined by the voltage \( V \) applied across the plates. Potential difference causes electric charges to move, creating an electric field between the plates.
  • This electric field is directly proportional to the potential difference \( V \): \( E = \frac{V}{d} \).
  • For a potential difference of 25.0 V and separation \( d = 0.005 \) m, \( E \) describes how strongly the electric force acts on charges between the plates.
The work required to move a charge through this potential difference determines the energy available for charge movement. This energy compels electrons to travel across the field, aligning with the concept of electron velocity.
Electron Velocity
When an electron is acted upon by an electric field, it accelerates, gaining kinetic energy in the process. The final velocity of an electron ejected from a negative plate and reaching a positive plate is derived from the energy gained by the electron.
  • The electron's kinetic energy \( KE \) upon reaching the positive plate is given by \( eV \), where \( e = 1.6 \times 10^{-19} \) C is the electron's charge.
  • This energy is equal to \( \frac{1}{2}mv^2 \), \( m \) being the mass of the electron \( 9.11 \times 10^{-31} \) kg.
  • Solving the equation \( eV = \frac{1}{2}mv^2 \) gives us the velocity \( v \) of the electron.
This concept helps to understand how potential difference and electron charge affect an electron's motion across a capacitor, ultimately resulting in measurable speed upon reaching the opposite plate.
Electric Charge
Electric charge is a basic property of matter responsible for electric force between objects. The charge \( Q \) held by each plate in a capacitor arises from the voltage applied across the plates, calculated by the relationship \( Q = CV \).
  • Here \( C \) is the capacitance of the plates, and \( V \) is the potential difference.
  • Charge on one plate is positive and the other equal and opposite, maintaining neutrality in the system.
By understanding how charge builds up in response to a potential difference, one gains insight into how energy is stored and manipulated in electric fields. This principle is fundamental in applications encompassing electronics, such as oscilloscopes managing signal alterations effectively.

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Most popular questions from this chapter

A very long insulating cylinder of charge of radius 2.50 cm carries a uniform linear density of 15.0 nC\(/\)m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175 V?

(a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 3.75 kV? (b) What is the potential of the sphere 's surface relative to infinity?

(a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00 \(\mu\)C and the other a charge of \(-\)3.50 \(\mu\)C, with their centers separated by a distance of 0.180 m. Assume that \(U = 0\) when the charges are infinitely separated. (b) Suppose that one sphere is held in place; the other sphere, with mass 1.50 g, is shot away from it. What minimum initial speed would the moving sphere need to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it is infinitely far from the fixed sphere.)

Certain sharks can detect an electric field as weak as 1.0 \(\mu\)V\(/\)m. To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5V AA battery across these plates, how far apart would the plates have to be?

A thin insulating rod is bent into a semicircular arc of radius \(a\), and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.
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