Problem 64 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(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 1: 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 \( A = (0.03)^2 \) m², since the side of the plate is 3.0 cm. The separation \( d = 0.005 \) m. Substitute these into the formula to find \( C \).

Step 2: 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 \( Q \).

Step 3: 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 \( E \).

Step 4: 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.

Key Concepts

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: \( A = (0.03)^2 \) m² 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.