Problem 68 - Physics

Question

An electron is released at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing), (a) As the electron gains kinetic energy, does its electric potential energy increase or decrease? Why? (b) The difference in the electron's electric potential energy between the positive and negative plates is EPE positive \(-\mathrm{EPE}_{\text {negative. }}\) How is this difference related to the charge on the electron \((-e)\) and to the difference \(V\) positive \(-V_{\text {negative in the electric potential between }}\) the plates? (c) How is the potential difference \(V\) positive \(-V_{\text {negative }}\) related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plate? The plates of a parallel plate capacitor are separated by a distance of \(1.2 \mathrm{~cm}\), and the electric field within the capacitor has a magnitude of \(2.1 \times 10^{6} \mathrm{~V} / \mathrm{m}\). An electron starts from rest at the negative plate and accelerates to the positive plate. What is the kinetic energy of the electron just as the electron reaches the positive plate?

Step-by-Step Solution

Answer

(a) Decrease; (b) \( \Delta \text{EPE} = -e \cdot \Delta V \); (c) \( \Delta V = E \times d \). KE = \( 4.03 \times 10^{-15} \) J.

Step 1: Understanding Electric Potential Energy Change

As the electron accelerates towards the positive plate, it loses electric potential energy and converts it into kinetic energy. This is because electric potential energy is highest when the electron is at the negative plate and decreases as it moves towards the positive plate, gaining speed.

Step 2: Relating Electric Potential Energy Difference to Charge and Voltage

The change in electric potential energy (\( ext{ΔEPE} \)) between the plates is related to the electron's charge (\(-e\)) and the potential difference (\( ext{∆V} \)) between the plates through the equation: \[ \Delta ext{EPE} = -e \times \Delta V \].The negative sign indicates that the potential energy decreases.

Step 3: Relating Potential Difference to Electric Field and Plate Separation

The potential difference (\( ext{∆V} \)) between two plates in a capacitor can be related to the electric field (\( E \)) and the separation (\( d \)) between the plates by the equation: \[ \Delta V = E \times d \].This relation shows how the field strength and the distance affect the potential difference.

Step 4: Calculating Kinetic Energy of the Electron

Given the separation of the plates (\( d = 1.2 imes 10^{-2} \) m) and the electric field (\( E = 2.1 \times 10^{6} \) V/m), calculate the potential difference: \[ \Delta V = E \times d = 2.1 \times 10^{6} \times 1.2 \times 10^{-2} = 25,200 \text{ V} \].The kinetic energy (\( ext{KE} \)) gained by the electron as it reaches the positive plate is equal to the work done on it, which is the change in its electric potential energy: \[ \text{KE} = e \times \Delta V \].Since the charge of an electron is \( e = 1.6 \times 10^{-19} \) C, the kinetic energy is: \[ \text{KE} = 1.6 \times 10^{-19} \times 25,200 = 4.03 \times 10^{-15} \text{ J} \].

Key Concepts

Parallel Plate Capacitor

A parallel plate capacitor consists of two large, flat conductive plates separated by a distance, usually filled with a dielectric material. The setup creates a uniform electric field between the plates, which can store electrical energy. When a voltage is applied across the plates, it causes positive charges to gather on one plate and negative charges on the opposite plate. This charge separation generates an electric field between the plates.

Key characteristics of parallel plate capacitors include:

• Uniform Electric Field: Between the plates, the electric field is uniform. This uniformity simplifies calculations involving the capacitor.
• Potential Difference: A voltage difference between the plates results in the accumulation of electric charge on the plates, proportional to the capacitance.
• Storage of Electric Charge: They can store electric charge, which can later be released.

Electron Acceleration

When an electron is released from the negative plate of a capacitor, it is accelerated towards the positive plate due to the electric field present between them. This acceleration occurs because the electric field exerts a force on the electron. As the electron moves under this force, it gains speed, translating the electric potential energy initially present into kinetic energy.

Key points about electron acceleration:

• Electric Force: The force on the electron is described by the equation \( F = eE \), where \( e \) is the charge of an electron and \( E \) is the electric field.
• Conversion of Energy: As the electron travels towards the positive plate, electric potential energy is converted into kinetic energy.

Electric Field

The electric field in a parallel plate capacitor is a region in which an electron experiences force due to the shifted charges on the plates. It is a vector quantity, pointing from the positive to the negative plate, and can be calculated using the potential difference divided by the separation distance.

Understanding electric fields:

• Direction and Magnitude: The electric field \( E \) is determined by the voltage difference \( \Delta V \) across the plates and their separation \( d \). The formula is \( E = \frac{\Delta V}{d} \).
• Impact on Charges: A positive charge will move in the direction of the field, while a negative charge, such as an electron, will move against the field.

Kinetic Energy Calculation

The kinetic energy of an electron accelerated across a capacitor can be determined by relating the work done on it by the electric field to its kinetic energy. As the electron moves from the negative to the positive plate, its potential energy is converted entirely into kinetic energy.

Calculating kinetic energy:

• Energy Conversion: The work done by the electric field is equivalent to the change in electric potential energy, which then equals the kinetic energy gained by the electron.
• Formula Application: The electron's kinetic energy \( KE \) is given by \( KE = e \times \Delta V \), where \( e \) is the electron charge and \( \Delta V \) is the potential difference calculated across the capacitor plates.
• Example Calculation: Given that \( e = 1.6 \times 10^{-19} \) C and \( \Delta V = 25,200 \) V, the electron's kinetic energy just before reaching the positive plate is \( 4.03 \times 10^{-15} \) J.