Problem 23 - University Physics with Modern Physics

Question

(a) An electron is to be accelerated from 3.00 \(\times 10^{6} \mathrm{m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s} .\) Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a halt?

Step-by-Step Solution

Answer

(a) 156 V; (b) -145 V

Step 1: Identify the Concept

The problem involves calculating the potential difference required to change the speed of an electron. This involves using the concepts of kinetic energy and potential energy. We will use the relationship that the change in kinetic energy of the electron is equal to the work done by the electric field, which is the charge of the electron times the change in potential.

Step 2: Use the Formula for Change in Kinetic Energy

The change in kinetic energy \[\Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \]where, \( m \) is the mass of the electron \( (9.11 \times 10^{-31} \text{ kg}) \), \( v_f \) and \( v_i \) are the final and initial velocities respectively.

Step 3: Calculate Change in Kinetic Energy for Part (a)

For part (a), to accelerate from \( v_i = 3.00 \times 10^{6} \text{ m/s} \) to \( v_f = 8.00 \times 10^{6} \text{ m/s} \), calculate \[\Delta KE = \frac{1}{2} \times 9.11 \times 10^{-31} \times ((8.00 \times 10^{6})^2 - (3.00 \times 10^{6})^2)\]Simplifying this gives\(\Delta KE = \frac{1}{2} \times 9.11 \times 10^{-31} \times (64.00 \times 10^{12} - 9.00 \times 10^{12})\)\(= \frac{1}{2} \times 9.11 \times 10^{-31} \times 55.00 \times 10^{12}\)Calculate it to find the change in kinetic energy.

Step 4: Calculate Potential Difference for Part (a)

We know the relationship \( q \Delta V = \Delta KE \), where \( q \) is the charge of an electron \( (-1.602 \times 10^{-19} \text{ C}) \).Therefore, \( \Delta V = \frac{\Delta KE}{e} \).Substitute the value obtained from Step 3 to find \( \Delta V \).

Step 5: Calculate Change in Kinetic Energy for Part (b)

For part (b), the electron is slowed from \( v_i = 8.00 \times 10^{6} \text{ m/s} \) to \( v_f = 0 \text{ m/s} \). Calculate \[\Delta KE = \frac{1}{2} \times 9.11 \times 10^{-31} \times (0^2 - (8.00 \times 10^{6})^2)\]\(= \frac{1}{2} \times 9.11 \times 10^{-31} \times (-64.00 \times 10^{12})\)Calculate it to find the change in kinetic energy.

Step 6: Calculate Potential Difference for Part (b)

Using the same method as in Step 4, where \( q \Delta V = \Delta KE \), substitute the value from Step 5 to find \( \Delta V \) for the slowing process.

Step 7: Solve and Substitute Values

- For (a), \( \Delta KE = 2.50 \times 10^{-17} \text{ J} \)\( \Delta V = \frac{2.50 \times 10^{-17} \text{ J}}{1.602 \times 10^{-19} \text{ C}} = 156 \text{ V} \)- For (b), \( \Delta KE = -2.33 \times 10^{-17} \text{ J} \)\( \Delta V = \frac{2.33 \times 10^{-17} \text{ J}}{1.602 \times 10^{-19} \text{ C}} = -145 \text{ V} \)

Key Concepts

Kinetic Energy

Kinetic energy is a measure of an object's energy due to its motion. It is calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity. For an electron, even though it has a tiny mass of \( 9.11 \times 10^{-31} \) kg, its velocity can have a significant impact on its kinetic energy since it moves very fast.
Understanding kinetic energy is crucial for analyzing electron motion within electric fields, as changes in speed result in changes in kinetic energy, which are pivotal when studying potential differences.

Potential Difference

The potential difference, often known as voltage, is the difference in electric potential energy between two points in space. It dictates how much energy is required to move a charge from one point to another. In the context of electrons, potential difference is critical for accelerating or decelerating them in an electric field.
The potential difference drives the kinetic energy change through the relationship \( \Delta V = \frac{\Delta KE}{q} \), where \( \Delta KE \) is the change in kinetic energy of the electron and \( q \) is the charge of the electron. This concept is fundamental in the analysis of electron acceleration and braking.

Electric Fields

Electric fields are regions where charged particles experience a force. They are created by electric charges and influence other charges nearby. The direction of an electric field is conventionally from positive to negative charges.
Electrons, bearing negative charge, move opposite to the direction of the electric field, which affects their kinetic energy. A significant concept here is that the electric field does work on the electron, changing its kinetic and potential energies. This interplay is vital for calculating how much potential difference is needed to alter an electron's velocity.

Work-Energy Principle

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy: \( W = \Delta KE \). In the case of electrons within electric fields, the work done by the electric force is expressed as \( q \Delta V \).
This principle connects potential difference with kinetic energy changes, showcasing how voltage can convert electrical potential energy into kinetic energy or vice versa. This relationship is essential for solving problems involving the acceleration or deceleration of charged particles, such as electrons, through electric fields.