Problem 31 - 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) Accelerated: Use the potential difference of 1500 V. (b) Slowed to halt: Use the potential difference of 1825 V.

Step 1: Identify the Formula

For both parts of the problem, we need to use the formula for electric potential energy change which is related to kinetic energy change by \[ \Delta KE = e \Delta V \]where \( \Delta KE \) is the change in kinetic energy, \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \text{ C} \)), and \( \Delta V \) is the potential difference.

Step 2: Calculate Change in Kinetic Energy for Part (a)

The initial kinetic energy \( KE_i \) is given by\[ KE_i = \frac{1}{2} m v_i^2 \] where \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \text{ kg} \)) and \( v_i = 3.00 \times 10^6 \text{ m/s} \).The final kinetic energy \( KE_f \) is\[ KE_f = \frac{1}{2} m v_f^2 \]where \( v_f = 8.00 \times 10^6 \text{ m/s} \).

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

Compute the change in kinetic energy:\[ \Delta KE = KE_f - KE_i \]Substitute the kinetic energy equations to find \( \Delta KE \):\[ \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 = \frac{1}{2} m (v_f^2 - v_i^2) \]Plug in \( m = 9.11 \times 10^{-31} \text{ kg} \), \( v_f = 8.00 \times 10^6 \text{ m/s} \), and \( v_i = 3.00 \times 10^6 \text{ m/s} \) to calculate \( \Delta KE \).

Step 4: Solve for Potential Difference for Part (a)

Use the relationship \( \Delta KE = e \Delta V \) to solve for \( \Delta V \):\[ \Delta V = \frac{\Delta KE}{e} \]Substitute the calculated \( \Delta KE \) from Step 3 and \( e = 1.6 \times 10^{-19} \text{ C} \) to find \( \Delta V \).

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

For part (b), the electron slows from \( v_i = 8.00 \times 10^6 \text{ m/s} \) to \( v_f = 0 \text{ m/s} \). The initial kinetic energy is \( \frac{1}{2} m v_i^2 \) and the final kinetic energy is \( 0 \). The change in kinetic energy is:\[ \Delta KE = 0 - \frac{1}{2} m v_i^2 = -\frac{1}{2} m v_i^2 \].

Step 6: Solve for Potential Difference for Part (b)

Use \( \Delta KE = e \Delta V \) and solve for \( \Delta V \):\[ \Delta V = \frac{-\frac{1}{2} m v_i^2}{e} \]Substitute \( m = 9.11 \times 10^{-31} \text{ kg} \), \( v_i = 8.00 \times 10^6 \text{ m/s} \), and \( e = 1.6 \times 10^{-19} \text{ C} \) to find the negative of \( \Delta V \).

Key Concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics that relates to the energy of motion. For any moving object, such as an electron, the kinetic energy is calculated with the formula:\[ KE = \frac{1}{2} m v^2 \]where:

• \( m \) is the mass of the object (for an electron, it's approximately \( 9.11 \times 10^{-31} \text{ kg} \))
• \( v \) is the velocity of the object

When an electron accelerates, its speed increases, resulting in an increase in kinetic energy. Conversely, if it decelerates or slows down, its kinetic energy decreases. In problems involving changes in speed from one value to another, calculating the change in kinetic energy \( \Delta KE \) helps us understand how much energy is gained or lost.
This change can then be related to electric potential difference, which governs how much energy is needed or released during the process.

Electron Acceleration

Electron acceleration involves an increase in the speed of the electron as it moves through an electric field. This is typically achieved by passing the electron through a potential difference, which provides the necessary energy:\[ \Delta KE = e \Delta V \]where:

• \( e \) is the charge of the electron \( (1.6 \times 10^{-19} \text{ C}) \)
• \( \Delta V \) is the potential difference

During acceleration, the electron's velocity goes from an initial speed \( v_i \) to a higher final speed \( v_f \). The kinetic energy increases, and the potential difference is what supplies this energy difference. The problem involves determining what this potential difference must be for a given change in the electron's speed.
Understanding this relationship helps in designing circuits and devices that control electron flow efficiently.

Electron Deceleration

Electron deceleration is the process of slowing down an electron, which implies a decrease in its kinetic energy. To bring an electron to a halt from a high speed, a negative potential difference is needed. This creates an electric field that opposes the electron's motion.
The calculation for deceleration involves:\[ \Delta KE = e \Delta V \]However, since the kinetic energy decreases, the potential difference \( \Delta V \) is also negative.
By using the change in kinetic energy formula:\[ \Delta KE = -\frac{1}{2} m v_i^2 \]we can determine the negative potential difference necessary to stop the electron. This concept is crucial when examining how particles lose energy as they interact with other fields or barriers.

Potential Energy Change

The potential energy change for an electron moving between two points with different electric potentials is directly related to the change in kinetic energy. If the electron gains speed, potential energy decreases; if the electron slows down, potential energy increases.
In any electric field, potential energy change can be computed using:\[ \Delta PE = -\Delta KE \]This relationship reflects the conservation of energy principle in an electric field. When an electron speeds up, the energy comes from converting potential to kinetic energy, and the opposite occurs when the electron slows.
This understanding enables problem-solving related to electron motion in electromagnetic fields, highlighting how potential differences translate into kinetic energy changes, crucial for applications in electronics and fundamental physics.