A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 2.5 cm away, in a time 1.8 × 108 s. (a) What is the speed of the electron as it strikes the second plate? (b) What is the magnitude of the electric field?

To solve this problem, we need to use the principles of motion under constant acceleration, as well as the relationship between electric force and acceleration of the electron in a uniform electric field. Here are the steps:

### (a) Speed of the electron as it strikes the second plate

1. **Given Data:**
   - Distance between plates: \( d = 2.5 \, \text{cm} = 0.025 \, \text{m} \)
   - Time taken to strike the second plate: \( t = 1.8 \times 10^8 \, \text{s} \)
   - Initial speed \( v_0 = 0 \, \text{m/s} \) (the electron is released from rest)
   - Electron mass: \( m = 9.11 \times 10^{-31} \, \text{kg} \)

2. Since the electron is subject to a constant acceleration due to the electric field, we can use the kinematic equation:
   \[
   d = v_0 t + \frac{1}{2} a t^2
   \]
   Since \( v_0 = 0 \), this simplifies to:
   \[
   d = \frac{1}{2} a t^2
   \]
   Solving for \( a \) (acceleration):
   \[
   a = \frac{2d}{t^2}
   \]

   Substituting the values:
   \[
   a = \frac{2 \times 0.025}{(1.8 \times 10^8)^2} \, \text{m/s}^2
   \]

3. Once we find the acceleration, we can calculate the final speed using the equation:
   \[
   v = v_0 + a t
   \]
   Again, since \( v_0 = 0 \), it simplifies to:
   \[
   v = a t
   \]
   Now, let's compute these values.

### (b) Magnitude of the Electric Field

1. The electric force on the electron is given by:
   \[
   F = e E
   \]
   where \( e = 1.6 \times 10^{-19} \, \text{C} \) is the charge of the electron, and \( E \) is the magnitude of the electric field.

2. By Newton's second law, we know:
   \[
   F = m a
   \]
   Therefore, equating the two forces:
   \[
   e E = m a
   \]
   Solving for \( E \):
   \[
   E = \frac{m a}{e}
   \]
   Using the acceleration \( a \) from part (a), we can compute the electric field.

Let's calculate the acceleration, final velocity, and the electric field magnitude now.### (a) Speed of the electron as it strikes the second plate:
The speed of the electron when it strikes the second plate is approximately \( 2.78 \times 10^{-10} \, \text{m/s} \).

### (b) Magnitude of the electric field:
The magnitude of the electric field is approximately \( 8.79 \times 10^{-30} \, \text{N/C} \).

These values suggest a very small acceleration and electric field, likely due to the extremely long time given for the electron to travel across a small distance.

Do you have any questions or would you like further clarification?

Here are 5 related questions to deepen understanding:
1. How does the acceleration of the electron depend on the electric field strength?
2. What would happen if the electron were initially given some velocity?
3. How does the distance between the plates affect the time taken for the electron to strike the opposite plate?
4. How would the electric field change if the charge of the electron were doubled?
5. What would happen to the electron's speed if the time were halved?

**Tip**: Electric fields exert constant force on charged particles, leading to uniform acceleration.

A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 2.5 cm away, in a time 1.8 × 10^8 s. (a) What is the speed of the electron as it strikes the second plate? (b) What is the magnitude of the electric field?

Explore the motion of an electron released from rest between two oppositely charged plates. Learn how to calculate the final speed of the electron as it strikes the opposite plate and determine the magnitude of the electric field using kinematic equations and Newton's second law. Key physics concepts are explained, including electric force and motion under constant acceleration.

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