Step 1: Understanding the Question:
The problem asks for the time an electron takes to fall from rest through a specific distance in a uniform electric field. This is a kinematics problem where the acceleration is provided by the electric force.
Step 2: Key Formula or Approach:
1. Electric force: \( F = qE \)
2. Newton's second law: \( F = ma \Rightarrow a = \frac{qE}{m} \)
3. Kinematics equation for constant acceleration (from rest): \( s = \frac{1}{2}at^2 \Rightarrow t = \sqrt{\frac{2s}{a}} \)
Step 3: Detailed Explanation:
Given values:
\( s = 1.5 \text{ cm} = 0.015 \text{ m} \)
\( E = 2.0 \times 10^4 \text{ N/C} \)
\( e = 1.6 \times 10^{-19} \text{ C} \)
\( m_e = 9.11 \times 10^{-31} \text{ kg} \)
First, calculate the acceleration (\( a \)):
\[ a = \frac{eE}{m_e} = \frac{(1.6 \times 10^{-19}) \times (2.0 \times 10^4)}{9.11 \times 10^{-31}} \]
\[ a = \frac{3.2 \times 10^{-15}}{9.11 \times 10^{-31}} \approx 3.51 \times 10^{15} \text{ m/s}^2 \]
Now, find the time (\( t \)):
\[ t = \sqrt{\frac{2 \times 0.015}{3.51 \times 10^{15}}} \]
\[ t = \sqrt{\frac{0.03}{3.51 \times 10^{15}}} \approx \sqrt{0.00854 \times 10^{-15}} \]
\[ t = \sqrt{8.54 \times 10^{-18}} \approx 2.92 \times 10^{-9} \text{ s} \]
Step 4: Final Answer:
The time taken is approximately 2.9 \(\times\) 10\(^{-9}\) seconds.