Question

An electron falls through a distance 1.5 cm in a uniform electric field of magnitude 2.0 \(\times\) 10\(^4\) N/C from rest. The time taken to cover this distance in second is \textunderscore\textunderscore\textunderscore\textunderscore\textunderscore.
(e = 1.6 \(\times\) 10\(^{-19}\) C, m\(_e\) = 9.11 \(\times\) 10\(^{-31}\) kg)

• A. 2.9 \(\times\) 10\(^{-9}\)
• B. 2.9 \(\times\) 10\(^9\)
• C. 4 \(\times\) 10\(^{-6}\)
• D. 4 \(\times\) 10\(^6\)

Hint:

In problems involving microscopic particles in electric fields, the effect of gravity is usually negligible compared to electric forces because the mass is extremely small. Focus only on the electric acceleration \( a = \frac{qE}{m} \).

Answer 1

The Correct Option is A

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.