An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude `2 xx10^(4) N//C`. The time taken by the electron to fall this distance is

Options

A. `1.3 xx10^(2) s`
B. `2.1 xx10^(-2) s`
C. `1.6xx10^(-10) s`
D. `2.9xx10^(-9) s`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have an electron falling in a uniform electric field. The distance it falls is given as \( s = 1.5 \, \text{cm} = 0.015 \, \text{m} \), and the magnitude of the electric field is \( E = 2 \times 10^4 \, \text{N/C} \).

### Step 2: Calculate the Force on the Electron
The force \( F \) acting on the electron can be calculated using the formula:
\[ F = qE \]
where \( q \) is the charge of the electron, which is approximately \( q = -1.6 \times 10^{-19} \, \text{C} \).

Substituting the values:
\[ F = (-1.6 \times 10^{-19} \, \text{C})(2 \times 10^4 \, \text{N/C}) = -3.2 \times 10^{-15} \, \text{N} \]
(Note: The negative sign indicates the direction of the force, but we will use the magnitude for calculations.)

### Step 3: Calculate the Acceleration of the Electron
Using Newton's second law, \( F = ma \), we can find the acceleration \( a \):
\[ a = \frac{F}{m} \]
where \( m \) is the mass of the electron, approximately \( m = 9.1 \times 10^{-31} \, \text{kg} \).

Substituting the values:
\[ a = \frac{3.2 \times 10^{-15} \, \text{N}}{9.1 \times 10^{-31} \, \text{kg}} \approx 3.51 \times 10^{15} \, \text{m/s}^2 \]

### Step 4: Use the Equation of Motion
Since the electron starts from rest, we can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( u = 0 \) (initial velocity), \( s = 0.015 \, \text{m} \), and \( a \) is the acceleration we just calculated.

This simplifies to:
\[ 0.015 = 0 + \frac{1}{2} (3.51 \times 10^{15}) t^2 \]
\[ 0.015 = 1.755 \times 10^{15} t^2 \]

### Step 5: Solve for Time \( t \)
Rearranging the equation gives:
\[ t^2 = \frac{0.015}{1.755 \times 10^{15}} \]
Calculating \( t^2 \):
\[ t^2 \approx 8.54 \times 10^{-18} \]
Taking the square root to find \( t \):
\[ t \approx \sqrt{8.54 \times 10^{-18}} \approx 2.92 \times 10^{-9} \, \text{s} \]

### Final Answer
The time taken by the electron to fall a distance of 1.5 cm in the electric field is approximately \( t \approx 2.92 \, \text{ns} \).

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