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 of an electron falling a distance of 1.5 cm in a uniform electric field of magnitude \(2 \times 10^4 \, \text{N/C}\), we can follow these steps:

### Step 1: Understand the forces acting on the electron
The electron, being negatively charged, will experience a force in the direction opposite to the electric field. The force acting on the electron can be calculated using the formula:
\[ F = eE \]
where:
- \(e\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\))
- \(E\) is the magnitude of the electric field (\(2 \times 10^4 \, \text{N/C}\))

### Step 2: Calculate the acceleration of the electron
Using Newton's second law, \(F = ma\), we can find the acceleration \(a\) of the electron:
\[ a = \frac{F}{m} = \frac{eE}{m_e} \]
where:
- \(m_e\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\))

### Step 3: Substitute the values into the acceleration formula
Now, substituting the values into the formula for acceleration:
\[ a = \frac{(1.6 \times 10^{-19} \, \text{C})(2 \times 10^4 \, \text{N/C})}{9.11 \times 10^{-31} \, \text{kg}} \]

### Step 4: Calculate the acceleration
Calculating the above expression:
\[ a = \frac{3.2 \times 10^{-15}}{9.11 \times 10^{-31}} \approx 3.51 \times 10^{15} \, \text{m/s}^2 \]

### Step 5: Use the kinematic equation to find the time taken
We can use the kinematic equation for uniformly accelerated motion:
\[ h = \frac{1}{2} a t^2 \]
Rearranging for time \(t\):
\[ t = \sqrt{\frac{2h}{a}} \]

### Step 6: Substitute the values for \(h\) and \(a\)
Given \(h = 1.5 \, \text{cm} = 1.5 \times 10^{-2} \, \text{m}\), we substitute:
\[ t = \sqrt{\frac{2 \times (1.5 \times 10^{-2})}{3.51 \times 10^{15}}} \]

### Step 7: Calculate the time taken
Calculating the above expression:
\[ t = \sqrt{\frac{3.0 \times 10^{-2}}{3.51 \times 10^{15}}} \approx \sqrt{8.55 \times 10^{-18}} \approx 2.93 \times 10^{-9} \, \text{s} \]

### Final Answer
The time taken by the electron to fall a distance of 1.5 cm in the electric field is approximately:
\[ t \approx 2.9 \times 10^{-9} \, \text{s} \]

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