An electron initially at rest falls a distance of 2 cm in a uniform electric field of magnitude `3 xx 10^(4) N C^(-1)`. The time taken by the electron to fall to this distance is

Options

A. `1.3 xx 10^(2) s`
B. `2.1 xx 10^(-12)s`
C. `1.6 xx 10^(-10)s`
D. `2.75 xx 10^(-9)s`

Answer

To solve the problem of an electron falling a distance of 2 cm in a uniform electric field of magnitude \(3 \times 10^4 \, \text{N/C}\), we will follow these steps:

### Step 1: Identify the forces acting on the electron
The force acting on the electron due to the electric field can be calculated using the formula:
\[ F = qE \]
where \(q\) is the charge of the electron and \(E\) is the electric field strength.

### Step 2: Calculate the force on the electron
The charge of an electron is approximately \(q = -1.6 \times 10^{-19} \, \text{C}\). The magnitude of the electric field is given as \(E = 3 \times 10^4 \, \text{N/C}\). Therefore, the magnitude of the force is:
\[ F = |q|E = (1.6 \times 10^{-19} \, \text{C})(3 \times 10^4 \, \text{N/C}) = 4.8 \times 10^{-15} \, \text{N} \]

### Step 3: Calculate the acceleration of the electron
Using Newton's second law, \(F = ma\), we can find the acceleration \(a\) of the electron:
\[ a = \frac{F}{m} \]
The mass of the electron is approximately \(m = 9.1 \times 10^{-31} \, \text{kg}\). Substituting the values:
\[ a = \frac{4.8 \times 10^{-15} \, \text{N}}{9.1 \times 10^{-31} \, \text{kg}} \approx 5.27 \times 10^{15} \, \text{m/s}^2 \]

### Step 4: Use kinematic equations to find the time taken
We will use the kinematic equation:
\[ S = ut + \frac{1}{2} a t^2 \]
where \(S\) is the distance fallen, \(u\) is the initial velocity (which is 0 since the electron starts from rest), and \(a\) is the acceleration. Since \(u = 0\), the equation simplifies to:
\[ S = \frac{1}{2} a t^2 \]
Rearranging for \(t\):
\[ t = \sqrt{\frac{2S}{a}} \]

### Step 5: Substitute the values
The distance \(S\) is given as \(2 \, \text{cm}\), which we convert to meters:
\[ S = 0.02 \, \text{m} \]
Now substituting the values:
\[ t = \sqrt{\frac{2 \times 0.02 \, \text{m}}{5.27 \times 10^{15} \, \text{m/s}^2}} = \sqrt{\frac{0.04}{5.27 \times 10^{15}}} \]
Calculating this gives:
\[ t \approx \sqrt{7.58 \times 10^{-17}} \approx 2.75 \times 10^{-9} \, \text{s} \]

### Final Answer
The time taken by the electron to fall a distance of 2 cm is approximately:
\[ t \approx 2.75 \times 10^{-9} \, \text{s} \]