To solve the problem of an electron moving a distance of 6.0 cm when accelerated from rest by an electric field of strength \(2 \times 10^{-4} \, \text{N/C}\), we will follow these steps:
### Step 1: Identify the given values
- Distance \( S = 6.0 \, \text{cm} = 6.0 \times 10^{-2} \, \text{m} \)
- Electric field strength \( E = 2 \times 10^{-4} \, \text{N/C} \)
- Charge of electron \( Q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)
### Step 2: Calculate the force acting on the electron
The force \( F \) acting on the electron due to the electric field is given by:
\[
F = Q \times E
\]
Substituting the values:
\[
F = (1.6 \times 10^{-19} \, \text{C}) \times (2 \times 10^{-4} \, \text{N/C}) = 3.2 \times 10^{-23} \, \text{N}
\]
### Step 3: Calculate the acceleration of the electron
Using Newton's second law, the acceleration \( A \) can be calculated as:
\[
A = \frac{F}{m}
\]
Substituting the values:
\[
A = \frac{3.2 \times 10^{-23} \, \text{N}}{9.1 \times 10^{-31} \, \text{kg}} \approx 3.51 \times 10^{7} \, \text{m/s}^2
\]
### Step 4: Use the kinematic equation to find the time of travel
Since the electron starts from rest, the initial velocity \( U = 0 \). The kinematic equation relating distance, initial velocity, acceleration, and time is:
\[
S = U T + \frac{1}{2} A T^2
\]
Substituting \( U = 0 \):
\[
S = \frac{1}{2} A T^2
\]
Rearranging for \( T \):
\[
T^2 = \frac{2S}{A}
\]
\[
T = \sqrt{\frac{2S}{A}}
\]
### Step 5: Substitute the values to find \( T \)
Substituting \( S = 6.0 \times 10^{-2} \, \text{m} \) and \( A \approx 3.51 \times 10^{7} \, \text{m/s}^2 \):
\[
T = \sqrt{\frac{2 \times (6.0 \times 10^{-2})}{3.51 \times 10^{7}}}
\]
Calculating the values:
\[
T = \sqrt{\frac{1.2 \times 10^{-1}}{3.51 \times 10^{7}}} \approx \sqrt{3.42 \times 10^{-9}} \approx 5.84 \times 10^{-5} \, \text{s}
\]
### Final Answer
The time of travel for the electron is approximately \( T \approx 5.84 \times 10^{-5} \, \text{s} \).
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