A proton is released from rest, 10 cm from a charged sheet carrying charged density of `-2.21xx10^-9C//m^2`. It will strike the sheet after the time (approximately)

Options

A. `4mus`
B. `2mus`
C. `2sqertmus`
D. `4sqrt2mus`

Explanation

To solve the problem of a proton being released from rest and moving towards a charged sheet, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Charge density of the sheet, \( \sigma = -2.21 \times 10^{-9} \, \text{C/m}^2 \)
- Distance from the sheet, \( s = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Charge of the proton, \( Q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of the proton, \( m = 1.67 \times 10^{-27} \, \text{kg} \)

2. **Calculate the Electric Field (E) due to the Charged Sheet:**
The electric field due to an infinite sheet with surface charge density \( \sigma \) is given by:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

Since the charge density is negative, the electric field will point towards the sheet.

3. **Substitute the Values to Find E:**
\[ E = \frac{-2.21 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}} = \frac{-2.21 \times 10^{-9}}{1.77 \times 10^{-11}} \approx -124.7 \, \text{N/C} \]

4. **Calculate the Force (F) on the Proton:**
The force acting on the proton can be calculated using:
\[ F = Q \cdot E \]
Substituting the values:
\[ F = (1.6 \times 10^{-19}) \cdot (-124.7) \approx -1.995 \times 10^{-17} \, \text{N} \]

5. **Calculate the Acceleration (a) of the Proton:**
Using Newton's second law, \( F = ma \):
\[ a = \frac{F}{m} = \frac{-1.995 \times 10^{-17}}{1.67 \times 10^{-27}} \approx -1.194 \times 10^{10} \, \text{m/s}^2 \]

6. **Use the Second Equation of Motion to Find Time (t):**
Since the proton is released from rest, we can use the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \), so:
\[ s = \frac{1}{2} a t^2 \implies t^2 = \frac{2s}{a} \]
Substituting the values:
\[ t^2 = \frac{2 \cdot 0.1}{1.194 \times 10^{10}} \approx 1.674 \times 10^{-11} \implies t \approx \sqrt{1.674 \times 10^{-11}} \approx 4.1 \times 10^{-6} \, \text{s} \]

7. **Final Result:**
The time taken for the proton to strike the sheet is approximately:
\[ t \approx 4.1 \, \mu s \]