An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet \( S \) having surface charge density \( +\sigma \). The electron at \( t = 0 \) is at a distance of 1 m from \( S \) and has a speed of 1 m/s. The maximum value of \( \sigma \) if the electron strikes \( S \) at \( t = 1 \, \text{s} \) is \( \alpha \left[ \frac{m \epsilon_0}{e} \right] \frac{\text{C}}{\text{m}^2} \). The value of \( \alpha \) is ______.
Correct Answer: 8
Approach Solution - 1
Step 1. Given Information and Variables:
Initial velocity u = 1 m/s - Acceleration a = - \( \frac{\sigma e}{2 \varepsilon_0 m} \) - Time t = 1 s - Displacement S = -1 m
Step 2. Use Kinematic Equation:
The kinematic equation for displacement S is:
S = ut + \(\frac{1}{2}\)at2
Substitute values:
-1 = 1 × 1 + \(\frac{1}{2}\) × \( - \frac{\sigma e}{2 \varepsilon_0 m} \) × (1)2
Step 3. Solve for σ:
-1 = 1 - \( \frac{\sigma e}{4 \varepsilon_0 m} \)
\( \frac{\sigma e}{4 \varepsilon_0 m} \) = 2
Therefore, σ = \( 8 \frac{\varepsilon_0 m}{e} \)
Step 4. Determine α:
Given σ = α \(\left[ \frac{m \varepsilon_0}{e} \right]\), we find α = 8.
Conclusion:
So, the correct answer is: α = 8
Approach Solution -2
Step 1: Electric field due to an infinite sheet of charge.
For a uniformly charged infinite plane sheet: \[ E = \frac{\sigma}{2\varepsilon_0} \]
Step 2: Force and acceleration on the electron.
Force on the electron: \[ F = eE = e \frac{\sigma}{2\varepsilon_0} \] Since the field is directed away from the sheet and the electron has negative charge, it will accelerate toward the sheet with acceleration: \[ a = \frac{F}{m} = \frac{e\sigma}{2m\varepsilon_0} \]
Step 3: Kinematic relation for motion toward the sheet.
Let the distance to the sheet be \( s = 1\,\text{m} \), initial speed \( u = 1\,\text{m/s} \), time \( t = 1\,\text{s} \), acceleration \( a = \frac{e\sigma}{2m\varepsilon_0} \).
Using equation of motion: \[ s = ut + \frac{1}{2}at^2 \] Substitute values: \[ 1 = 1(1) + \frac{1}{2}a(1)^2 \] \[ 1 = 1 + \frac{a}{2} \] \[ \frac{a}{2} = 0 \implies a = 0? \] Wait—that can’t be right because \( s \) is decreasing. The electron is moving toward the sheet (so displacement toward the sheet is taken as **negative**).
Step 4: Take direction properly.
Let the direction toward the sheet be negative. So, displacement \( s = -1\,\text{m} \), initial velocity \( u = -1\,\text{m/s} \), and acceleration \( a \) is also negative (toward the sheet). Equation: \[ -1 = -1(1) + \frac{1}{2}a(1)^2 \] \[ -1 = -1 + \frac{a}{2} \] \[ 0 = \frac{a}{2} \] Still gives \( a = 0 \). Let's fix the sign: the electron is **accelerating**, so the **magnitude of displacement** must equal the sum of distances due to initial velocity and acceleration effect.
Step 5: Use magnitude relation.
\[ 1 = ut + \frac{1}{2} a t^2 \] \[ 1 = 1(1) + \frac{1}{2} a (1)^2 \] \[ 1 = 1 + \frac{a}{2} \] So, in order to reach the sheet at 1 m distance, the **extra acceleration distance** must cover **1 m**, not **add to it**. Hence we consider displacement **toward** the sheet: \[ 1 = -u t + \frac{1}{2} a t^2 \] \[ 1 = -1(1) + \frac{1}{2}a(1)^2 \] \[ 1 = -1 + \frac{a}{2} \] \[ a = 4 \, \text{m/s}^2 \]
Step 6: Substitute value of \( a \).
\[ a = \frac{e\sigma}{2m\varepsilon_0} = 4 \] \[ \sigma = \frac{8m\varepsilon_0}{e} \]
Final Answer:
\[ \boxed{\alpha = 8} \]
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