An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density +σ. 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 σ if the electron strikes S at t = 1 s is α[(mε₀)/(e)]C/m², the value of α is ___________. Numerical answer type. Work it out, then open the answer.

Answer: 8. The sheet attracts the electron with constant acceleration a = eE/m = (eσ)/(2ε₀ m). Largest σ needs the largest a: the electron must first move away from S at 1 m/s. Taking distance from S: 0 = 1 + (1)(1) - 1/2a(1)² ⇒ a = 4 m/s² σ = (2ε₀ m a)/(e) = 8[(mε₀)/(e)] So α = 8.

An electron is moving under the influence of the electric field of a… | JEE Main 2024 (29 Jan, Shift 1)

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Motion of Charges in an Electric Field

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 $1\text{ m}$ from $S$ and has a speed of $1\text{ m/s}$ . The maximum value of $\sigma$ if the electron strikes $S$ at $t = 1\text{ s}$ is $\alpha\left[\frac{m\epsilon_0}{e}\right]\frac{C}{m^2}$ , the value of $\alpha$ is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 8

The sheet attracts the electron with constant acceleration $a = \frac{eE}{m} = \frac{e\sigma}{2\varepsilon_0 m}$ .
Largest $\sigma$ needs the largest $a$ : the electron must first move away from $S$ at 1 m/s.
Taking distance from $S$ : $0 = 1 + (1)(1) - \frac{1}{2}a(1)^2 \Rightarrow a = 4\text{ m/s}^2$
$\sigma = \frac{2\varepsilon_0 m a}{e} = 8\left[\frac{m\varepsilon_0}{e}\right]$
So $\alpha = 8$ .

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