Two insulating plates are both uniformly charged in such a way that the potential difference between them is V₂ - V₁ = 20 V. (i.e., plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e = 1.6 × 10⁻¹⁹ C, me = 9.11 × 10⁻³¹ kg) (a) 2.65 × 10⁶ m/s (b) 7.02 × 10¹² m/s (c) 1.87 × 10⁶ m/s (d) 32 × 10⁻¹⁹ m/s

Answer: (a). The electron gains energy eV moving to the higher-potential plate: 1/2mv² = eV. v = √(2eV/m) = √2 × 1.6 × 10⁻¹⁹ × 209.11 × 10⁻³¹ = √7.02 × 10¹² v ≈ 2.65 × 10⁶ m/s So the answer is (a) 2.65 × 10⁶ m/s.

Two insulating plates are both uniformly charged in such a way that… | JEE Main 2006 (AIEEE)

Q42JEE Main 2006 · AIEEEMedium

Motion of Charges in an Electric Field

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2 - V_1 = 20\text{ V}.$ (i.e., plate $2$ is at a higher potential). The plates are separated by $d = 0.1\text{ m}$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $1.$ What is its speed when it hits plate $2$ ?

$\left(e = 1.6 \times 10^{-19}\text{ C},\ m_e = 9.11 \times 10^{-31}\text{ kg}\right)$

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Answer: (a)

The electron gains energy $eV$ moving to the higher-potential plate: $\frac{1}{2}mv^2 = eV$ .
$v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 20}{9.11 \times 10^{-31}}} = \sqrt{7.02 \times 10^{12}}$
$v \approx 2.65 \times 10^{6}\text{ m/s}$
So the answer is (a) $2.65 \times 10^{6}\text{ m/s}$ .

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